W8 · Toolkit— gates26–30 hrs

Week 8 of 28 · Toolkit · due 2026-07-18 · 26–30 hrs

Analysis: Power Series + Algebra: Polynomial Advanced + LA: Applications + Comb: Recurrences + NT: Valuations + Hidden Tools: Feynman's Trick + Root-of-Unity Filter + Valuations

Analysis: Power Series + Algebra: Polynomial Advanced + LA: Applications + Comb: Recurrences + NT: Valuations + Hidden Tools: Feynman's Trick + Root-of-Unity Filter + Valuations gate complete

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This week's lesson · 10 min read

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Power series, valuations, and tricks named after physicists

A power series is a polynomial that refused to stop, and almost everything you love about polynomials survives the refusal — inside the radius of convergence. This week closes the first toolkit lap: analytic machinery for series, the p-adic valuation that turns divisibility into arithmetic you can add, and two signature tricks (Feynman's parameter, the root-of-unity filter) that will each be worth a full problem by December.

Local information is global power: a series' behavior at one point, a prime's exponent in one factorization, one evaluation of a generating function — each tiny probe determines far more than it should.

Key ideas — the week on one card

  1. Inside its radius of convergence a power series is as nice as a polynomial: differentiate and integrate term by term, and matching coefficients is legal.
  2. $v_p$ is fully additive on products, and $v_p(a+b) = \min(v_p a, v_p b)$ WHEN the two valuations differ — the equality clause is the weapon.
  3. Feynman's trick: embed the integral in a family $I(t)$, differentiate in $t$, and integrate back from an anchor value you know outright.
  4. The roots-of-unity filter: for a primitive $k$-th root $\omega$, $\sum_{j=0}^{k-1} \omega^{jn}$ is $k$ when $k \mid n$ and $0$ otherwise — average $F(\omega^j)$ to extract every $k$-th coefficient.

1Power series: polynomials with infinite patience

Every power series $\sum a_n x^n$ owns a radius of convergence $R$ (possibly $0$ or $\infty$): absolute convergence inside, divergence outside, anarchy on the boundary. Compute $R$ by the root test ($1/R = \limsup |a_n|^{1/n}$) or ratio test when it exists. Inside $R$, the series is as nice as you could ask: continuous, differentiable term by term, integrable term by term — the M-test from last week working silently on every compact subinterval.

The five series you must produce from memory, instantly: $e^x = \sum \frac{x^n}{n!}$, $\frac{1}{1-x} = \sum x^n$, $\ln(1+x) = \sum \frac{(-1)^{n+1} x^n}{n}$, $\sin x, \cos x$ (alternating odd/even factorials), and the binomial series $(1+x)^\alpha = \sum \binom{\alpha}{n} x^n$. Most 'evaluate this sum' problems are one of these five, differentiated, integrated, or evaluated somewhere clever.

Uniqueness is the quiet superpower: if two power series agree on an interval, all their coefficients match. This licenses the method of undetermined coefficients for ODEs and functional equations — propose $\sum a_n x^n$, grind out a recurrence for $a_n$, and the answer is forced.

Worked example

Evaluate $\sum_{n=1}^{\infty} \frac{n^2}{2^n}$.

  1. Nudge 1

    The operator $x\frac{d}{dx}$ turns $x^n$ into $n x^n$ — apply it to the geometric series.

    Reveal step 1

    Start from the geometric series $\frac{1}{1-x} = \sum_{n \ge 0} x^n$ and apply the operator $x\frac{d}{dx}$ (differentiate, multiply by $x$) — it converts $x^n$ into $n x^n$: one application gives $\sum n x^n = \frac{x}{(1-x)^2}$.

  2. Nudge 2

    Apply the same operator a second time to get $n^2$ upstairs.

    Reveal step 2

    Apply $x \frac{d}{dx}$ again: $\sum n^2 x^n = x \frac{d}{dx} \frac{x}{(1-x)^2} = \frac{x(1+x)}{(1-x)^3}$ (quotient rule, then simplify).

  3. Nudge 3

    Check $x = \frac12$ is inside the radius before plugging in — that check is doing legal work.

    Reveal step 3

    Evaluate at $x = \frac12$ (inside $R = 1$, so everything is licensed): $\frac{\frac12 \cdot \frac32}{(\frac12)^3} = \frac{3/4}{1/8} = 6$.

  4. Answer

    $6$ — two rounds of $x\frac{d}{dx}$ on the geometric series, evaluated at $x = 1/2$.

Pitfall. Evaluating a manipulated series ON the boundary of convergence, where term-by-term operations lose their license. Check $|x| < R$ before plugging in — the half in $x = 1/2$ is doing quiet legal work.

2p-adic valuation: divisibility you can add

Define $v_p(n)$ as the exponent of prime $p$ in $n$'s factorization. The rules are additive and merciless: $v_p(ab) = v_p(a) + v_p(b)$, and $v_p(a + b) \ge \min(v_p(a), v_p(b))$ WITH EQUALITY when the two valuations differ. That equality clause is the weapon — in any sum where one term is strictly divisible by less of $p$ than every other, the whole sum inherits exactly that valuation and in particular cannot vanish.

Legendre's formula counts $p$'s in factorials: $v_p(n!) = \sum_{i \ge 1} \lfloor n/p^i \rfloor = \frac{n - s_p(n)}{p-1}$, where $s_p(n)$ is the digit sum of $n$ base $p$. Every 'how many zeros does $1000!$ end in' is $v_5$; every 'show this binomial coefficient is divisible by $p$' is Legendre applied to $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ — Kummer's theorem repackages it as counting CARRIES when adding $k$ and $n-k$ base $p$.

The classic demonstration: $H_n = 1 + \frac12 + \cdots + \frac1n$ is never an integer for $n \ge 2$. Take the largest power $2^k \le n$: the term $\frac{1}{2^k}$ has strictly lower $v_2$ than every other term (each other denominator contains fewer 2s), so $v_2(H_n) = -k < 0$. One term dominates; the sum cannot be an integer.

Worked example

How many trailing zeros does $100!$ have, and what is $v_2\binom{100}{50}$?

  1. Nudge 1

    Trailing zeros come from the SCARCER prime in $10 = 2 \cdot 5$ — count fives with Legendre.

    Reveal step 1

    Trailing zeros are $\min(v_2, v_5)$ of $100!$, and 5s are scarcer: $v_5(100!) = \lfloor 100/5 \rfloor + \lfloor 100/25 \rfloor = 20 + 4 = 24$. So 24 zeros.

  2. Nudge 2

    Kummer: the valuation of the binomial counts CARRIES when adding $50 + 50$ in base 2.

    Reveal step 2

    For the binomial: Kummer says $v_2\binom{100}{50}$ equals the number of CARRIES when adding $50 + 50$ in base 2. In binary $50 = 110010$; adding it to itself is a left shift, and every 1-bit produces a carry.

  3. Nudge 3

    How many 1-bits does 50 have in binary? Each one produces a carry when doubled.

    Reveal step 3

    $50$ has three 1-bits ($32 + 16 + 2$), so there are 3 carries: $v_2\binom{100}{50} = 3$. Check via Legendre: $v_2(100!) - 2v_2(50!) = 97 - 2\cdot 47 = 3$ ✓ (using $v_2(n!) = n - s_2(n)$).

  4. Answer

    24 trailing zeros; $v_2\binom{100}{50} = 3$ (three binary carries in $50 + 50$, or $97 - 94$ by Legendre).

Pitfall. Using $v_p(a+b) = \min(v_p a, v_p b)$ when the valuations are EQUAL — then you only get $\ge$, and the sum can gain extra factors ($v_2(2 + 2) = 2$, not 1). The equality clause requires distinct valuations; check before you cash it.

3Feynman's trick: differentiate the parameter you invented

Differentiation under the integral sign becomes an evaluation ENGINE when you introduce the parameter yourself. The pattern: embed the target integral $I$ into a family $I(t)$, compute $I'(t)$ (which is easier because differentiating in $t$ simplifies the integrand), then integrate back in $t$ using one known anchor value $I(t_0)$.

Anchor selection is the underrated half. You need one $t_0$ where $I(t_0)$ is known outright — usually $t = 0$ (integrand collapses) or $t \to \infty$ (integrand dies). Without the anchor, integrating $I'$ leaves an unknown constant and the method stalls at the finish line.

The licensing conditions (continuity of $\partial f/\partial t$, domination) are genuine but rarely the point on a contest; a one-line citation — 'differentiation under the integral is justified since the integrand and its $t$-derivative are continuous on the compact domain' — is what the rubric wants.

Worked example

Evaluate $I = \int_0^1 \frac{x - 1}{\ln x}\,dx$.

  1. Nudge 1

    Put the parameter one level below the $\ln x$: try $x^t$ upstairs.

    Reveal step 1

    Embed: define $I(t) = \int_0^1 \frac{x^t - 1}{\ln x}\,dx$ so the target is $I(1)$, and the anchor is free: $I(0) = 0$.

  2. Nudge 2

    Differentiate under the integral — the logarithm should cancel outright.

    Reveal step 2

    Differentiate in $t$: $\frac{\partial}{\partial t} \frac{x^t - 1}{\ln x} = \frac{x^t \ln x}{\ln x} = x^t$ — the logarithm cancels, which is the entire miracle. So $I'(t) = \int_0^1 x^t dx = \frac{1}{t+1}$.

  3. Nudge 3

    You know $I(0)$ for free; integrate $I'(t)$ from that anchor to $t = 1$.

    Reveal step 3

    Integrate back from the anchor: $I(1) = I(0) + \int_0^1 \frac{dt}{t+1} = \ln 2$.

  4. Answer

    $\ln 2$ — parameterize the exponent, watch $\ln x$ cancel, integrate $\frac{1}{t+1}$ back from $I(0) = 0$.

Pitfall. Forgetting the anchor. $I'(t) = \frac{1}{t+1}$ determines $I$ only up to a constant; the collapse at $t = 0$ is what nails it. Choose the parameterization so SOME value is free.

4Roots-of-unity filter: sampling a generating function

To extract every $k$-th coefficient of a polynomial or series $A(x) = \sum a_n x^n$, average over the $k$-th roots of unity: $\sum_{n \equiv r \pmod k} a_n = \frac{1}{k} \sum_{j=0}^{k-1} \omega^{-jr} A(\omega^j)$, where $\omega = e^{2\pi i/k}$. The engine is Week 4's fact that $\sum_j \omega^{jn}$ is $k$ when $k \mid n$ and $0$ otherwise — the roots of unity vote, and only multiples of $k$ survive the election.

The $k = 2$ case is one you already know: even-index coefficients are $\frac{A(1) + A(-1)}{2}$. The filter is that idea industrialized. Its favorite contest costume: 'how many subsets of $\{1, \dots, n\}$ have size divisible by 3' — apply the filter to $(1+x)^n$ at the cube roots of unity, and the answer emerges with a small correction term of modulus $\le \frac{2}{3}$, which also proves the count is nearly exactly $2^n/3$.

1ω¹ω²ω³ω⁴ω⁵Σ ωᵏⁿ = 6 when 6 | nΣ ωᵏⁿ = 0 otherwisethe vote every filterargument casts
The filter's engine: equally spaced unit vectors cancel perfectly — except when the exponent lines them all up.

Worked example

Count the subsets of $\{1, 2, \dots, 10\}$ whose size is divisible by 3.

  1. Nudge 1

    Subset sizes live in $(1+x)^{10}$ — filter its coefficients by residue mod 3.

    Reveal step 1

    The size generating function is $(1+x)^{10}$: the coefficient of $x^s$ counts size-$s$ subsets. Filter with $k = 3$, $r = 0$, $\omega = e^{2\pi i/3}$: count $= \frac{1}{3}\left[(1+1)^{10} + (1+\omega)^{10} + (1+\omega^2)^{10}\right]$.

  2. Nudge 2

    Compute $(1 + \omega)^{10}$ by writing $1 + \omega$ in polar form — it is a unit-length rotation.

    Reveal step 2

    Simplify using $1 + \omega = -\omega^2 = e^{i\pi/3}$ (a unit-length rotation!): $(1+\omega)^{10} = e^{10 i \pi/3} = e^{-2\pi i/3} = \omega^2$, and conjugately $(1+\omega^2)^{10} = \omega$.

  3. Nudge 3

    Average the three evaluations; the two conjugate terms nearly cancel.

    Reveal step 3

    So count $= \frac{1}{3}\left[1024 + \omega + \omega^2\right] = \frac{1024 - 1}{3} = 341$, using $1 + \omega + \omega^2 = 0$. Sanity: $341 \approx 1024/3$ ✓, and it is exact arithmetic, not an estimate.

  4. Answer

    $341 = \frac{2^{10} - 1}{3}$ subsets — the filter at cube roots of unity, with $(1+\omega)^{10} + (1+\omega^2)^{10} = -1$.

Pitfall. Grinding $(1 + \omega)^{10}$ by binomial expansion. Convert $1 + \omega$ to polar form FIRST ($e^{i\pi/3}$) — powers become rotations and the arithmetic is two lines instead of twenty.

Before you open the gates

  • Know the five mother series cold; most sum evaluations are one of them under $x\frac{d}{dx}$, integration, or a clever evaluation point.
  • Check the evaluation point is strictly inside the radius before any term-by-term manipulation.
  • In any sum of fractions, hunt the term with the strictly smallest $v_p$ — it controls the whole sum.
  • Feynman's trick needs an anchor value; choose the parameter so $t = 0$ or $t \to \infty$ collapses the integral.
  • Counting by residue class of size/weight → roots-of-unity filter; convert $1 + \omega$ to polar before taking powers.

Check yourself

1. $\sum_{n=1}^{\infty} \frac{n}{3^n}$ equals:

2. $v_3(45) + v_3(18) - v_3(45 \cdot 18)$ equals:

3. The number of subsets of an $n$-set with EVEN size is:

Practice ladder — three rungs, rising

Each rung: attempt cold, one hint if stuck, worked resolution only after a real try.

Rung 1 (power series, mother series). Evaluate $\sum_{n=1}^{\infty} \frac{1}{n\,2^n}$.

One hint

Which of the five mother series has $\frac{x^n}{n}$ as its terms? Evaluate it at a point strictly inside its radius of convergence.

Worked resolution

The mother series $-\ln(1-x) = \sum_{n\ge1} \frac{x^n}{n}$ has radius of convergence $1$. Evaluating at $x = \tfrac12$ (strictly inside), $\sum_{n\ge1} \frac{1}{n\,2^n} = \sum_{n\ge1} \frac{(1/2)^n}{n} = -\ln\!\left(1-\tfrac12\right) = -\ln\tfrac12 = \ln 2$. The check $\tfrac12 < 1$ is what licenses evaluating the series term by term. [Source: this week's Track A — Analysis: Rudin Ch. 8 (power series and the five mother series; evaluate strictly inside the radius).]

Rung 2 (roots-of-unity filter). How many subsets of $\{1, 2, \dots, 9\}$ (including the empty set) have a sum divisible by $3$?

One hint

Encode subset sums as $F(x) = \prod_{k=1}^{9}(1+x^k)$ and filter mod $3$: the count is $\tfrac13\big(F(1)+F(\omega)+F(\omega^2)\big)$ with $\omega = e^{2\pi i/3}$. Group the factors by $k \bmod 3$.

Worked resolution

Subset sums are the exponents appearing in $F(x) = \prod_{k=1}^{9}(1+x^k)$, so the count is $\tfrac13\big(F(1)+F(\omega)+F(\omega^2)\big)$ with $\omega = e^{2\pi i/3}$. Here $F(1) = 2^9 = 512$. For $F(\omega)$, group $\{1,\dots,9\}$ by residue mod $3$: the three multiples of $3$ contribute $(1+1)=2$ each ($2^3 = 8$); the three $k\equiv1$ contribute $(1+\omega)$ each; the three $k\equiv2$ contribute $(1+\omega^2)$ each. Thus $F(\omega) = 8\,[(1+\omega)(1+\omega^2)]^3 = 8\cdot 1^3 = 8$, using $(1+\omega)(1+\omega^2) = 1+\omega+\omega^2+1 = 1$. By conjugation $F(\omega^2) = 8$ as well. Hence the count is $\tfrac13(512 + 8 + 8) = 176$. (Sanity: $176 \approx 512/3$.) [Source: this week's Hidden Tool — Root-of-Unity Filter (MIT 18.A34 roots handout; PnB section 2.2.4); the assigned classic of counting subsets with sum divisible by $n$.]

Rung 3 (Feynman's trick). For $a > 0$, evaluate $I(a) = \int_0^{\pi/2} \ln\!\big(a^2\sin^2\theta + \cos^2\theta\big)\,d\theta$.

One hint

Differentiate under the integral in $a$; the substitution $t = \tan\theta$ turns $I'(a)$ into a rational integral. Anchor at $a = 1$, where $I(1) = 0$.

Worked resolution

Differentiate under the integral sign — justified since the integrand and its $a$-derivative are continuous in $(a,\theta)$ for $a$ near any $a_0 > 0$ over the compact interval $[0,\tfrac{\pi}{2}]$: $I'(a) = \int_0^{\pi/2} \frac{2a\sin^2\theta}{a^2\sin^2\theta+\cos^2\theta}\,d\theta$. Substitute $t = \tan\theta$ (so $d\theta = \frac{dt}{1+t^2}$ and $\sin^2\theta = \frac{t^2}{1+t^2}$): $I'(a) = \int_0^{\infty} \frac{2a\,t^2}{(a^2 t^2+1)(t^2+1)}\,dt$. For $a \ne 1$, partial fractions give $\frac{t^2}{(a^2t^2+1)(t^2+1)} = \frac{1}{a^2-1}\!\left(\frac{1}{t^2+1} - \frac{1}{a^2t^2+1}\right)$, and with $\int_0^\infty \frac{dt}{t^2+1} = \frac{\pi}{2}$, $\int_0^\infty \frac{dt}{a^2t^2+1} = \frac{\pi}{2a}$ this yields $I'(a) = \frac{2a}{a^2-1}\cdot\frac{\pi}{2}\!\left(1 - \frac1a\right) = \frac{\pi(a-1)}{a^2-1} = \frac{\pi}{a+1}$. Integrating, $I(a) = \pi\ln(a+1) + C$; the anchor $I(1) = \int_0^{\pi/2}\ln 1\,d\theta = 0$ forces $C = -\pi\ln 2$. Therefore $I(a) = \pi\ln\frac{a+1}{2}$ (the value at $a=1$ following by continuity). [Source: this week's Hidden Tool — Feynman's Trick (differentiation under the integral sign); Nahin, Inside Interesting Integrals, Ch. 3, and MIT 18.A34 sum_integrals.pdf.]

Prove it — constructed response

Prove that for every integer $n \ge 2$, the harmonic sum $H_n = \sum_{k=1}^{n} \frac{1}{k}$ is not an integer. Use the $2$-adic valuation, and make the uniqueness of the dominating term explicit.

The gates

Asymptotics & Estimation — estimate under pressure (study before the analysis problems)

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Estimation toolkit (recognize on sight): Taylor with explicit remainder; isolate the dominant term; integral comparison ∑ vs ∫; Stirling n! ~ (n/e)^n√(2πn); log-transform products into sums; big-O / squeeze bounds; Abel summation for ∑aₙbₙ. This shows up in analysis, sums, products, number theory, and combinatorics — most A-session analysis problems are an estimate in disguise.

sources & assignments (5)
  • Source Michael Penn — Real Analysis (playlist) — Limit/series estimation, Taylor with remainder, bounding arguments
  • Source Maths 505 — Asymptotic integral/series estimates, Stirling-type scaling, clever bounds
  • Source Radulescu — Problems in Real Analysis — Rădulescu *Problems in Real Analysis* Ch. 3 (limits/series) — do 5 estimation problems (integral comparison, Stirling-type bounds)
  • Source Anulus Smaragdinus — Polynomials (playlist) — Intuition companion for “Asymptotics & Estimation — estimate under pressure (study before the analysis problems)” — the picture behind the machinery; afterwards write one sentence connecting the visual to this gate's exercises. Watch-for: olympiad polynomial-problem walkthroughs (titled by problem number, not topic) — attempt each problem cold before watching the solution.
  • Problems Asymptotics notation + estimation ladder (self-contained) — Written, ~30 min: (1) State the definitions of big-O, little-o, and Θ (Theta) precisely (with the limit or constant-witness form), then rank the growth hierarchy log n ≺ n^a (a>0) ≺ n^a (log n)^b ≺ c^n ≺ n! ≺ n^n and justify two adjacent gaps by an explicit limit. (2) Harmonic estimate: prove H_n = Σ_{k=1}^n 1/k = ln n + γ + O(1/n) at the O(1) level by the integral comparison ∫_1^{n+1} dx/x ≤ H_n ≤ 1 + ∫_1^n dx/x; conclude H_n = ln n + Θ(1). (3) Dominant-term + squeeze: find lim_{n→∞} (2^n + n^5)^{1/n} by squeezing between 2 and 2(1 + n^5/2^n)^{1/n}; name the dominant term first. (4) Recurrence asymptotics: for a_n = 2a_{n/2} + n, state the Θ from the recursion-tree intuition (Θ(n log n)) and say in one line which term dominates. (5) Central binomial coefficient: bound 4ⁿ/(2n+1) ≤ C(2n,n) ≤ 4ⁿ — the lower bound because C(2n,n) is the largest of the 2n+1 terms summing to Σₖ C(2n,k) = 4ⁿ, the upper bound trivially; then state the Stirling-sharp form C(2n,n) ~ 4ⁿ/√(πn) and check the bounds are consistent. (Restores the central-binomial bound displaced by the W20 AbA swap.)Putnam

AMC/AIME Speed Warmup

AMC/AIME Speed Warmup (build pattern recognition + speed)

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- AMC/AIME speed warmup — fast pattern-recognition reps; not full Putnam difficulty.

sources & assignments (5)
  • Source Rudin — Principles of Mathematical Analysis (PMA) — Week reference — this gate applies this week's lead material (“Track A — Analysis: Rudin Ch. 8 (1.5 hrs/day): Load”). If an attempt stalls, the repair source is here.
  • Source Silver — Integration Bee Training (playlist) — Week reference — this gate applies this week's lead material (“Track A — Analysis: Rudin Ch. 8 (1.5 hrs/day): Load”). If an attempt stalls, the repair source is here.
  • Source Maths 505 — Cool Integrals (Feynman tricks, playlist) — Week reference — this gate applies this week's lead material (“Track A — Analysis: Rudin Ch. 8 (1.5 hrs/day): Load”). If an attempt stalls, the repair source is here.
  • Source Anulus Smaragdinus — Polynomials (playlist) — Intuition companion for “AMC/AIME Speed Warmup (build pattern recognition + speed)” — the picture behind the machinery; afterwards write one sentence connecting the visual to this gate's exercises. Watch-for: olympiad polynomial-problem walkthroughs (titled by problem number, not topic) — attempt each problem cold before watching the solution.
  • Problems Archive (hand-picked) — 3 AMC/AIME number theory speed reps (open each below) · ~8 min eachAMC/AIME · archive: AMC 12 BAMC 12 BAMC 12 A

Track A — Analysis

Track A — Analysis: Rudin Ch. 8 (1.5 hrs/day): Load

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- Read: Rudin *PMA* Ch. 8 section section 8.1–8.5 (~30 pp.) — power series, Abel's theorem, rigorous definitions of eˣ, log, sin, cos; Fourier series section section 8.9–8.13 (read results only) - Watch: Silver Integration Bee Advanced — 3 videos - Watch: Maths 505 Feynman integration — 3 videos - Do: Rudin Ch. 8 ex 1, 4, 7, 10, 14 - Do: MIT sum_integrals.pdf probs 7–8 — Abel/summation-by-parts structure - Do: Stanford 06wk3 probs 5–8

sources & assignments (7)

Track A — Analysis: Rudin Ch. 8 (1.5 hrs/day): Drill

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unlocks after: Track A — Analysis: Rudin Ch. 8 (1.5 hrs/day): Load

- Read: Rudin *PMA* Ch. 8 section section 8.1–8.5 (~30 pp.) — power series, Abel's theorem, rigorous definitions of eˣ, log, sin, cos; Fourier series section section 8.9–8.13 (read results only) - Watch: Silver Integration Bee Advanced — 3 videos - Watch: Maths 505 Feynman integration — 3 videos - Do: Rudin Ch. 8 ex 1, 4, 7, 10, 14 - Do: MIT sum_integrals.pdf probs 7–8 — Abel/summation-by-parts structure - Do: Stanford 06wk3 probs 5–8

sources & assignments (7)
  • Source Rudin — Principles of Mathematical Analysis (PMA) — Drill reference — this gate trains the material of “Track A — Analysis: Rudin Ch. 8 (1.5 hrs/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source Silver — Integration Bee Training (playlist) — Drill reference — this gate trains the material of “Track A — Analysis: Rudin Ch. 8 (1.5 hrs/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source Maths 505 — Cool Integrals (Feynman tricks, playlist) — Drill reference — this gate trains the material of “Track A — Analysis: Rudin Ch. 8 (1.5 hrs/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source Anulus Smaragdinus — Polynomials (playlist) — Intuition companion for “Track A — Analysis: Rudin Ch. 8 (1.5 hrs/day): Drill” — the picture behind the machinery; afterwards write one sentence connecting the visual to this gate's exercises. Watch-for: olympiad polynomial-problem walkthroughs (titled by problem number, not topic) — attempt each problem cold before watching the solution.
  • Problems Rudin — Rudin Ch. 8 ex 1, 4, 7, 10, 14Putnam
  • Problems MIT sum_integrals.pdf — MIT 18.A34 sum_integrals.pdf — problems 5–8 (Feynman parameter integrals)Putnam
  • Problems Stanford 06wk3 — Stanford 06wk3 probs 5–8Putnam

Hidden Tool

Hidden Tool: Feynman's Trick (Differentiation Under the Integral Sign)

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*Prerequisites: FTC (Week 6) + Leibniz rule (Week 5) + power series (Track A above)* - Do — 3 targeted problems: 1. Evaluate ∫₀¹ (xᵃ−1)/ln(x) dx by differentiating with respect to a. Set I(a) = ∫₀¹ xᵃ dx/(... ), compute I'(a) = 1/(a+1), integrate → I(a) = ln(a+1). Verify I(0) = 0. 2. Jago Alexander hard integrals playlist — 1 problem using Feynman's trick. 3. MIT sum_integrals.pdf — solve one Feynman-technique problem from scratch without hints. - Feynman: - Introduce parameter α: replace a constant in the integrand with α - Differentiate: I'(α) = ∫ ∂f/∂α dx (justified by uniform convergence/Leibniz) - Solve simpler integral: I'(α) is often elementary - Integrate back: I(α) from I'(α); fix constant using known value I(α₀) - If I(α) satisfies an ODE in α: solve via ODE card instead of direct integration - **Derive it cold (required, self-attested — no notes):** before you tick this gate, reconstruct differentiation under the integral sign blank-page: for I(t)=∫_a^b f(x,t) dx, form the difference quotient [I(t+h)−I(t)]/h = ∫_a^b ([f(x,t+h)−f(x,t)]/h) dx, apply the Mean Value Theorem in t to the inner ratio (it equals ∂_t f(x,ξ) for some ξ between t and t+h), and pass h→0 inside the integral — clean sufficient license: ∂_t f continuous on the compact rectangle [a,b]×(t-interval); more generally f(·,t) differentiable in t with |∂_t f| ≤ a fixed integrable g(x) independent of t (dominated-convergence) — giving I′(t)=∫_a^b ∂_t f(x,t) dx. Then the METHOD: plant a parameter t so I(t_0) is your target, differentiate to a solvable I′(t) (an elementary integral, or a first-order ODE in t), integrate back, and pin the constant with an anchor value I(t_1) you can compute directly. Self-check: (1) you *established* the swap from the difference quotient plus a continuity/domination license, didn't quote "Leibniz"; (2) you can say WHY the license is needed — without it the limit and the integral need not commute. Can't rebuild it blank-page → not learned yet.

why this gate: the lesson's §3 Feynman's trick: differentiate the parameter you invented is what it trains

sources & assignments (6)
  • Problems Targeted drill (built on this gate's reading) — (1) Evaluate ∫₀¹ (xᵃ−1)/ln(x) dx by differentiating with respect to a. Set I(a) = ∫₀¹ xᵃ dx/(... ), compute I'(a) = 1/(a+1), integrate → I(a) = ln(a+1). Verify I(0) = 0. (2) Jago Alexander hard integrals playlist — 1 problem using Feynman's trick. (3) MIT sum_integrals.pdf — solve one Feynman-technique problem from scratch without hints.Drill

Track B — Algebra

Track B — Algebra: Polynomial Advanced (1.5 hrs/day): Load

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- Read: PnB section 2.2.3–2.2.6 — Newton's identities, integer polynomials, roots of unity advanced - Read: CMU 11-Integer-Polynomials — Vandermonde, Lagrange interpolation - Watch: Anulus Smaragdinus — Vieta's formulas advanced + Newton's sums videos - Do: MIT polynomials.pdf probs 13–17 - Do: CMU 11-Integer-Poly exercise sheet probs 1–2 - Do: 117 Polynomial Problems #50, #60, #70 - Archive: 3 Putnam A2/A3 · Algebra/Polynomial · any slot · 20 min each

sources & assignments (5)

Track B — Algebra: Polynomial Advanced (1.5 hrs/day): Drill

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unlocks after: Track B — Algebra: Polynomial Advanced (1.5 hrs/day): Load

- Read: PnB section 2.2.3–2.2.6 — Newton's identities, integer polynomials, roots of unity advanced - Read: CMU 11-Integer-Polynomials — Vandermonde, Lagrange interpolation - Watch: Anulus Smaragdinus — Vieta's formulas advanced + Newton's sums videos - Do: MIT polynomials.pdf probs 13–17 - Do: CMU 11-Integer-Poly exercise sheet probs 1–2 - Do: 117 Polynomial Problems #50, #60, #70 - Archive: 3 Putnam A2/A3 · Algebra/Polynomial · any slot · 20 min each

sources & assignments (7)
  • Source Putnam and Beyond — Drill reference — this gate trains the material of “Track B — Algebra: Polynomial Advanced (1.5 hrs/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source Anulus Smaragdinus — Polynomials (playlist) — Drill reference — this gate trains the material of “Track B — Algebra: Polynomial Advanced (1.5 hrs/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing. Watch-for: olympiad polynomial-problem walkthroughs (titled by problem number, not topic) — attempt each problem cold before watching the solution.
  • Source Michael Penn — Newton's Sums — Drill reference — this gate trains the material of “Track B — Algebra: Polynomial Advanced (1.5 hrs/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source Michael Penn — Putnam Exam Solutions (playlist) — Intuition companion for “Track B — Algebra: Polynomial Advanced (1.5 hrs/day): Drill” — the picture behind the machinery; afterwards write one sentence connecting the visual to this gate's exercises.
  • Problems MIT polynomials.pdf — MIT polynomials.pdf probs 13–17Putnam
  • Problems CMU 11-Integer-Poly exercise sheet — CMU 11-Integer-Poly exercise sheet probs 1–2Putnam
  • Problems 117 Polynomial — 117 Polynomial Problems #50, #60, #70Putnam

Track B — Algebra: Polynomial Advanced (1.5 hrs/day): Archive bridge

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unlocks after: Track B — Algebra: Polynomial Advanced (1.5 hrs/day): Drill

- Read: PnB section 2.2.3–2.2.6 — Newton's identities, integer polynomials, roots of unity advanced - Read: CMU 11-Integer-Polynomials — Vandermonde, Lagrange interpolation - Watch: Anulus Smaragdinus — Vieta's formulas advanced + Newton's sums videos - Do: MIT polynomials.pdf probs 13–17 - Do: CMU 11-Integer-Poly exercise sheet probs 1–2 - Do: 117 Polynomial Problems #50, #60, #70 - Archive: 3 Putnam A2/A3 · Algebra/Polynomial · any slot · 20 min each

sources & assignments (5)
  • Source Putnam and Beyond — Drill reference — this gate trains the material of “Track B — Algebra: Polynomial Advanced (1.5 hrs/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source Anulus Smaragdinus — Polynomials (playlist) — Drill reference — this gate trains the material of “Track B — Algebra: Polynomial Advanced (1.5 hrs/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing. Watch-for: olympiad polynomial-problem walkthroughs (titled by problem number, not topic) — attempt each problem cold before watching the solution.
  • Source Michael Penn — Newton's Sums — Drill reference — this gate trains the material of “Track B — Algebra: Polynomial Advanced (1.5 hrs/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source Michael Penn — Putnam Exam Solutions (playlist) — Intuition companion for “Track B — Algebra: Polynomial Advanced (1.5 hrs/day): Archive bridge” — the picture behind the machinery; afterwards write one sentence connecting the visual to this gate's exercises.
  • Problems Archive Browser — Putnam 2020 A3, 2019 A3 · Algebra · 20 min eachPutnam

archive pull: 2 problems · Algebra · A2/A3 · Challenge · 20 min →

Hidden Tool

Hidden Tool: Root-of-Unity Filter: Load

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*Prerequisites: complex numbers (Week 4) + polynomial advanced reading (Track B above)* - Read: MIT 18.A34 roots handout (PUTNAM_SEMINAR_SPINE.mit2018.roots) - Do: 2 problems from the handout using the filter (1/n)Σⱼf(ωʲ) = sum of coefficients at positions divisible by n - Root-of-Unity Filter: - ω = e^{2πi/n}; then (1/n)Σⱼ₌₀ⁿ⁻¹ f(ωʲ) = sum of [xᵏ]f(x) for k ≡ 0 mod n - Use when: extracting every n-th term from a polynomial or generating function - Classic: "how many subsets of {1,...,m} have sum divisible by n?" → apply to (1+x)(1+x²)···(1+xᵐ) at roots of unity Prereq refresher: the filter sums binomial coefficients via (1+omega^k)^n — if the binomial theorem is rusty, re-derive (x+y)^n by induction first (5 min, assumed from Week 2 AMC fluency). - **Derive it cold (required, self-attested — no notes):** before you tick this gate, reconstruct the filter from scratch: with ω = e^{2πi/n} (n a positive integer), evaluate Σ_{j=0}^{n−1} ω^{jk} — if n | k then ω^k=1 and all n terms are 1, giving n; otherwise ω^k≠1 and the geometric sum is (ω^{kn}−1)/(ω^k−1)=0 since ω^{kn}=(ω^n)^k=1 — hence (1/n)Σ_{j=0}^{n−1} ω^{jk} = [n | k]. Apply it to a polynomial F(x)=Σ_k a_k x^k (finite support, so the interchange below is a finite reordering): (1/n)Σ_j ω^{−jr} F(ω^j) = Σ_k a_k·(1/n)Σ_j ω^{j(k−r)} = Σ_{k ≡ r (mod n)} a_k, isolating one residue class of coefficients. Self-check: (1) you *established* the indicator from the two-case geometric sum, didn't quote it; (2) you can say WHY the off-class terms vanish — 1+ζ+⋯+ζ^{n−1}=0 for any n-th root of unity ζ≠1, since (ζ−1)(1+⋯+ζ^{n−1})=ζ^n−1=0. Can't rebuild it blank-page → not learned yet.

why this gate: the lesson's §4 Roots-of-unity filter: sampling a generating function is what it trains

sources & assignments (6)

Hidden Tool: Root-of-Unity Filter: Drill

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unlocks after: Hidden Tool: Root-of-Unity Filter: Load

*Prerequisites: complex numbers (Week 4) + polynomial advanced reading (Track B above)* - Read: MIT 18.A34 roots handout (PUTNAM_SEMINAR_SPINE.mit2018.roots) - Do: 2 problems from the handout using the filter (1/n)Σⱼf(ωʲ) = sum of coefficients at positions divisible by n - Root-of-Unity Filter: - ω = e^{2πi/n}; then (1/n)Σⱼ₌₀ⁿ⁻¹ f(ωʲ) = sum of [xᵏ]f(x) for k ≡ 0 mod n - Use when: extracting every n-th term from a polynomial or generating function - Classic: "how many subsets of {1,...,m} have sum divisible by n?" → apply to (1+x)(1+x²)···(1+xᵐ) at roots of unity

why this gate: the lesson's §4 Roots-of-unity filter: sampling a generating function is what it trains

sources & assignments (4)
  • Source Anulus Smaragdinus — Polynomials (playlist) — Re-watch the worked-solution segment only AFTER attempting this gate's drill problems; note where your route diverged. Watch-for: olympiad polynomial-problem walkthroughs (titled by problem number, not topic) — attempt each problem cold before watching the solution.
  • Source MIT 18.01SC — Single Variable Calculus (OCW) — Problem-session companion for “Hidden Tool: Root-of-Unity Filter: Drill” — watch one worked problem, stop, finish it yourself on paper, then compare.
  • Source 3Blue1Brown — Essence of Calculus — Intuition companion for “Hidden Tool: Root-of-Unity Filter: Drill” — the picture behind the machinery; afterwards write one sentence connecting the visual to this gate's exercises.
  • Problems 2 — 2 problems from the handout using the filter (1/n)Σⱼf(ωʲ) = sum of coefficients at positions divisible by nPutnam

Track D — Linear Algebra

Track D — Linear Algebra: Putnam Applications (1 hr/day): Load

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- Read: Yufei Zhao LA tricks handout (yufeizhao.com/olympiad/putnam_linear_algebra.pdf) - Read: PnB section 2.3.4 (Cayley-Hamilton pp. 70–73) - Watch: Sheldon Axler LADR lectures — lectures 17–18 - Do: PnB section 2.3.4 probs 1–2 · section 2.3.7 probs 1–2 - Do: CMU 09-Linear exercise sheet probs 1–2 - Archive: 3 Putnam A1/B1 · Linear Algebra · any slot · 15 min each

sources & assignments (5)

Track D — Linear Algebra: Putnam Applications (1 hr/day): Drill

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unlocks after: Track D — Linear Algebra: Putnam Applications (1 hr/day): Load

- Read: Yufei Zhao LA tricks handout (yufeizhao.com/olympiad/putnam_linear_algebra.pdf) - Read: PnB section 2.3.4 (Cayley-Hamilton pp. 70–73) - Watch: Sheldon Axler LADR lectures — lectures 17–18 - Do: PnB section 2.3.4 probs 1–2 · section 2.3.7 probs 1–2 - Do: CMU 09-Linear exercise sheet probs 1–2 - Archive: 3 Putnam A1/B1 · Linear Algebra · any slot · 15 min each

sources & assignments (6)
  • Source Yufei Zhao — Linear Algebra Tricks for the Putnam — Drill reference — this gate trains the material of “Track D — Linear Algebra: Putnam Applications (1 hr/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source Putnam and Beyond — Drill reference — this gate trains the material of “Track D — Linear Algebra: Putnam Applications (1 hr/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source Sheldon Axler — Linear Algebra Done Right (lecture videos) — Drill reference — this gate trains the material of “Track D — Linear Algebra: Putnam Applications (1 hr/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source Michael Penn — Putnam Exam Solutions (playlist) — Intuition companion for “Track D — Linear Algebra: Putnam Applications (1 hr/day): Drill” — the picture behind the machinery; afterwards write one sentence connecting the visual to this gate's exercises.
  • Problems PnB — PnB section 2.3.4 probs 1–2 · section 2.3.7 probs 1–2Putnam
  • Problems CMU 09-Linear exercise sheet — CMU 09-Linear exercise sheet probs 1–2Putnam

Track D — Linear Algebra: Putnam Applications (1 hr/day): Archive bridge

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unlocks after: Track D — Linear Algebra: Putnam Applications (1 hr/day): Drill

- Read: Yufei Zhao LA tricks handout (yufeizhao.com/olympiad/putnam_linear_algebra.pdf) - Read: PnB section 2.3.4 (Cayley-Hamilton pp. 70–73) - Watch: Sheldon Axler LADR lectures — lectures 17–18 - Do: PnB section 2.3.4 probs 1–2 · section 2.3.7 probs 1–2 - Do: CMU 09-Linear exercise sheet probs 1–2 - Archive: 3 Putnam A1/B1 · Linear Algebra · any slot · 15 min each

sources & assignments (5)
  • Source Yufei Zhao — Linear Algebra Tricks for the Putnam — Drill reference — this gate trains the material of “Track D — Linear Algebra: Putnam Applications (1 hr/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source Putnam and Beyond — Drill reference — this gate trains the material of “Track D — Linear Algebra: Putnam Applications (1 hr/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source Sheldon Axler — Linear Algebra Done Right (lecture videos) — Drill reference — this gate trains the material of “Track D — Linear Algebra: Putnam Applications (1 hr/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source Michael Penn — Putnam Exam Solutions (playlist) — Intuition companion for “Track D — Linear Algebra: Putnam Applications (1 hr/day): Archive bridge” — the picture behind the machinery; afterwards write one sentence connecting the visual to this gate's exercises.
  • Problems Archive Browser — Putnam 1966 B1 · Linear Algebra · 15 min eachPutnam

archive pull: 1 problems · Linear Algebra · A1/B1 · Standard · 15 min →

Track C — Combinatorics

Track C — Combinatorics: Recurrences + Game Theory (1 hr/day): Load

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- Read: Stanford 07wk3 (Recurrences) — full handout - Watch: MIT 6.042J Combinatorial Games lecture (OCW, ~45 min) - Do: Stanford 07wk3 probs 1–3 · Stanford 06wk2 probs 7–10 - Do: MIT ss8.pdf prob 99 (coin-rotation game) · MIT ss9.pdf prob 104 (P/N positions) - Do: Engel Ch. 8 game-type probs 1–5 - Recurrence Methods: - Direct/closed form - Telescoping - Characteristic polynomial - Generating function - Matrix method - Combinatorial interpretation - Game Strategy: - P/N positions (backward induction) - Pairing strategy - Strategy stealing - Game invariant - Mod-cycle - Symmetry

sources & assignments (6)

Track C — Combinatorics: Recurrences + Game Theory (1 hr/day): Drill

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unlocks after: Track C — Combinatorics: Recurrences + Game Theory (1 hr/day): Load

- Read: Stanford 07wk3 (Recurrences) — full handout - Watch: MIT 6.042J Combinatorial Games lecture (OCW, ~45 min) - Do: Stanford 07wk3 probs 1–3 · Stanford 06wk2 probs 7–10 - Do: MIT ss8.pdf prob 99 (coin-rotation game) · MIT ss9.pdf prob 104 (P/N positions) - Do: Engel Ch. 8 game-type probs 1–5 - Recurrence Methods: - Direct/closed form - Telescoping - Characteristic polynomial - Generating function - Matrix method - Combinatorial interpretation - Game Strategy: - P/N positions (backward induction) - Pairing strategy - Strategy stealing - Game invariant - Mod-cycle - Symmetry

sources & assignments (7)
  • Source Stanford 07wk3 (Recurrences) — Drill reference — this gate trains the material of “Track C — Combinatorics: Recurrences + Game Theory (1 hr/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source MIT 6.042J Combinatorial Games lecture (OCW, ~45 min) — Drill reference — this gate trains the material of “Track C — Combinatorics: Recurrences + Game Theory (1 hr/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source Generating-functions lecture series — Drill reference — this gate trains the material of “Track C — Combinatorics: Recurrences + Game Theory (1 hr/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source Maths 505 — Complex Analysis Lectures (playlist) — Intuition companion for “Track C — Combinatorics: Recurrences + Game Theory (1 hr/day): Drill” — the picture behind the machinery; afterwards write one sentence connecting the visual to this gate's exercises.
  • Problems Stanford 07wk3 — Stanford 07wk3 probs 1–3 · Stanford 06wk2 probs 7–10Putnam
  • Problems MIT ss8.pdf — MIT ss8.pdf prob 99 (coin-rotation game) · MIT ss9.pdf prob 104 (P/N positions)Putnam
  • Problems Engel — Engel Ch. 8 game-type probs 1–5Putnam

Track E — Number Theory

Track E — Number Theory: Valuations + Advanced (0.75 hrs/day): Load

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- Read: PnB section 5.1.3 (valuations pp. 252–256) + section 5.2.4–5.2.7 (Euler, Wilson, quadratic residues) - Read: CMU 03-NT advanced section - Do: 104 NT Problems #51, 55 - Do: MIT ss7.pdf probs 77–79

why this gate: the lesson's §2 p-adic valuation: divisibility you can add is what it trains

sources & assignments (4)

Track E — Number Theory: Valuations + Advanced (0.75 hrs/day): Drill

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unlocks after: Track E — Number Theory: Valuations + Advanced (0.75 hrs/day): Load

- Read: PnB section 5.1.3 (valuations pp. 252–256) + section 5.2.4–5.2.7 (Euler, Wilson, quadratic residues) - Read: CMU 03-NT advanced section - Do: 104 NT Problems #51, 55 - Do: MIT ss7.pdf probs 77–79

why this gate: the lesson's §2 p-adic valuation: divisibility you can add is what it trains

sources & assignments (6)
  • Source Putnam and Beyond — Drill reference — this gate trains the material of “Track E — Number Theory: Valuations + Advanced (0.75 hrs/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source CMU 03-Number Theory (Putnam seminar) — Drill reference — this gate trains the material of “Track E — Number Theory: Valuations + Advanced (0.75 hrs/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source Michael Penn — Number Theory v2 (playlist) — Drill reference — this gate trains the material of “Track E — Number Theory: Valuations + Advanced (0.75 hrs/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source Michael Penn — Modular Arithmetic and Linear Congruences (playlist) — Intuition companion for “Track E — Number Theory: Valuations + Advanced (0.75 hrs/day): Drill” — the picture behind the machinery; afterwards write one sentence connecting the visual to this gate's exercises.
  • Problems 104 NT — 104 NT Problems #51, 55 Also, recognition-level: (a) derive the primitive Pythagorean-triple parametrization (m²−n², 2mn, m²+n²) from x²+y²=z² with the gcd/parity conditions on m, n; (b) define the Möbius function μ(n) and the divisor functions d(n) (number of divisors) and σ(n) (sum of divisors), and state Σ_{d|n} μ(d) = [n = 1] as the inclusion–exclusion recognition trigger behind Möbius inversion.Putnam
  • Problems MIT ss7.pdf — MIT ss7.pdf probs 77–79Putnam

Hidden Tool

Hidden Tool: Valuations

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*Prerequisites: NT foundations (Weeks 5–7)* - Valuation: - vₚ(n) = largest k such that pᵏ | n - vₚ(ab) = vₚ(a) + vₚ(b) · vₚ(a+b) ≥ min(vₚ(a), vₚ(b)) with equality when vₚ(a) ≠ vₚ(b) - Legendre: vₚ(n!) = Σ⌊n/pᵏ⌋ = (n−sₚ(n))/(p−1) - LTE preview: if p | a+b (odd p), then vₚ(aⁿ+bⁿ) = vₚ(a+b) + vₚ(n) - **Derive it cold (required, self-attested — no notes):** before you tick this gate, reconstruct the valuation laws and LTE blank-page. (i) v_p(ab)=v_p(a)+v_p(b): write a=p^α a′, b=p^β b′ with p∤a′,b′; then ab=p^{α+β}(a′b′) and p∤a′b′, so the p-exponent adds. (ii) v_p(a+b) ≥ min(v_p a, v_p b): pull p^m out, m=min, leaving an integer bracket; equality holds when v_p a≠v_p b because, taking m=v_p a<v_p b, a+b=p^m(a′+p^{β−m}b′) with a′ a unit mod p and the second summand ≡0, so the bracket is a unit and the valuation is exactly m. (iii) LTE — hypotheses: p an ODD prime, integers a,b with a≠b, p | a−b, p∤a, p∤b, and n a positive integer. From a^n−b^n=(a−b)·S with S=Σ_{k=0}^{n−1} a^{n−1−k} b^k, use a≡b (mod p) so every term ≡b^{n−1}, giving S≡n·b^{n−1} (mod p); if p∤n then p∤S, so v_p(a^n−b^n)=v_p(a−b). The extra +v_p(n) comes from the lifting step v_p(a^p−b^p)=v_p(a−b)+1: writing a=b+d with p | d, expand S mod p² — every d²-and-higher term dies (p | d ⇒ p² | d²), leaving S≡p·b^{p−1}+b^{p−2}·d·(p(p−1)/2), whose second term ≡0 mod p² ONLY because odd p makes p(p−1)/2 a multiple of p — so v_p(S)=1; then induct over the p-part of n. Self-check: (1) you *established* each law, didn't quote it; (2) you can name every LTE hypothesis AND say WHY odd p is required — for p=2 the p(p−1)/2 step fails and instead v_2(a^n−b^n)=v_2(a−b)+v_2(a+b)+v_2(n)−1 for even n (a,b odd). Can't rebuild it blank-page → not learned yet.

why this gate: the lesson's §2 p-adic valuation: divisibility you can add is what it trains

sources & assignments (6)
  • Problems Valuation drill (standard examples) — (1) Compute v_5(1000!) with Legendre's formula; (2) prove v_p(a+b) >= min(v_p(a), v_p(b)) with equality whenever v_p(a) != v_p(b); (3) re-prove the irrationality of sqrt(2) purely via the parity of v_2. Write all three as complete proofs.Putnam

Hidden Tool

Hidden Tool: Periodicity mod m / Finite-State Sequences

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*Hidden Tool — recognise when a sequence lives in a finite state space.* - Finite state ⇒ eventual periodicity: if xₙ ∈ S finite and xₙ₊₁ = f(xₙ), two indices must repeat (pigeonhole) ⇒ the tail is periodic; purely periodic iff f is injective on the reachable set. - Mod m: any integer recurrence reduced mod m has finite state ⇒ eventually periodic; invertible/linear ⇒ purely periodic. - Fibonacci mod m: purely periodic, Pisano period π(m); π(10)=60. - Use it to: force a value, bound a term, or prove a sequence hits / avoids a residue. - **Derive it cold (required, self-attested — no notes):** before you tick this gate, reconstruct eventual periodicity blank-page: a depth-k recurrence x_{n+1}=f(x_n,…,x_{n−k+1}) taken mod m is driven by its state tuple s_n=(x_n,…,x_{n−k+1}) ∈ (ℤ/mℤ)^k, a set of exactly m^k elements; among s_0,…,s_{m^k} (that is m^k+1 tuples) two must coincide by pigeonhole, say s_i=s_j with i<j≤m^k; because the next state is a deterministic function of the current one, s_i=s_j forces s_{i+t}=s_{j+t} for all t≥0, so the sequence is periodic from index i with period dividing j−i ≤ m^k. It is PURELY periodic (from n=0, no pre-period) exactly when the state map is a bijection on the reachable states — e.g. a linear recurrence whose companion matrix is invertible mod m — since then no two distinct states can merge. Self-check: (1) you *established* it from pigeonhole + determinism, didn't quote "eventually periodic"; (2) you can say WHY finiteness forces a repeat and WHY determinism propagates that repeat forever forward. Can't rebuild it blank-page → not learned yet.

sources & assignments (4)
  • Problems Periodicity drill (standard examples) — (1) Prove any sequence with x_{n+1}=f(x_n), all x_n in a finite set S, is eventually periodic — and purely periodic when f is injective on the reachable set. (2) Show the Fibonacci sequence is purely periodic mod m and compute the Pisano period mod 10. (3) Prove a linear recurrence a_{n+1}≡c·a_n+d (mod m) is eventually periodic; identify exactly when it is purely periodic. Write all three as complete proofs.Putnam

archive pull: 3 problems · Number Theory/Combinatorics · A2/B1 · Standard,Hard · 15 min →

Track F — Geometry + Probability + Abstract

Track F — Geometry + Probability + Abstract (0.5 hrs/day): Load

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- Do: CMU 14-Geometry exercise sheet probs 1–2 - Read: CMU 12-Probability — lecture notes - Archive: 2 Putnam A1/B1 · Geometry · any slot · 12 min each

sources & assignments (6)
  • Source CMU 12-Probability — CMU 12-Probability — lecture notes
  • Source CMU 14-Geometry — CMU 14-geometry exercise sheet — coordinate/vector/area methods
  • Source OlympicMathTutor — Olympiad Geometry Theorems (playlist) — Evan Chen geometry notes — coordinate & complex methods (intro)
  • Source Cezar Lupu — TTU MATH 4000, Recitation 13 — Intuition companion for “Track F — Geometry + Probability + Abstract (0.5 hrs/day): Load” — the picture behind the machinery; afterwards write one sentence connecting the visual to this gate's exercises.
  • Source Michael Penn — Abstract Algebra (playlist) — Companion lecture series for this week's abstract-algebra rep — watch the relevant 1–2 videos, then do the written rep below. Watch-for: this playlist opens with set-theory / proof-writing videos — skip to the 'Abstract Algebra |' entries for the group/ring/field content this gate needs.
  • Problems Abstract algebra — Lagrange's theorem (self-contained) — Prove |H| divides |G| for a finite group G via the coset partition (left cosets are equal-sized and disjoint). Deduce that the order of every element divides |G|. Apply it to re-derive Fermat's little theorem a^(p−1) ≡ 1 (mod p) by working in the unit group (ℤ/pℤ)*. (Replaces the recognition-only Dummit/Foote reading with a proof rep.)Putnam

Track F — Geometry + Probability + Abstract (0.5 hrs/day): Drill

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unlocks after: Track F — Geometry + Probability + Abstract (0.5 hrs/day): Load

- Do: CMU 14-Geometry exercise sheet probs 1–2 - Read: CMU 12-Probability — lecture notes - Archive: 2 Putnam A1/B1 · Geometry · any slot · 12 min each

sources & assignments (6)
  • Source CMU 14-Geometry — Drill reference — this gate trains the material of “Track F — Geometry + Probability + Abstract (0.5 hrs/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source CMU 12-Probability — Drill reference — this gate trains the material of “Track F — Geometry + Probability + Abstract (0.5 hrs/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source OlympicMathTutor — Olympiad Geometry Theorems (playlist) — Drill reference — this gate trains the material of “Track F — Geometry + Probability + Abstract (0.5 hrs/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source Cezar Lupu — TTU MATH 4000, Recitation 13 — Intuition companion for “Track F — Geometry + Probability + Abstract (0.5 hrs/day): Drill” — the picture behind the machinery; afterwards write one sentence connecting the visual to this gate's exercises.
  • Problems CMU 14-Geometry exercise sheet — CMU 14-Geometry exercise sheet prob 1 (prob 2's slot displaced to the radical-axis rep)Putnam
  • Problems Radical-axis rep (self-contained) — State power of a point and define the radical axis of two circles (locus of equal power; a line perpendicular to the center line). Prove the three radical axes of three circles with non-collinear centers concur (the radical center) — two-line argument from the definition. Then verify with coordinates on one concrete triple of circles.Putnam

Track F — Geometry + Probability + Abstract (0.5 hrs/day): Archive bridge

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unlocks after: Track F — Geometry + Probability + Abstract (0.5 hrs/day): Drill

- Do: CMU 14-Geometry exercise sheet probs 1–2 - Read: CMU 12-Probability — lecture notes - Archive: 2 Putnam A1/B1 · Geometry · any slot · 12 min each

sources & assignments (5)
  • Source CMU 14-Geometry — Drill reference — this gate trains the material of “Track F — Geometry + Probability + Abstract (0.5 hrs/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source CMU 12-Probability — Drill reference — this gate trains the material of “Track F — Geometry + Probability + Abstract (0.5 hrs/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source OlympicMathTutor — Olympiad Geometry Theorems (playlist) — Drill reference — this gate trains the material of “Track F — Geometry + Probability + Abstract (0.5 hrs/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source Cezar Lupu — TTU MATH 4000, Recitation 13 — Intuition companion for “Track F — Geometry + Probability + Abstract (0.5 hrs/day): Archive bridge” — the picture behind the machinery; afterwards write one sentence connecting the visual to this gate's exercises.
  • Problems Archive Browser — Putnam 1967 B1 · Geometry · 12 min eachPutnam

archive pull: 1 problems · Geometry · A1/B1 · Standard · 12 min →

Hidden Tool

Hidden Tool: Divisor Sums & Double-Counting (σ, τ, coprime pairs)

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Divisor-sum / Möbius / coprime double-counting (~2.5% of the archive) — was light; this adds a focused rep. - **Derive it cold (required, self-attested — no notes):** before you tick this gate, reconstruct the divisor formulas blank-page: for n=∏ p_i^{e_i}, a divisor d picks an exponent 0 ≤ a_i ≤ e_i independently at each prime, so divisors biject with tuples (a_1,…,a_r) — counting them gives τ(n)=∏(e_i+1), and summing d=∏ p_i^{a_i} factors across primes into a product of finite geometric series σ(n)=∏_i (1+p_i+⋯+p_i^{e_i})=∏_i (p_i^{e_i+1}−1)/(p_i−1). Multiplicativity (f(mn)=f(m)f(n) when gcd(m,n)=1) is the same bijection: each divisor of mn splits uniquely as (divisor of m)·(divisor of n). Then the double-counting swap: Σ_{n≤N} Σ_{d | n} g(d) counts pairs (d,n) with d | n, so reindex by n=d·q to get Σ_{d≤N} g(d)⌊N/d⌋ (the Dirichlet/hyperbola reordering). Self-check: (1) you *established* the formulas from the independent exponent-choice bijection, didn't quote them; (2) you can say WHY multiplicativity holds — coprime parts share no primes, so the per-prime exponent choices are independent. Can't rebuild it blank-page → not learned yet.

sources & assignments (4)
  • Source Divisor functions σ(n), τ(n) — multiplicativity — Compute τ and σ from the prime factorization; double-count divisor pairs; coprime-pair counting via Euler φ.
  • Source 104 Number Theory Problems — Divisor-function / coprime-counting problems.
  • Source Michael Penn — the divisor function — σ and τ multiplicativity + a worked divisor-sum problem.
  • Problems 104 NT — 104 Number Theory Problems — divisor-function (σ/τ) or coprime-pair counting, 1–2 problemsPutnam

archive pull: 1 problems · Number Theory · Putnam · 20 min →

FE Micro-Spine — iteration and finite-state periodicity

FE Micro-Spine — iteration and finite-state periodicity

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Added because FE cannot be a one-week topic; pattern exposure must recur after the W6 load.

Ritual: Write the substitution table before any solution check. Every FE rep must name the first move and the trap it avoided.

sources & assignments (2)
  • Source 100 Functional Equations / Putnam FE repair spine — Micro-rep source for repeated FE pattern exposure: special values, symmetry, injective/surjective forcing, Cauchy/Jensen, iteration, and polynomial-degree comparison.
  • Problems Functional equations micro-rep — 100 Functional Equations #38–39 — solve as an iteration/periodicity rep; explicitly write f(f(x)) or the repeated substitution chain.Putnam

Problem-Solving Reps — weekly homework

Problem-Solving Reps — weekly homework

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Standing weekly homework — the problem-solving book stack, every week. Solve ALL listed; write ONE full clean solution (the rest may stay scratch). Up to 2 due re-solves from your review queue surface first.

sources & assignments (3)
  • Problems Engel — Engel Ch. 9 probs 1–8 (sequences/recurrences)Putnam
  • Problems Concrete Math — Concrete Mathematics Ch. 1 ex 1–4 · Ch. 7 ex 1–3Putnam
  • Problems 104 NT — 104 Number Theory Problems — Introductory probs 9–12 (valuations)Putnam

Reflect

Reflect: reconstruct and log the pattern

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- Reconstruct: 1 Rudin Ch. 8 exercise → close → rewrite → pattern note - Reflect: power series/Taylor card · recurrence method decision card · vₚ(n!) Legendre formula from memory - Verify: one LA problem — every rank/eigenvalue claim cited from a named theorem? ---

Ritual: 1 Rudin Ch. 8 exercise → close → rewrite → pattern note | power series/Taylor card · recurrence method decision card · vₚ(n!) Legendre formula from memory

sources & assignments (4)

Verify: audit one proof before closing

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unlocks after: Reflect: reconstruct and log the pattern

- Reconstruct: 1 Rudin Ch. 8 exercise → close → rewrite → pattern note - Reflect: power series/Taylor card · recurrence method decision card · vₚ(n!) Legendre formula from memory - Verify: one LA problem — every rank/eigenvalue claim cited from a named theorem? ---

Ritual: one LA problem — every rank/eigenvalue claim cited from a named theorem?

sources & assignments (4)

Exit contract — Week 8

Verification remaining

  • reading progress…

Carry-forward repairs

  • reading queue…

Next week opens with

W9 · All Tracks: Archive Sprint + Hidden Tools: Abel Summation + LA in Combinatorics
the exam follows the final taper week

Train Week 8 in the trainer →