HMMT 二月 2026 · 冲刺赛 · 第 34 题
HMMT February 2026 — Guts Round — Problem 34
题目详情
- [20] A positive integer is called good if it has no prime factors larger than 10 . Sarunyu picks two odd 4 positive integers a and b , both between 1 and 10 (inclusive), independently and uniformly at random. 2 2 Estimate the expected number of good divisors of a + b . 2 − 400(1 − E/A ) Submit a real number E . If the correct answer is A , you will receive ⌊ 20 . 05 e ⌋ points.
解析
- [20] A positive integer is called good if it has no prime factors larger than 10 . Sarunyu picks two odd 4 positive integers a and b , both between 1 and 10 (inclusive), independently and uniformly at random. 2 2 Estimate the expected number of good divisors of a + b . 2 − 400(1 − E/A ) Submit a real number E . If the correct answer is A , you will receive ⌊ 20 . 05 e ⌋ points. Proposed by: Rohan Das Answer: 13.12557328 2 2 Solution: Recall that the number of good divisors of a + b is comptued by ∏ 2 2 2 2 τ ( a + b ) = ( ν ( a + b ) + 1) . good p 4 p ≤ 10 In what follows, any product indexed by p is understood to run across only prime numbers. We 2 2 2 2 estimate each random variable ν ( a + b ) as independent. In particular, we estimate E [ τ ( a + b )] p good as ( ) ∏ 2 2 T = E [ ν ( a + b )] + 1 . p 4 p ≤ 10 We estimate T by separating into cases of p . 2 2 2 2 • If p = 2 , then since a and b are always odd, we get that a + b is always 2 (mod 4) , so ν ( a + b ) = 2 1 always, implying that 2 2 E [ ν ( a + b )] + 1 = 2 . 2 ©2026 HMMT k 2 2 ⌈ k/ 2 ⌉ • If p ≡ 3 (mod 4) , we claim that p | a + b if and only if p divides both a and b . In particular, 2 2 ν ( a + b ) is always even. Proof TBD. Thus, we get that p ∑ 2 2 2 2 E [ ν ( a + b )] + 1 = 1 + 2 P [ ν ( a + b ) ≥ 2 k ] p p k ≥ 1 ∑ 1 ≈ 1 + 2 2 k p k ≥ 1 2 2 p + 1 = 1 + = , 2 2 p − 1 p − 1 2 p +1 contributing into the product. 2 p − 1 k k 2 2 • If p ≡ 1 (mod 4) , then first count the number of ( a, b ) such that 1 ≤ a, b ≤ p and p | a + b . Let this number be N . Then we have two cases. k ′ a ′ b ′ ′ k − 1 – If p divides both a and b , then let a = and b = . Then we need 1 ≤ a , b ≤ p and p p k − 2 2 2 2 p | a + b , so there are p N solution. k − 2 – If p does not divide either a or b , then by a routine application of Hensel’s lemma, fixing any k − 1 b not divisible by p gives exactly two solutions for a . Thus, there are 2 p ( p − 1) solutions. This establishes the recurrence relation 2 k − 1 N = p N + 2 p ( p − 1) . k k − 2 By a straightforward induction, with N = 1 and N = 2 p − 1 , we get that 0 1 k k − 1 N = ( k + 1) p − kp . k Thus, we get that N k + 1 k k 2 2 P ( ν ( a + b ) ≥ k ) ≈ = − , p 2 k k k +1 p p p so we can approximate ∞ ∑ 2 2 2 2 E [ ν ( a + b ) + 1] = 1 + P ( ν ( a + b ) ≥ k ) p p k =1 ∞ ∑ k + 1 k ≈ 1 + − k k +1 p p k =1 ( ) ( ) ( ) 2 1 3 2 4 3 = 1 + − + − + − + . . . 2 2 3 3 4 p p p p p p 2 2 2 = 1 + + + + . . . 2 3 p p p 2 p + 1 = 1 + = , p − 1 p − 1 p +1 contributing to the product. p − 1 Therefore, we have 2 ∏ ∏ p + 1 p + 1 T ≈ 2 · · . 2 p − 1 p − 1 p ≡ 1 mod 4 p ≡ 3 mod 4 4 4 p ≤ 10 p ≤ 10 To compute this, we estimate each term separately. First, we will attempt to estimate ∏ p + 1 p − 1 4 p ≤ 10 ©2026 HMMT 4 4 over all primes p ≤ 10 . Let N = 10 . We rewrite this product as ( ) ( ) − 2 2 2 ∏ ∏ ∏ ( p − 1) /p 1 1 = 1 − · 1 − . 2 2 2 ( p − 1) /p p p p ≤ N p ≤ N p ≤ N Note that ( ) ( ) ∞ − 1 2 ∏ ∏ ∑ 1 1 1 1 π 1 − = 1 + + + . . . = = . 2 2 4 2 p p p n 6 p p n =1 Since this product is convergent, restricting to p ≤ N won’t change the product meaningfully when N is large. To evaluate the other term, we use Mertens’ theorem, which states ( ) ∏ 1 γ 1 − ∼ e ln N, p p ≤ N x where γ ≈ 0 . 577 is the Euler-Mascheroni constant. Using the Taylor series expansion of e , we can approximate 2 3 γ γ γ e ≈ 1 + γ + + ≈ 1 . 78 . 2 6 Also, ln 10000 = 4 ln 10 ≈ 4 · 2 . 3 = 9 . 2 , so 2 ∏ p + 1 (1 . 78 · 9 . 2) ≈ ≈ 163 . 2 p − 1 π / 6 p ≤ N We can approximate the product restricted to p ≡ 1 mod 4 by noting that large primes tend to be evenly split between 1 mod 4 and 3 mod 4 , though since the smallest primes contribute the largest terms, we should consider them separately: √ ∏ ∏ p + 1 5 + 1 p + 1 ≈ · p − 1 5 − 1 p − 1 p ≡ 1 mod 4 13 ≤ p ≤ N 4 p ≤ 10 √ 6 1 2 4 6 10 ≈ · 163 · · · · · 4 3 4 6 8 12 √ 6 1 = · 407 . 5 4 6 ( ) 1 7 . 5 ≈ 20 + 4 40 ≈ 5 . 047 . As for the other term, note that ( ) ( ) 2 − 2 2 2 ∏ ∏ p + 1 1 π ≤ 1 − = 2 2 p − 1 p 6 p p converges, so any product of such terms should be completely dominated by the first several terms. 10 50 Restricting to p ≡ 3 mod 4 , the first two terms are and , giving us a final estimate of 8 48 10 50 2 · 5 . 047 · · ≈ 13 . 143 , 8 48 an estimate close enough for all 20 points, though we note that this estimate makes many rough ap- proximations and its accuracy is not entirely indicative of its soundness. ©2026 HMMT