HMMT 二月 2024 · 冲刺赛 · 第 28 题

HMMT February 2024 — Guts Round — Problem 28

[14] Given that the 32 -digit integer 64 312 311 692 944 269 609 355 712 372 657 is the product of 6 consecutive primes, compute the sum of these 6 primes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT February 2024, February 17, 2024 — GUTS ROUND Organization Team Team ID#

解析

[14] Given that the 32 -digit integer 64 312 311 692 944 269 609 355 712 372 657 is the product of 6 consecutive primes, compute the sum of these 6 primes. Proposed by: Derek Liu Answer: 1200974 30 Solution: Because the product is approximately 64 · 10 , we know the primes are all around 200000 . Say they are 200000 + x for i = 1 , . . . , 6 . i ∏ 6 By expanding (200000 + x ) as a polynomial in 200000 , we see that i i =1 25 5 31231 · 10 = 200000 ( x + · · · + x ) 1 6 plus the carry from the other terms. Note that 31231 = 975 · 32 + 31 , so x + · · · + x ≤ 975 . 1 6 Thus, 5 20 2 2 5 16( x x + x x + · · · + x x ) ≤ 16 · ( x + · · · + x ) < · 1000 < 67 · 10 , 1 2 1 3 5 6 1 6 12 3 4 25 so the carry term from 200000 ( x x + · · · + x x ) is at most 67 · 10 . The other terms have negligible 1 2 5 6 carry, so it is pretty clear x + · · · + x > 972 , otherwise the carry term would have to be at least 1 6 25 5 25 31231 · 10 − 200000 (972) = 127 · 10 . It follows that x + · · · + x lies in [973 , 975] , so the sum of the primes, 6 · 200000 + ( x + · · · + x ) , lies 1 6 1 6 in [1200973 , 1200975] . As these primes are all greater than 2 , they are all odd, so their sum is even. Thus it must be 1200974 .