HMMT 十一月 2010 · 冲刺赛 · 第 34 题
HMMT November 2010 — Guts Round — Problem 34
题目详情
英文原题
- [ 25 ] Estimate the sum of all the prime numbers less than 1 , 000 , 000. If the correct answer is X and
( )
25 X 25 Ayou write down A , your team will receive min b c , b c points, where b x c is the largest integer
A Xless than or equal to x .
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解析
英文解析
- [ 25 ] Estimate the sum of all the prime numbers less than 1 , 000 , 000. If the correct answer is X and
( )
25 X 25 Ayou write down A , your team will receive min b c , b c points, where b x c is the largest integer
A Xless than or equal to x .
Answer: 37550402023 A decent approximation to the sum of all the primes can be obtained withnthe following two facts. First, there are approximately primes less than n and second, the nthln nprime is approximately n ln n . We’ll approximate ln 1000000 as 15 (the actual number is 13.8), so
∑106
15106
there are approximately primes. Then we want n ln n . If you know calculus, this can ben =1
∫ 10615
1 2 1 2
approximated by the integral x ln x dx , which has the antiderivative x ln x − x , giving an 15
1 2 4
12 12
1 10 1 10
answer of around · · (15 − ln 15) − · . Estimating ln 15 as about 3, this is approximately
2 225 4 225
23 · 1012 = 2 . 5 · 10 = 25 , 000 , 000 , 000. The actual answer is 37,550,402,023, so an approximation of this 10
accuracy would get 16 points. We can arrive at the same answer without calculus by approximating 900
6 6 6 6
10 10 10 10 212
∑ ∑ ∑ ∑ ∑ ∑10 − kn n
1 n n 1
15 15 15 15
ln n as , so that our sum becomes = ≈ · =225
n =1
k =1 k =1 k =1 n = k k =1
k k k 2 k
∑106
12 6 12 6 12
1 10 10 1 1 10 10 1 10 10
· · ln( ) − k ≈ · · ln( ) − ≈ 2 . 5 · 10 .15
k =1
2 225 15 2 2 225 15 4 225
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