Mock AIME 5 2005-2006 Problems/Problem 1

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Problem

Suppose $n$ is a positive integer. Let $f(n)$ be the sum of the distinct positive prime divisors of $n$ less than $50$ (e.g. $f(12) = 2+3 = 5$ and $f(101) = 0$ ). Evaluate the remainder when $f(1)+f(2)+\cdots+f(99)$ is divided by $1000$ .

Solution

So all of the prime numbers less than 50 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, and 47. So we just need to find the number of numbers that are divisible by 2, the number of numbers divisible by 3, etc.

$\lfloor 99/2\rfloor =49$

$\lfloor 99/3\rfloor =33$

$\lfloor 99/5\rfloor =19$

$\lfloor 99/7\rfloor =14$

$\lfloor 99/11\rfloor =9$

$\lfloor 99/13\rfloor =7$

$\lfloor 99/17\rfloor =5$

$\lfloor 99/19\rfloor =5$

$\lfloor 99/23\rfloor =4$

$\lfloor 99/29\rfloor =3$

$\lfloor 99/31\rfloor =3$

$\lfloor 99/37\rfloor =2$

$\lfloor 99/41\rfloor =2$

$\lfloor 99/43\rfloor =2$

$\lfloor 99/47\rfloor =2$

So we compute

49*2+33*3+19*5+14*7+9*11+7*13+5*17+5*19+4*23+3*29+3*31+2*37+2*41+2*43+2*47

=98+99+95+98+99+91+85+95+92+87+93+74+82+86+94

=197+193+190+180+179+167+168+94=390+370+346+262

$=760+608=1\boxed{368}$

Mock AIME 5 2005-2006 (Problems, Source)
Preceded by First Question	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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Mock AIME 5 2005-2006 Problems/Problem 1

From AoPSWiki

Problem

Solution

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