Difference between revisions of "Mock AIME 5 2005-2006 Problems/Problem 1"

Revision as of 12:25, 13 February 2008

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

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

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

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

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

@@ Line 45: / Line 45: @@
-<cmath>=197+193+190+180+179+167+168+94=390+370+346+262=760+608=1\boxed{368}</cmath>
+<cmath>=197+193+190+180+179+167+168+94=390+370+346+262</cmath>
+<cmath>=760+608=1\boxed{368}</cmath>
 == See also ==

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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Difference between revisions of "Mock AIME 5 2005-2006 Problems/Problem 1"

Revision as of 12:25, 13 February 2008

Problem

Solution

See also