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

(testing template)
 
(could someone post a SHORTER solution?)
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== Solution ==
 
== Solution ==
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{{solution}}
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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.
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<math>\lfloor 99/2\rfloor =49</math>
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<math>\lfloor 99/3\rfloor =33</math>
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<math>\lfloor 99/5\rfloor =19</math>
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<math>\lfloor 99/7\rfloor =14</math>
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<math>\lfloor 99/11\rfloor =9</math>
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<math>\lfloor 99/13\rfloor =7</math>
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<math>\lfloor 99/17\rfloor =5</math>
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<math>\lfloor 99/19\rfloor =5</math>
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<math>\lfloor 99/23\rfloor =4</math>
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<math>\lfloor 99/29\rfloor =3</math>
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<math>\lfloor 99/31\rfloor =3</math>
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<math>\lfloor 99/37\rfloor =2</math>
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<math>\lfloor 99/41\rfloor =2</math>
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<math>\lfloor 99/43\rfloor =2</math>
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<math>\lfloor 99/47\rfloor =2</math>
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So we compute
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<cmath>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</cmath>
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<cmath>=197+193+190+180+179+167+168+94=390+370+346+262=760+608=1\boxed{368}</cmath>
  
 
== See also ==
 
== See also ==
 
{{Mock AIME box|year=2005-2006|n=5|source=76847|before=First Question|num-a=2}}
 
{{Mock AIME box|year=2005-2006|n=5|source=76847|before=First Question|num-a=2}}
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[[Category:Introductory Number Theory Problems]]

Revision as of 13:24, 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

\[49*2+33*3+19*5+14*7+9*11+7*13+5*17+5*19+4*23+3*29+3*31+2*37+2*41+2*43+2*47=98+99+95+98+99+91+85+95+92+87+93+74+82+86+94\]

\[=197+193+190+180+179+167+168+94=390+370+346+262=760+608=1\boxed{368}\]

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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