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

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== Solution ==
 
== 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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So all of the prime numbers less than <math>50</math> are <math>2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43,</math> and <math>47</math>. So we just need to find the number of numbers that are divisible by <math>2</math>, the number of numbers divisible by <math>3</math>, etc.
  
 
<math>\lfloor 99/2\rfloor =49</math>
 
<math>\lfloor 99/2\rfloor =49</math>
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So we compute
 
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</cmath>
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<cmath>49\times2+33\times3+19\times5+14\times7+9\times11+7\times13+5\times17+5\times19+4\times23+3\times29+3\times31+2\times37+2\times41+2\times43+2\times47</cmath>
  
  
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<cmath>=197+193+190+180+179+167+168+94=390+370+346+262=760+608=1\boxed{368}</cmath>
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<cmath>=197+193+190+180+179+167+168+94=390+370+346+262</cmath>
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 +
 
 +
 
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<cmath>=760+608=1368</cmath>
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Our desired answer then is <cmath>\boxed{368}</cmath>
  
 
== See also ==
 
== See also ==

Latest revision as of 11:43, 10 August 2019

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

\[49\times2+33\times3+19\times5+14\times7+9\times11+7\times13+5\times17+5\times19+4\times23+3\times29+3\times31+2\times37+2\times41+2\times43+2\times47\]


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


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


\[=760+608=1368\]

Our desired answer then is \[\boxed{368}\]

See also

Mock AIME 5 2005-2006 (Problems, Source)
Preceded by
First Question 		Followed by
Problem 2
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