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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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{{solution}}
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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.
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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\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>=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</cmath>
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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 ==
 
{{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]]

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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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