Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
m (→‎Solution)
 
(4 intermediate revisions by 3 users not shown)
Line 3: Line 3:
  
 
== Solution ==
 
== 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.
+
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>
Line 37: Line 37:
 
So we compute
 
So we compute
  
−
<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>
+
<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>
  
  
  
−
<cmath>=197+193+190+180+179+167+168+94=390+370+346+262=760+608=1\boxed{368}</cmath>
+
<cmath>=98+99+95+98+99+91+85+95+92+87+93+74+82+86+94</cmath>
 +
 
 +
 
 +
 
 +
<cmath>=197+193+190+180+179+167+168+94=390+370+346+262</cmath>
 +
 
 +
 
 +
 
 +
<cmath>=760+608=1368</cmath>
 +
 
 +
Our desired answer then is <cmath>\boxed{368}</cmath>
  
 
== See also ==
 
== See also ==

Latest revision as of 12:43, 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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_5_2005-2006_Problems/Problem_1&oldid=108702"