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 "1966 AHSME Problems/Problem 29"

(→‎Solution)
(→‎Solution)
 
(One intermediate revision by the same user not shown)
Line 5: Line 5:
  
 
== Solution ==
 
== Solution ==
−
The number of numbers under <math>1000</math> that are divisible by <math>5</math> is <math>\lfloor\frac{999}{5}\rfloor=199</math>. The number of numbers under <math>1000</math> that are divisible by <math>7</math> is <math>\lfloor\frac{1000}{7}\rfloor=142</math>. Adding them together, we get <math>341</math>. However, we have over counted the numbers which are divisible by <math>35</math>. There are <math>\lfloor\frac{999}{35}\rfloor=28</math> of these. So, the number of numbers divisible be <math>2</math> or <math>5</math> under <math>1000</math> is <math>341-28=313</math>. We can conclude that the number of numbers divisible by neither <math>5</math> or <math>7</math> is <math>999-313=686</math> or answer choice <math>\fbox{B}</math>.
+
The number of positive integers under <math>1000</math> that are divisible by <math>5</math> is <math>\lfloor\frac{999}{5}\rfloor=199</math>. The number of positive integers under <math>1000</math> that are divisible by <math>7</math> is <math>\lfloor\frac{1000}{7}\rfloor=142</math>. Adding them together, we get <math>341</math>. However, we have over counted the numbers which are divisible by <math>35</math>. There are <math>\lfloor\frac{999}{35}\rfloor=28</math> of these. So, the number of positive integers divisible be <math>7</math> or <math>5</math> under <math>1000</math> is <math>341-28=313</math>. We can conclude that the number of positive integers divisible by neither <math>5</math> or <math>7</math> is <math>999-313=686</math> or answer choice <math>\fbox{B}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 07:44, 1 January 2024

Problem

The number of positive integers less than $1000$ divisible by neither $5$ nor $7$ is:

$\text{(A) } 688 \quad \text{(B) } 686 \quad \text{(C) } 684 \quad \text{(D) } 658 \quad \text{(E) } 630$

Solution

The number of positive integers under $1000$ that are divisible by $5$ is $\lfloor\frac{999}{5}\rfloor=199$. The number of positive integers under $1000$ that are divisible by $7$ is $\lfloor\frac{1000}{7}\rfloor=142$. Adding them together, we get $341$. However, we have over counted the numbers which are divisible by $35$. There are $\lfloor\frac{999}{35}\rfloor=28$ of these. So, the number of positive integers divisible be $7$ or $5$ under $1000$ is $341-28=313$. We can conclude that the number of positive integers divisible by neither $5$ or $7$ is $999-313=686$ or answer choice $\fbox{B}$.

See also

1966 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 28 		Followed by
Problem 30
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1966_AHSME_Problems/Problem_29&oldid=209332"