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Difference between revisions of "1966 AHSME Problems/Problem 29"

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== Solution ==
 
== Solution ==
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The number of numbers under <math>1000</math> that are divisible by <math>5</math> is <math>\lfloor\frac{1000}{5}\rfloor=200</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>342</math>. However, we have over counted the numbers which are divisible by <math>35</math>. There are <math>\lfloor\frac{1000}{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>342-28=314</math>. We can conclude that the number of numbers divisible by neither <math>5</math> or <math>7</math> is <math>1000-314=686</math> or answer choice <math>\fbox{B}</math>.
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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>.
  
 
== See also ==
 
== See also ==

Revision as of 07:28, 13 April 2023

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 numbers under $1000$ that are divisible by $5$ is $\lfloor\frac{999}{5}\rfloor=199$. The number of numbers 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 numbers divisible be $2$ or $5$ under $1000$ is $341-28=313$. We can conclude that the number of numbers 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

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