Difference between revisions of "1966 AHSME Problems/Problem 29"
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== Solution == | == Solution == | ||
| − | <math>\fbox{B}</math> | + | 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 answe choice <math>\fbox{B}</math> |
== See also == | == See also == | ||
Revision as of 18:12, 5 July 2017
Problem
The number of positive integers less than 
divisible by neither 
nor 
is:
Solution
The number of numbers under that are divisible by is 
. The number of numbers under that are divisible by is 
. Adding them together, we get 
. However, we have over counted the numbers which are divisible by 
. There are 
of these. So, the number of numbers divisible be 
or under is 
. We can conclude that the number of numbers divisible by neither or is 
or answe choice 
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. 
