Difference between revisions of "1991 AJHSME Problems/Problem 9"
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There are <math>\left\lfloor \frac{46}{3}\right\rfloor =15</math> numbers divisible by <math>3</math>, <math>\left\lfloor\frac{46}{5}\right\rfloor =9</math> numbers divisible by <math>5</math>, so at first we have <math>15+9=24</math> numbers that are divisible by <math>3</math> or <math>5</math>, except we counted the multiples of <math>\text{LCM}(3,5)=15</math> twice, once for <math>3</math> and once for <math>5</math>. | There are <math>\left\lfloor \frac{46}{3}\right\rfloor =15</math> numbers divisible by <math>3</math>, <math>\left\lfloor\frac{46}{5}\right\rfloor =9</math> numbers divisible by <math>5</math>, so at first we have <math>15+9=24</math> numbers that are divisible by <math>3</math> or <math>5</math>, except we counted the multiples of <math>\text{LCM}(3,5)=15</math> twice, once for <math>3</math> and once for <math>5</math>. | ||
| − | There are <math>\left\lfloor \frac{46}{15}\right\rfloor =3</math> numbers divisible by <math>15</math>, so there are <math>24-3=21</math> numbers divisible by <math>3</math> or <math>5</math>. <math>\rightarrow \boxed{\text{B | + | There are <math>\left\lfloor \frac{46}{15}\right\rfloor =3</math> numbers divisible by <math>15</math>, so there are <math>24-3=21</math> numbers divisible by <math>3</math> or <math>5</math>. <math>\rightarrow \boxed{\text{(B)21</math>}}$ |
==See Also== | ==See Also== | ||
Revision as of 18:33, 11 October 2024
Problem
How many whole numbers from 
through 
are divisible by either 
or 
or both?
Solution
There are 
numbers divisible by , 
numbers divisible by , so at first we have 
numbers that are divisible by or , except we counted the multiples of 
twice, once for and once for .
There are 
numbers divisible by 
, so there are 
numbers divisible by or . $\rightarrow \boxed{\text{(B)21$ (Error compiling LaTeX. Unknown error_msg)}}$
See Also
| 1991 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
