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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>.
  
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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>}}$
+
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}}</math>
  
 
==See Also==
 
==See Also==

Latest revision as of 18:33, 11 October 2024

Problem

How many whole numbers from $1$ through $46$ are divisible by either $3$ or $5$ or both?

$\text{(A)}\ 18 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 24 \qquad \text{(D)}\ 25 \qquad \text{(E)}\ 27$

Solution

There are $\left\lfloor \frac{46}{3}\right\rfloor =15$ numbers divisible by $3$, $\left\lfloor\frac{46}{5}\right\rfloor =9$ numbers divisible by $5$, so at first we have $15+9=24$ numbers that are divisible by $3$ or $5$, except we counted the multiples of $\text{LCM}(3,5)=15$ twice, once for $3$ and once for $5$.

There are $\left\lfloor \frac{46}{15}\right\rfloor =3$ numbers divisible by $15$, so there are $24-3=21$ numbers divisible by $3$ or $5$. $\rightarrow \boxed{\text{B}}$

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

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