AoPSWiki
Try our innovative online adaptive learning system, Alcumus.
Over 1100 problems and 60+ video lessons. FREE!
Personal tools

1991 AJHSME Problems/Problem 9

From AoPSWiki

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 (ProblemsResources)
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
Stay informed about new Art of Problem Solving developments.
Click here to join our mailing lists.
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • FoundationPrivacyContact Us