Problem:
The least common multiple of a positive integer n and 18 is 180 , and the greatest common divisor of n and 45 is 15. What is the sum of the digits of n ?
Answer Choices:
A. 3
B. 6
C. 8
D. 9
E. 12
Solution:
The positive integer n must be a multiple of 15 and a divisor of 180 . There are 6 such integers, as shown in the following table.
n1530456090180lcm(n,18)90909018090180gcd(n,45)151545154545
The only choice that satisfies the conditions is n=60, and the requested sum of digits is (B)6.
OR
Because 18=2⋅32 and 180=22⋅32⋅5, it follows from lcm(n,18)=180 that n is divisible by 22 but not 23, that n is divisible by 5 but not 52, and that n can have no prime factors other than 2 , 3 , or 5 . Because 45=32⋅5 and 15=3⋅5, it follows from gcd(n,45)=15 that n is divisible by 3 but not 32. Therefore n=22⋅3⋅5=60, and the sum of its digits is (B)6.
The problems on this page are the property of the MAA's American Mathematics Competitions