Maths Olympiad Prep

Track / Stage 5 / 36 of 400 #636 of 1964

Problem 636

AIME late
Number theory Difficulty 5.1 Multiple choice

Five positive integers are listed in increasing order. The difference between any two consecutive numbers in the list is three. The fifth number is a multiple of the first number. How many different such lists of five integers are there?

Pick one

Official solution

Suppose the first integer in the list is nn.

Then the remaining four integers are n+3,n+6,n+9n+3, n+6, n+9, and n+12n+12.

Since the fifth number is a multiple of the first, then n+12n=nn+12n=1+12n\frac{n+12}{n}=\frac{n}{n}+\frac{12}{n}=1+\frac{12}{n} is an integer. Since 1+12n1+\frac{12}{n} is an integer, then 12n\frac{12}{n} is an integer, or nn is a positive divisor of 12 .

The positive divisors of 12 are 1,2,3,4,61,2,3,4,6, and 12 , so there are 6 possible values of nn and so 6 different lists.

(We can check that each of 6 values of nn produces a different list, each of which has the required property.)

ANSwER: (D)

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.