Maths Olympiad Prep

Track / Stage 3 / 112 of 260 #112 of 1964

Problem 112

AMC 10/12, early questions
Number theory Difficulty 3.4 Multiple choice

How many sets of two or more consecutive positive integers have a sum of 1515?

Pick one

Official solution

Notice that if the consecutive positive integers have a sum of 1515, then their average (which could be a fraction) must be a divisor of 1515. If the number of integers in the list is odd, then the average must be either 1,3,1, 3, or 55, and 11 is clearly not possible. The other two possibilities both work:

1+2+3+4+5=151 + 2 + 3 + 4 + 5 = 15
4+5+6=154 + 5 + 6 = 15
If the number of integers in the list is even, then the average will have a 12\frac{1}{2}. The only possibility is 152\frac{15}{2}, from which we get:

15=7+815 = 7 + 8
Thus, the correct answer is (C) 3.\boxed{\textbf{(C) }3}.

Question: (RealityWrites) Is it possible that the answer is 44, because 0+1+2+3+4+50+1+2+3+4+5 should technically count, right?
Answer: (IMGROOT2) It isn't possible because the question asks for positive integers, and this means that negative integers or zero aren't allowed.
Note to readers: make sure to always read the problem VERY carefully before attempting; it could mean the difference of making the cutoff.

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