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Problem 566

AMC 12 late, AIME early

Number theory Difficulty 5.0 Find the answer

If $x$ is a positive integer less than 100 , how many values of $x$ make $\sqrt{1+2+3+4+x}$ an integer?
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

Your answer

Official solution

To make $\sqrt{1+2+3+4+x}$ an integer, we need $1+2+3+4+x=10+x$ to be a perfect square.

So we can rephrase the question as "For how many positive integers $x$ less than 100 is $10+x$ a perfect square?".

Since $x$ is between 1 and 99 , then $10+x$ is between 11 and 109 .

There are 7 perfect squares in this interval: $16,25,36,49,64,81$ , and 100 , so there are 7 possible values of $x: 6,15,26,39,54,71$ , and 90 .

ANSWER: (B)

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