Difference between revisions of "2002 AMC 12B Problems/Problem 13"

Revision as of 14:54, 2 July 2019

Problem

The sum of $18$ consecutive positive integers is a perfect square. The smallest possible value of this sum is

$\mathrm{(A)}\ 169 \qquad\mathrm{(B)}\ 225 \qquad\mathrm{(C)}\ 289 \qquad\mathrm{(D)}\ 361 \qquad\mathrm{(E)}\ 441$

Solution

Solution 1

Let $a, a+1, \ldots, a + 17$ be the consecutive positive integers. Their sum, $18a + \frac{17(18)}{2} = 9(2a+17)$ , is a perfect square. Since $9$ is a perfect square, it follows that $2a + 17$ is a perfect square. The smallest possible such perfect square is $25$ when $a = 4$ , and the sum is $225 \Rightarrow \mathrm{(B)}$ .

Solution 2

Notice that all five choices given are perfect squares.

Let $a$ be the smallest number, we have $a+(a+1)+(a+2)+...+(a+17)=17a+\sum{^17}{_1}$ $

2002 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 16: / Line 16: @@
 Notice that all five choices given are perfect squares.
-Let <math>a</math> be the smallest number, we have <math>a+(a+1)+(a+2)+...+(a+17)=17a+\sum{17}{1}</math>$
+Let <math>a</math> be the smallest number, we have <math>a+(a+1)+(a+2)+...+(a+17)=17a+\sum{^17}{_1}</math>$
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