2002 AMC 12B Problems/Problem 13

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 $a+(a+1)+(a+2)+...+(a+17)=18a+\sum_{k=1}^{17}k=18a+153$

Subtract $153$ from each of the choices and then check its divisibility by $18$ , we have $225$ as the smallest possible sum. $\mathrm {(B)}$

~ Nafer

Solution 1.1

the normal sequence can be described as N^2+N divided by 2.

Since have 18 terms adding 18n will increase the consective sequence startering number by 1

(N^2+N)/2 +18n

now subsitute 18 as N

we get (18^2+18)/2 = 171

put I^2 which is integer square and plug in all our results 171 + 18n = I^2

18n = I^2-171 I^2-171 = 0 mod(18)

subsitute the answer choices starting with B because 169 is less than 171 and results in a neagtive number

225-177 = mod(18) 54 = mod(18)

54 is dividsble by 18 and is therefore the smallest number possible.

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.

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