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

(Solution 2)
(Solution 2)
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Let <math>a</math> be the smallest number, we have <cmath>a+(a+1)+(a+2)+...+(a+17)=18a+\sum_{k=1}^{17}k=18a+153</cmath>
 
Let <math>a</math> be the smallest number, we have <cmath>a+(a+1)+(a+2)+...+(a+17)=18a+\sum_{k=1}^{17}k=18a+153</cmath>
  
Subtract 153 from each of the choices and then check the divisibility by 18, we have 225 as the smallest possible sum. <math>\mathrm {(B)}</math>
+
Subtract 153 from each of the choices and then check its divisibility by 18, we have 225 as the smallest possible sum. <math>\mathrm {(B)}</math>
  
 
~ Nafer
 
~ Nafer

Revision as of 15:59, 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 \[a+(a+1)+(a+2)+...+(a+17)=18a+\sum_{k=1}^{17}k=18a+153\]

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

See also

2002 AMC 12B (ProblemsAnswer KeyResources)
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

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