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

(Solution 2)
(Solution 2)
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Notice that all five choices given are perfect squares.
 
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>$
  
 
== See also ==
 
== See also ==

Revision as of 15:54, 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}$$

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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