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

(Solution 2)
(Solution 2)
 
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== Solution ==
 
== Solution ==
  
=== Solution 1 ===
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== Solution 1 ==
 
Let <math>a, a+1, \ldots, a + 17</math> be the consecutive positive integers. Their sum, <math>18a + \frac{17(18)}{2} = 9(2a+17)</math>, is a perfect square. Since <math>9</math> is a perfect square, it follows that <math>2a + 17</math> is a perfect square. The smallest possible such perfect square is <math>25</math> when <math>a = 4</math>, and the sum is <math>225 \Rightarrow \mathrm{(B)}</math>.
 
Let <math>a, a+1, \ldots, a + 17</math> be the consecutive positive integers. Their sum, <math>18a + \frac{17(18)}{2} = 9(2a+17)</math>, is a perfect square. Since <math>9</math> is a perfect square, it follows that <math>2a + 17</math> is a perfect square. The smallest possible such perfect square is <math>25</math> when <math>a = 4</math>, and the sum is <math>225 \Rightarrow \mathrm{(B)}</math>.
  
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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</math>$
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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>
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Subtract <math>153</math> from each of the choices and then check its divisibility by <math>18</math>, we have <math>225</math> as the smallest possible sum. <math>\mathrm {(B)}</math>
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~ Nafer
  
 
== See also ==
 
== See also ==

Latest revision as of 21:24, 31 August 2020

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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