Difference between revisions of "2002 AMC 12B Problems/Problem 13"
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\qquad\mathrm{(E)}\ 441</math> | \qquad\mathrm{(E)}\ 441</math> | ||
== Solution == | == Solution == | ||
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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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+ | === Solution 2 === | ||
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+ | Notice that all five choices given are perfect squares. | ||
+ | |||
+ | 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 == | ||
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 21:24, 31 August 2020
Problem
The sum of consecutive positive integers is a perfect square. The smallest possible value of this sum is
Solution
Solution 1
Let be the consecutive positive integers. Their sum, , is a perfect square. Since is a perfect square, it follows that is a perfect square. The smallest possible such perfect square is when , and the sum is .
Solution 2
Notice that all five choices given are perfect squares.
Let be the smallest number, we have
Subtract from each of the choices and then check its divisibility by , we have as the smallest possible sum.
~ Nafer
See also
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.