Difference between revisions of "2002 AMC 12B Problems/Problem 13"
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\qquad\mathrm{(D)}\ 361 | \qquad\mathrm{(D)}\ 361 | ||
\qquad\mathrm{(E)}\ 441</math> | \qquad\mathrm{(E)}\ 441</math> | ||
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== Solution 1 == | == 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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| − | the normal sequence can be described as N^2+N divided by 2. | + | the normal sequence can be described as <math>N^2+N</math> divided by 2. |
Since have 18 terms adding 18n will increase the consective sequence startering number by 1 | Since have 18 terms adding 18n will increase the consective sequence startering number by 1 | ||
| − | (N^2+N) | + | <math>\frac{(N^2+N)}{2} +18n</math> |
now subsitute 18 as N | now subsitute 18 as N | ||
| − | we get (18^2+18) | + | we get <math>\frac{(18^2+18)}{2} = 171</math> |
| − | put I^2 which is integer square and plug in all our results | + | put <math>I^2</math> which is integer square and plug in all our results |
| − | 171 + 18n = I^2 | + | <math>171 + 18n = I^2</math> |
| − | 18n = I^2-171 | + | <math>18n = I^2-171 |
| − | I^2-171 = 0 | + | I^2-171 = 0 \pmod{18}</math> |
subsitute the answer choices starting with B because 169 is less than 171 and results in a neagtive number | subsitute the answer choices starting with B because 169 is less than 171 and results in a neagtive number | ||
| − | 225-177 = | + | <math>225-177 = \pmod{18}</math> |
| − | 54 = | + | <math>54 = \pmod{18}</math> |
| − | 54 is | + | 54 is divisble by 18 and is therefore the smallest number possible. |
== See also == | == See also == | ||
Latest revision as of 11:02, 19 October 2025
Problem
The sum of 
consecutive positive integers is a perfect square. The smallest possible value of this sum is
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 ![\[a+(a+1)+(a+2)+...+(a+17)=18a+\sum_{k=1}^{17}k=18a+153\]](http://latex.artofproblemsolving.com/2/4/b/24ba9a08041a00d74638a06c236e5ee189a8420e.png)
Subtract 
from each of the choices and then check its divisibility by , we have 
as the smallest possible sum. 
~ Nafer
Solution 1.1
the normal sequence can be described as 
divided by 2.
Since have 18 terms adding 18n will increase the consective sequence startering number by 1
now subsitute 18 as N
we get 
put 
which is integer square and plug in all our results
subsitute the answer choices starting with B because 169 is less than 171 and results in a neagtive number
54 is divisble by 18 and is therefore the smallest number possible.
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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
