Difference between revisions of "2002 AMC 12B Problems/Problem 13"
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== Solution == | == Solution == | ||
| − | + | == 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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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 | + | 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> |
~ Nafer | ~ Nafer | ||
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| + | == Solution 1.1 == | ||
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| + | the normal sequence can be described as N^2+N divided by 2. | ||
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| + | Since have 18 terms adding 18n will increase the consective sequence startering number by 1 | ||
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| + | (N^2+N)/2 +18n | ||
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| + | now subsitute 18 as N | ||
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| + | we get (18^2+18)/2 = 171 | ||
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| + | put I^2 which is integer square and plug in all our results | ||
| + | 171 + 18n = I^2 | ||
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| + | 18n = I^2-171 | ||
| + | I^2-171 = 0 mod(18) | ||
| + | |||
| + | subsitute the answer choices starting with B because 169 is less than 171 and results in a neagtive number | ||
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| + | 225-177 = mod(18) | ||
| + | 54 = mod(18) | ||
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| + | 54 is dividsble by 18 and is therefore the smallest number possible. | ||
== See also == | == See also == | ||
Latest revision as of 22:34, 18 March 2023
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 ![\[a+(a+1)+(a+2)+...+(a+17)=18a+\sum_{k=1}^{17}k=18a+153\]](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 N^2+N divided by 2.
Since have 18 terms adding 18n will increase the consective sequence startering number by 1
(N^2+N)/2 +18n
now subsitute 18 as N
we get (18^2+18)/2 = 171
put I^2 which is integer square and plug in all our results
171 + 18n = I^2
18n = I^2-171 I^2-171 = 0 mod(18)
subsitute the answer choices starting with B because 169 is less than 171 and results in a neagtive number
225-177 = mod(18) 54 = mod(18)
54 is dividsble 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
