2002 AMC 12B Problems/Problem 13
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
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 |
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