Difference between revisions of "2016 AMC 10B Problems/Problem 18"
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==Solution== | ==Solution== | ||
− | + | Factorize <math>345=3\cdot 5\cdot 23</math>. | |
− | + | Suppose we take an odd number <math>k</math> of consecutive integers, centred on <math>m</math>. Then <math>mk=345</math> with <math>\tfrac12k<m</math>. | |
− | + | Looking at the factors of <math>345</math>, the possible values of <math>k</math> are <math>3,5,15,23</math> centred on <math>115,69,23,15</math> respectively. | |
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− | + | Suppose instead we take an even number <math>2k</math> of consecutive integers, centred on <math>m</math> and <math>m+1</math>. Then <math>k(2m+1)=345</math> with <math>k\le m</math>. | |
− | + | Looking again at the factors of <math>345</math>, the possible values of <math>k</math> are <math>1,3,5</math> centred on <math>(172,173),(57,58),(34,35)</math> respectively. | |
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+ | Thus the answer is <math>\textbf{(E) }7</math>. | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=B|num-b=17|num-a=19}} | {{AMC10 box|year=2016|ab=B|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:37, 21 February 2016
Problem
In how many ways can be written as the sum of an increasing sequence of two or more consecutive positive integers?
Solution
Factorize .
Suppose we take an odd number of consecutive integers, centred on . Then with . Looking at the factors of , the possible values of are centred on respectively.
Suppose instead we take an even number of consecutive integers, centred on and . Then with . Looking again at the factors of , the possible values of are centred on respectively.
Thus the answer is .
See Also
2016 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 10 Problems and Solutions |
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