Difference between revisions of "1968 AHSME Problems/Problem 26"
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== Solution == | == Solution == | ||
Note that <math>S = 2(1+2+3+...+N) = N(N +10)</math>. It follows that <math>N = 1000</math>, so the sum of the digits of <math>N</math> is <math>\fbox{E}</math>. | Note that <math>S = 2(1+2+3+...+N) = N(N +10)</math>. It follows that <math>N = 1000</math>, so the sum of the digits of <math>N</math> is <math>\fbox{E}</math>. | ||
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== See also == | == See also == |
Revision as of 01:17, 27 January 2019
Problem
Let , where is the smallest positive integer such that . Then the sum of the digits of is:
Solution
Note that . It follows that , so the sum of the digits of is .
FrostFox
See also
1968 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 25 |
Followed by Problem 27 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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