Difference between revisions of "1968 AHSME Problems/Problem 26"
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== Solution == | == Solution == | ||
| − | Note that S = 2(1+2+3+...+N) = N(N +1). It follows that N = 1000, so the sum of the digits of N is <math>\fbox{E}</math>. | + | Note that <math>S = 2(1+2+3+...+N) = N(N +1)</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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| + | FrostFox | ||
== See also == | == See also == | ||
| − | {{AHSME box|year=1968|num-b=25|num-a=27}} | + | {{AHSME 35p box|year=1968|num-b=25|num-a=27}} |
[[Category: Intermediate Algebra Problems]] | [[Category: Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 00:53, 16 August 2023
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 AHSC (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 • 31 • 32 • 33 • 34 • 35 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
