1968 AHSME Problems/Problem 26

Revision as of 00:34, 18 September 2018 by Frostfox (talk | contribs) (Solution)

Problem

Let $S=2+4+6+\cdots +2N$, where $N$ is the smallest positive integer such that $S>1,000,000$. Then the sum of the digits of $N$ is:

$\text{(A) } 27\quad \text{(B) } 12\quad \text{(C) } 6\quad \text{(D) } 2\quad \text{(E) } 1$

Solution

Note that $S = 2(1+2+3+...+N) = N(N +10)$. It follows that N = 1000, so the sum of the digits of N is $\fbox{E}$. (Solution Done By FrostFox)

See also

1968 AHSME (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png