Difference between revisions of "1968 AHSME Problems/Problem 26"

Revision as of 23:34, 17 September 2018

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:

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

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)

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

@@ Line 10: / Line 10: @@
 == 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 +10)</math>. It follows that N = 1000, so the sum of the digits of N is <math>\fbox{E}</math>.
 (Solution Done By FrostFox)