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

Latest revision as of 00:53, 16 August 2023

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

FrostFox

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

@@ Line 10: / Line 10: @@
 == 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 +1)</math>. It follows that <math>N = 1000</math>, so the sum of the digits of <math>N</math> is <math>\fbox{E}</math>.
-(Solution Done By FrostFox)
+FrostFox
 == 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]]
 {{MAA Notice}}