Problem: Let S=2+4+6−…−2N, where N is the smallest positive integer such that S>1,000,000. Then the sum of the digits of N is:
Answer Choices:
A. 27
B. 12
C. 6
D. 2
E. 1
Solution:
S=2+4+6+…+2N=2(1+2+3+…+N)=N(N+1). Since S>106, N(N+1)>106, N≈103. When N=103−1, S<106.
∴N≥103∴ N (smallest) =103=1000; the sum of its digits is 1 .