1971 AHSME Problems/Problem 1

Problem

The number of digits in the number $N=2^{12}\times 5^8$ is

$\textbf{(A) }9\qquad \textbf{(B) }10\qquad \textbf{(C) }11\qquad \textbf{(D) }12\qquad  \textbf{(E) }20$

Solution

$N=2^{12}\cdot5^8=2^4\cdot2^8\cdot5^8=2^4\cdot(2\cdot5)^8=2^4\cdot10^8=16\cdot100000000$, which has $2+8=10$ digits, so the answer is $\boxed{\textbf{(B) }10}$.

See Also

1971 AHSC (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
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. AMC logo.png