1999 AHSME Problems/Problem 11

Revision as of 13:24, 4 June 2011 by JSGandora (talk | contribs) (Solution)

Problem

The student locker numbers at Olympic High are numbered consecutively beginning with locker number $1$. The plastic digits used to number the lockers cost two cents apiece. Thus, it costs two cents to label locker number $9$ and four centers to label locker number $10$. If it costs $137.94 to label all the lockers, how many lockers are there at the school?

$\textbf{(A)}\ 2001 \qquad  \textbf{(B)}\ 2010 \qquad  \textbf{(C)}\ 2100 \qquad  \textbf{(D)}\ 2726 \qquad  \textbf{(E)}\ 6897$

Solution

The locker labeling requires $\frac{137.94}{0.02}=6897$ digits. Lockers $1$ through $9$ require $9$ digits total, lockers $10$ through $99$ require $2 \times 90=180$ digits, and lockers $100$ through $999$ require $3 \times 900=2700$ digits. Thus, the remaining lockers require $6897-2700-180-9=4008$ digits, so there must be $\frac{4008}{4}=1002$ more lockers, because they each use $4$ digits. Thus, there are $1002+999=2001$ student lockers, or answer choice $\boxed{\textbf{(A)}}$.

See Also

1999 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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