# Difference between revisions of "1999 AHSME Problems/Problem 11"

## 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

### Solution 1

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 $\boxed{\textbf{(A)}}$.

### Solution 2

Since all answers are over $2000$, work backwards and find the cost of the first $1999$ lockers. The first $9$ lockers cost $0.18$ dollars, while the next $90$ lockers cost $0.04\cdot 90 = 3.60$. Lockers $100$ through $999$ cost $0.06\cdot 900 = 54.00$, and lockers $1000$ through $1999$ inclusive cost $0.08\cdot 1000 = 80.00$.

This gives a total cost of $0.18 + 3.60 + 54.00 + 80.00 = 137.78$. There are $137.94 - 137.78 = 0.16$ dollars left over, which is enough for $8$ digits, or $2$ more four digit lockers. These lockers are $2000$ and $2001$, leading to answer $\boxed{\textbf{(A)}}$.