Difference between revisions of "1999 AHSME Problems/Problem 11"
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==Solution== | ==Solution== | ||
===Solution 1=== | ===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 | + | 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=== | ===Solution 2=== | ||
Revision as of 14:21, 11 July 2022
Problem
The student locker numbers at Olympic High are numbered consecutively beginning with locker number 
. The plastic digits used to number the lockers cost two cents apiece. Thus, it costs two cents to label locker number 
and four centers to label locker number 
. If it costs $137.94 to label all the lockers, how many lockers are there at the school?
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 
, work backwards and find the cost of the first 
lockers. The first lockers cost 
dollars, while the next 
lockers cost 
. Lockers 
through 
cost 
, and lockers 
through inclusive cost 
.
This gives a total cost of 
. There are 
dollars left over, which is enough for 
digits, or 
more four digit lockers. These lockers are and 
, leading to answer 
.
See Also
| 1999 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
