Difference between revisions of "1999 AHSME Problems/Problem 11"
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==Solution== | ==Solution== | ||
| − | + | The locker labeling requires <math> \frac{137.94}{0.02}=6897</math> digits. Lockers <math> 1</math> through <math> 9</math> require <math> 9</math> digits total, lockers <math> 10</math> through <math> 99</math> require <math> 2 \times 90=180</math> digits, and lockers <math> 100</math> through <math> 999</math> require <math> 3 \times 900=2700</math> digits. Thus, the remaining lockers require <math> 6897-2700-180-9=4008</math> digits, so there must be <math> \frac{4008}{4}=1002</math> more lockers, because they each use <math> 4</math> digits. Thus, there are <math> 1002+999=2001</math> student lockers, or answer choice <math> \boxed{\textbf{(A)}}</math>. | |
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==See Also== | ==See Also== | ||
{{AHSME box|year=1999|num-b=10|num-a=12}} | {{AHSME box|year=1999|num-b=10|num-a=12}} | ||
Revision as of 13:24, 4 June 2011
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
The locker labeling requires 
digits. Lockers through require digits total, lockers through 
require 
digits, and lockers 
through 
require 
digits. Thus, the remaining lockers require 
digits, so there must be 
more lockers, because they each use 
digits. Thus, there are 
student lockers, or answer choice 
.
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 | ||
