Difference between revisions of "1999 AHSME Problems/Problem 11"
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\textbf{(E)}\ 6897</math> | \textbf{(E)}\ 6897</math> | ||
| − | ==Solution== | + | ==Solution 1== |
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>. | 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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| + | ==Solution 2== | ||
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| + | Since all answers are over <math>2000</math>, work backwards and find the cost of the first <math>1999</math> lockers. The first <math>9</math> lockers cost <math>0.18</math> dollars, while the next <math>90</math> lockers cost <math>0.04\cdot 90 = 3.60</math>. Lockers <math>100</math> through <math>999</math> cost <math>0.06\cdot 900 = 54.00</math>, and lockers <math>1000</math> through <math>1999</math> inclusive cost <math>0.08\cdot 1000 = 80.00</math>. | ||
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| + | This gives a total cost of <math>0.18 + 3.60 + 54.00 + 80.00 = 137.78</math>. There are <math>137.94 - 137.78 = 0.16</math> dollars left over, which is enough for <math>8</math> digits, or <math>2</math> more four digit lockers. These lockers are <math>2000</math> and <math>2001</math>, leading to answer <math> \boxed{\textbf{(A)}}</math>. | ||
==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 11:19, 8 August 2011
Contents
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 1
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 
.
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 | ||
