Difference between revisions of "1999 AHSME Problems/Problem 11"
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\textbf{(E)}\ 6897</math> | \textbf{(E)}\ 6897</math> | ||
− | ==Solution 1== | + | ==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>. | ||
− | ==Solution 2== | + | ===Solution 2=== |
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>. | 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>. |
Revision as of 11:33, 12 July 2014
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 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 |
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