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

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\textbf{(E)}\ 6897</math>
 
\textbf{(E)}\ 6897</math>
  
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==Solution==
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==Solution 1==
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===Solution 1===
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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)}}.
 
  
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===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>.
  
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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>.  
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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==

Latest revision as of 06:19, 25 August 2024

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 1

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

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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