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>.

Latest revision as of 10:33, 12 July 2014

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

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 $2000$, 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AHSME Problems and Solutions

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