Difference between revisions of "2003 AMC 12B Problems/Problem 7"
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| + | ==Problem== | ||
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Penniless Pete's piggy bank has no pennies in it, but it has 100 coins, all nickels,dimes, and quarters, whose total value is $8.35. It does not necessarily contain coins of all three types. What is the difference between the largest and smallest number of dimes that could be in the bank? | Penniless Pete's piggy bank has no pennies in it, but it has 100 coins, all nickels,dimes, and quarters, whose total value is $8.35. It does not necessarily contain coins of all three types. What is the difference between the largest and smallest number of dimes that could be in the bank? | ||
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\text {(A) } 0 \qquad \text {(B) } 13 \qquad \text {(C) } 37 \qquad \text {(D) } 64 \qquad \text {(E) } 83 | \text {(A) } 0 \qquad \text {(B) } 13 \qquad \text {(C) } 37 \qquad \text {(D) } 64 \qquad \text {(E) } 83 | ||
</math> | </math> | ||
| + | |||
| + | ==Solution== | ||
| + | Where <math>a,b,c</math> is the number of nickels, dimes, and quarters, respectively, we can set up two equations: | ||
| + | |||
| + | <cmath>(1)\ 5a+10b+25c=835\ \ \ \ (2)\ a+b+c=100</cmath> | ||
| + | |||
| + | Eliminate <math>a</math> by subtracting <math>(2)</math> from <math>(1)/5</math> to get <math>b+4c=67</math>. Of the integer solutions <math>(b,c)</math> to this equation, the number of dimes <math>b</math> is least in <math>(3,16)</math> and greatest in <math>(67,0)</math>, yielding a difference of <math>67-3=\boxed{\textbf{(D)}\ 64}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC12 box|year=2003|ab=B|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 00:30, 5 January 2014
Problem
Penniless Pete's piggy bank has no pennies in it, but it has 100 coins, all nickels,dimes, and quarters, whose total value is $8.35. It does not necessarily contain coins of all three types. What is the difference between the largest and smallest number of dimes that could be in the bank?
Solution
Where 
is the number of nickels, dimes, and quarters, respectively, we can set up two equations:
![\[(1)\ 5a+10b+25c=835\ \ \ \ (2)\ a+b+c=100\]](http://latex.artofproblemsolving.com/0/9/2/092d5d4a43045893d55007f55221c8ed84eb2059.png)
Eliminate 
by subtracting 
from 
to get 
. Of the integer solutions 
to this equation, the number of dimes 
is least in 
and greatest in 
, yielding a difference of 
.
See Also
| 2003 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 6 |
Followed by Problem 8 |
| 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 | |
| All AMC 12 Problems and Solutions | |
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