Difference between revisions of "2004 AMC 10B Problems/Problem 13"
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== Solution == | == Solution == | ||
| − | All numbers in this solution will be in | + | All numbers in this solution will be in hundredths of a millimeter. |
The thinnest coin is the dime, with thickness <math>135</math>. A stack of <math>n</math> dimes has height <math>135n</math>. | The thinnest coin is the dime, with thickness <math>135</math>. A stack of <math>n</math> dimes has height <math>135n</math>. | ||
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If <math>n=10</math> the possible stack heights are <math>1350,1370,1390,\dots</math>, with the remaining ones exceeding <math>1400</math>. | If <math>n=10</math> the possible stack heights are <math>1350,1370,1390,\dots</math>, with the remaining ones exceeding <math>1400</math>. | ||
| − | Therefore there are <math>\boxed{8}</math> coins in the stack. | + | Therefore there are <math>\boxed{\mathrm{(B)\ }8}</math> coins in the stack. |
Using the above observation we can easily construct such a stack. A stack of <math>8</math> dimes would have height <math>8\cdot 135=1080</math>, thus we need to add <math>320</math>. | Using the above observation we can easily construct such a stack. A stack of <math>8</math> dimes would have height <math>8\cdot 135=1080</math>, thus we need to add <math>320</math>. | ||
Revision as of 23:42, 23 July 2014
Contents
Problem
In the United States, coins have the following thicknesses: penny, 
mm; nickel, 
mm; dime, 
mm; quarter, 
mm. If a stack of these coins is exactly 
mm high, how many coins are in the stack?
Solution
All numbers in this solution will be in hundredths of a millimeter.
The thinnest coin is the dime, with thickness 
. A stack of 
dimes has height 
.
The other three coin types have thicknesses 
, 
, and 
. By replacing some of the dimes in our stack by other, thicker coins, we can clearly create exactly all heights in the set 
.
If we take an odd , then all the possible heights will be odd, and thus none of them will be 
. Hence is even.
If 
the stack will be too low and if 
it will be too high. Thus we are left with cases 
and 
.
If the possible stack heights are 
, with the remaining ones exceeding .
Therefore there are 
coins in the stack.
Using the above observation we can easily construct such a stack. A stack of 
dimes would have height 
, thus we need to add 
.
This can be done for example by replacing five dimes by nickels (for 
), and one dime by a penny (for 
).
Note
We can easily add up 
and 
to get 
We multiply that by 
to get 
. Since this works and it requires 8 coins, the answer is clearly 
.
See also
| 2004 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
