Difference between revisions of "2004 AMC 10B Problems/Problem 13"
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We can easily add up <math>1.55\text{\ mm}</math> and <math>1.95\text{\ mm}</math> to get <math>3.50\text{\ mm}</math>. We multiply that by <math>4</math> to get <math>14\text{\ mm}</math>. Since this works and it requires 8 coins, the answer is clearly <math>\boxed{\mathrm{(B)\ }8}</math>. | We can easily add up <math>1.55\text{\ mm}</math> and <math>1.95\text{\ mm}</math> to get <math>3.50\text{\ mm}</math>. We multiply that by <math>4</math> to get <math>14\text{\ mm}</math>. Since this works and it requires 8 coins, the answer is clearly <math>\boxed{\mathrm{(B)\ }8}</math>. | ||
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== See also == | == See also == | ||
Revision as of 16:35, 1 July 2018
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. 
