Difference between revisions of "2004 AMC 10B Problems/Problem 13"
m (→Solution) |
|||
| Line 21: | Line 21: | ||
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>. | ||
| − | This can be done for example by replacing five dimes by nickels (for <math>+60\cdot 5 = +300</math>), and one dime by a penny (for <math>+20</math>). | + | This can be done for example by replacing five dimes by nickels (for <math>+60\cdot 5 = +300</math>), and one dime by a penny (for <math>+20</math>). |
| + | |||
| + | ==Note== | ||
| + | |||
| + | We can easily add up <math>1.55mm</math> and <math>1.95mm</math> to get <math>3.50mm</math> We multiply that by <math>4</math> to get <math>14mm</math>. Since this works and it requires 8 coins, the answer is clearly <math>\boxed{8}</math>. | ||
== See also == | == See also == | ||
Revision as of 17:29, 1 February 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 hundreds 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. 
