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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+ | Similarly, we can simply take <math>8</math> quarters to get <math>8\cdot 1.75=14</math>. | ||
== See also == | == See also == |
Revision as of 15:29, 20 July 2019
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 .
Similarly, we can simply take quarters to get .
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 |
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