Difference between revisions of "2006 AMC 10B Problems/Problem 22"
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Hence, <math>B=2,J=3</math>. | Hence, <math>B=2,J=3</math>. | ||
| − | Hence, the price of the Jam is <math>3\cdot11\cdot{0.05}\implies 1.65 \implies\boxed{ | + | Hence, the price of the Jam is <math>3\cdot11\cdot{0.05}\implies 1.65 \implies\boxed{D}</math>. |
== See Also == | == See Also == | ||
Revision as of 10:35, 20 June 2017
Contents
Problem
Elmo makes 
sandwiches for a fundraiser. For each sandwich he uses 
globs of peanut butter at 
per glob and 
blobs of jam at 
per blob. The cost of the peanut butter and jam to make all the sandwiches is 
. Assume that , , and are positive integers with 
. What is the cost of the jam Elmo uses to make the sandwiches?
Solution
The peanut butter and jam for each sandwich costs 
, so the peanut butter and jam for sandwiches costs 
.
Setting this equal to 
:
The only possible positive integer pairs 
whose product is 
are: 
The first pair violates and the third and fourth pair have no positive integer solutions for and .
So, 
and 
The only integer solutions for and are 
and 
Therefore the cost of the jam Elmo uses to make the sandwiches is 
Solution 2
Note as above, you get the equation 
Notice that we can multiply by 
on both sides to get whole numbers. Hence 
Note that the prime factorization of 
.
Hence, we want or 
Now, we have two cases to test.
Case 1:
Notice that we want 
or 
Taking 
Hence, 
.
Hence, the price of the Jam is 
.
See Also
| 2006 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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. 
