Difference between revisions of "2006 AMC 10B Problems/Problem 22"
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== Problem == | == Problem == | ||
− | == Solution == | + | Elmo makes <math>N</math> sandwiches for a fundraiser. For each sandwich he uses <math>B</math> globs of peanut butter at <math>4\cent</math> per glob and <math>J</math> blobs of jam at <math>5\cent</math> per blob. The cost of the peanut butter and jam to make all the sandwiches is <math> \$ 2.53</math>. Assume that <math>B</math>, <math>J</math>, and <math>N</math> are positive integers with <math>N>1</math>. What is the cost of the jam Elmo uses to make the sandwiches? |
+ | |||
+ | <math> \mathrm{(A) \ } \$ 1.05\qquad \mathrm{(B) \ } \$ 1.25\qquad \mathrm{(C) \ } \$ 1.45\qquad \mathrm{(D) \ } \$ 1.65\qquad \mathrm{(E) \ } \$ 1.85 </math> | ||
+ | |||
+ | == Solution 1 == | ||
+ | The peanut butter and jam for each sandwich costs <math>4B\cent+5J\cent</math>, so the peanut butter and jam for <math>N</math> sandwiches costs <math>N(4B+5J)\cent</math>. | ||
+ | |||
+ | Setting this equal to <math>253\cent</math>: | ||
+ | |||
+ | <math>N(4B+5J)=253=11\cdot23</math> | ||
+ | |||
+ | The only possible positive integer pairs <math>(N , 4B+5J)</math> whose product is <math>253</math> are: <math> (1,253) ; (11,23) ; (23,11) ; (253,1) </math> | ||
+ | |||
+ | The first pair violates <math>N>1</math> and the third and fourth pair have no positive integer solutions for <math>B</math> and <math>J</math>. | ||
+ | |||
+ | So, <math>N=11</math> and <math>4B+5J=23</math> | ||
+ | |||
+ | The only integer solutions for <math>B</math> and <math>J</math> are <math>B=2</math> and <math>J=3</math> | ||
+ | |||
+ | Therefore the cost of the jam Elmo uses to make the sandwiches is <math>3\cdot5\cdot11=165\cent</math> <math>= \$1.65 \Rightarrow D </math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Note as above, you get the equation <math>N(0.04B+0.05J=2.53)</math> | ||
+ | |||
+ | Notice that we can multiply by <math>100</math> on both sides to get whole numbers. Hence <math>\implies N(4B+5J)=253</math> | ||
+ | |||
+ | Note that the prime factorization of <math>253=11\cdot23</math>. | ||
+ | |||
+ | Hence, we want <math>4B+5J=23</math> or <math>11</math> | ||
+ | |||
+ | Now, we have two cases to test. | ||
+ | |||
+ | Case 1: <math>4B+5J=23</math> | ||
+ | |||
+ | Notice that we want <math>B\le5</math> or <math>J\le4</math> | ||
+ | |||
+ | Taking <math>\pmod{5}\implies 4B\equiv3\pmod{5}\implies B=2</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{D}</math>. | ||
+ | |||
+ | == Solution 3 (Video) == | ||
+ | Video solution: https://www.youtube.com/watch?v=7248rEcCSfM | ||
+ | |||
== See Also == | == See Also == | ||
− | + | {{AMC10 box|year=2006|ab=B|num-b=21|num-a=23}} | |
+ | |||
+ | [[Category:Introductory Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 16:53, 27 November 2020
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 1
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 .
Solution 3 (Video)
Video solution: https://www.youtube.com/watch?v=7248rEcCSfM
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 |
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