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Difference between revisions of "2006 AMC 10B Problems/Problem 22"

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== Solution 1 ==
 
== Solution 1 ==
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The peanut butter and jam for each sandwich costs <math>4B+5J\cent</math>, so the peanut butter and jam for <math>N</math> sandwiches costs <math>N(4B+5J)\cent</math>.  
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The peanut butter and jam for each sandwich costs <math>4B\cent</math>+5J\cent<math>, so the peanut butter and jam for </math>N<math> sandwiches costs </math>N(4B+5J)\cent<math>.  
  
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Setting this equal to <math>253\cent</math>:
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Setting this equal to </math>253\cent<math>:
  
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<math>N(4B+5J)=253=11\cdot23</math>
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</math>N(4B+5J)=253=11\cdot23<math>
  
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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>
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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>
  
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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>.  
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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>.  
  
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So, <math>N=11</math> and <math>4B+5J=23</math>
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So, </math>N=11<math> and </math>4B+5J=23<math>
  
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The only integer solutions for <math>B</math> and <math>J</math> are <math>B=2</math> and <math>J=3</math>
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The only integer solutions for </math>B<math> and </math>J<math> are </math>B=2<math> and </math>J=3<math>
  
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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>
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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 $
  
 
==Solution 2==
 
==Solution 2==

Revision as of 15:18, 26 July 2019

Problem

Elmo makes $N$ sandwiches for a fundraiser. For each sandwich he uses $B$ globs of peanut butter at $4\cent$ per glob and $J$ blobs of jam at $5\cent$ per blob. The cost of the peanut butter and jam to make all the sandwiches is $$ 2.53$. Assume that $B$, $J$, and $N$ are positive integers with $N>1$. What is the cost of the jam Elmo uses to make the sandwiches?

$\mathrm{(A) \ } $ 1.05\qquad \mathrm{(B) \ } $ 1.25\qquad \mathrm{(C) \ } $ 1.45\qquad \mathrm{(D) \ } $ 1.65\qquad \mathrm{(E) \ } $ 1.85$

Solution 1

The peanut butter and jam for each sandwich costs $4B\cent$+5J\cent$, so the peanut butter and jam for$N$sandwiches costs$N(4B+5J)\cent$.

Setting this equal to$ (Error compiling LaTeX. Unknown error_msg)253\cent$:$N(4B+5J)=253=11\cdot23$The only possible positive integer pairs$(N , 4B+5J)$whose product is$253$are:$ (1,253) ; (11,23) ; (23,11) ; (253,1) $The first pair violates$N>1$and the third and fourth pair have no positive integer solutions for$B$and$J$.

So,$ (Error compiling LaTeX. Unknown error_msg)N=11$and$4B+5J=23$The only integer solutions for$B$and$J$are$B=2$and$J=3$Therefore the cost of the jam Elmo uses to make the sandwiches is$3\cdot5\cdot11=165\cent$$ (Error compiling LaTeX. Unknown error_msg)= $1.65 \Rightarrow D $

Solution 2

Note as above, you get the equation $N(0.04B+0.05J=2.53)$

Notice that we can multiply by $100$ on both sides to get whole numbers. Hence $\implies N(4B+5J=253)$

Note that the prime factorization of $253=11\cdot23$.

Hence, we want $4B+5J=23$ or $11$

Now, we have two cases to test.

Case 1: $4B+5J=23$

Notice that we want $B\le5$ or $J\le4$

Taking $\pmod{5}\implies 4B\equiv3\pmod{5}\implies B=2$

Hence, $B=2,J=3$.

Hence, the price of the Jam is $3\cdot11\cdot{0.05}\implies 1.65 \implies\boxed{D}$.

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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