Art of Problem Solving

2006 AMC 10B Problems/Problem 22: Difference between revisions

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== See Also ==
== See Also ==
*[[2006 AMC 10B Problems]]
{{AMC10 box|year=2006|ab=B|num-b=21|num-a=23}}
 
*[[2006 AMC 10B Problems/Problem 21|Previous Problem]]
 
*[[2006 AMC 10B Problems/Problem 23|Next Problem]]


[[Category:Introductory Number Theory Problems]]
[[Category:Introductory Number Theory Problems]]

Revision as of 22:04, 7 September 2011

Problem

Elmo makes $N$ sandwiches for a fundraiser. For each sandwich he uses $B$ globs of peanut butter at $4$¢ per glob and $J$ blobs of jam at $5$¢ 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

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

Setting this equal to $253$¢:

$N(4B+5J)=253=11\cdot23$

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, $N=11$ 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 $3\cdot5\cdot11=165$¢ $=$ $1.65 \Rightarrow D$

See Also

2006 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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