Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2)
m (→‎Solution 1)
Line 13: Line 13:
 
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 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>.  
+
The first pair violates <math>N>1</math> and the third and fourth pairs have no positive integer solutions for <math>B</math> and <math>J</math>.  
  
 
So, <math>N=11</math> and <math>4B+5J=23</math>
 
So, <math>N=11</math> and <math>4B+5J=23</math>

Revision as of 17:21, 2 June 2021

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 $253\cent$:

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

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 pairs 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 $3\cdot5\cdot11=165\cent$ $= $1.65 \Rightarrow D$

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 $3\cdot11\cdot{0.05}\implies 1.65 \implies\boxed{D}$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10B_Problems/Problem_22&oldid=154857"