Difference between revisions of "2005 AMC 8 Problems/Problem 5"

 
Line 9: Line 9:
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2005|num-b=4|num-a=6}}
 
{{AMC8 box|year=2005|num-b=4|num-a=6}}
 +
{{MAA Notice}}

Latest revision as of 01:08, 5 July 2013

Problem

Soda is sold in packs of 6, 12 and 24 cans. What is the minimum number of packs needed to buy exactly 90 cans of soda?

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 15$

Solution

Start by buying the largest packs first. After three $24$-packs, $90-3(24)=18$ cans are left. After one $12$-pack, $18-12=6$ cans are left. Then buy one more $6$-pack. The total number of packs is $3+1+1=\boxed{\textbf{(B)}\ 5}$.

See Also

2005 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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 AJHSME/AMC 8 Problems and Solutions

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