Difference between revisions of "2007 AMC 12B Problems/Problem 18"

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<math>a+b+c = 16 \Rightarrow \mathrm{(C)}</math>
 
<math>a+b+c = 16 \Rightarrow \mathrm{(C)}</math>
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==See Also==
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{{AMC12 box|year=2007|ab=B|num-b=17|num-a=19}}

Revision as of 23:20, 21 February 2008

Problem 18

Let $a$, $b$, and $c$ be digits with $a\ne 0$. The three-digit integer $abc$ lies one third of the way from the square of a positive integer to the square of the next larger integer. The integer $acb$ lies two thirds of the way between the same two squares. What is $a+b+c$?

$\mathrm{(A)}\ 10 \qquad \mathrm{(B)}\ 13 \qquad \mathrm{(C)}\ 16 \qquad \mathrm{(D)}\ 18 \qquad \mathrm{(E)}\ 21$

Solution

The difference between $acb$ and $abc$ is given by

$(100a + 10c + b) - (100a + 10b + c) = 9(c-b)$

The difference between the two squares is three times this amount or

$27(c-b)$

The difference between two consecutive squares is always an odd number. The consecutive squares with common difference $27$ are $13^2=169$ and $14^2=196$. One third of the way between them is $178$ and two thirds of the way is $187$

This gives $a=1$, $b=7$, $c=8$

$a+b+c = 16 \Rightarrow \mathrm{(C)}$

See Also

2007 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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 12 Problems and Solutions