2007 AMC 12B Problems/Problem 18
Contents
Problem 18
Let 
, 
, and 
be digits with 
. The three-digit integer 
lies one third of the way from the square of a positive integer to the square of the next larger integer. The integer 
lies two thirds of the way between the same two squares. What is 
?
Solution
The difference between and is given by
The difference between the two squares is three times this amount or
The difference between two consecutive squares is always an odd number, therefore 
is odd. We will show that must be 1. Otherwise we would be looking for two consecutive squares that are at least 81 apart. But already the equation 
solves to 
, and 
has more than three digits.
The consecutive squares with common difference 
are 
and 
. One third of the way between them is 
and two thirds of the way is 
.
This gives 
, 
, 
.
Solution
One-third the distance from 
to 
is 
.
Since this must be an integer, 
is divisible by 
. Since 
is always divisible by , 
must be divisible by .
Therefore, x must be 
or 
. (1, 4, and 7 don't work because their squares are too small)
Guessing and checking, we find that 
works, so the integer is one-third of the way from 
to 
, which is . 
See Also
| 2007 AMC 12B (Problems • Answer Key • Resources) | |
| 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 | |
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