2007 AMC 12B Problems/Problem 18
Problem
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 1
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 2
One-third the distance from to is .
Since must be an integer, and therefore must be divisible by .
Therefore, x must be or . (1, 4, and 7 don't work because their squares are too small. Similarly if x is greater than 28, the squares are too large.)
Guessing and checking, we find that works, so the integer is one-third of the way from to , which is .
- JN5537 - edited by numerophile
Solution 3
Let be the lesser of the two integers. Then the squares of the integers are and , and the distance between them is . Let this be equivalent to , so that the one-third of the distance between the squares is equivalent to . The numbers and are one-third and two-thirds of the way between and . Therefore, the distance between these two numbers is also one-third the distance between the squares, or . Setting these equal to each other, we have
.
Notice that since and are digits, their difference is at most and at least . Also notice that since is greater than , . Representing this as an inequality, we have
.
Substituting , we have
.
However, we know that is a -digit number, and since is less than , must be at most , or . Therefore . Plugging this back into our inequality, we have
.
But (c-b) must be an integer, so now we have
is also an integer, so now we have
.
Once again, must be an integer, so we have
The two squares are and , or and . A third of the distance between them is , and . .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.