Difference between revisions of "2007 AMC 12B Problems/Problem 18"
Chickendude (talk | contribs) (Added "See Also") |
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<math>27(c-b)</math> | <math>27(c-b)</math> | ||
− | The difference between two consecutive squares is always an odd number | + | The difference between two consecutive squares is always an odd number, therefore <math>c-b</math> is odd. We will show that <math>c-b</math> must be 1. Otherwise we would be looking for two consecutive squares that are at least 81 apart. But already the equation <math>(x+1)^2-x^2 = 27\cdot 3</math> solves to <math>x=40</math>, and <math>40^2</math> has more than three digits. |
− | This gives <math>a=1</math>, <math>b=7</math>, <math>c=8</math> | + | The consecutive squares with common difference <math>27</math> are <math>13^2=169</math> and <math>14^2=196</math>. One third of the way between them is <math>178</math> and two thirds of the way is <math>187</math>. |
+ | |||
+ | This gives <math>a=1</math>, <math>b=7</math>, <math>c=8</math>. | ||
<math>a+b+c = 16 \Rightarrow \mathrm{(C)}</math> | <math>a+b+c = 16 \Rightarrow \mathrm{(C)}</math> |
Revision as of 13:51, 5 January 2009
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 ,
,
.
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 |