Difference between revisions of "2007 AMC 8 Problems/Problem 19"
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Let the smaller of the two numbers be <math>x</math>. Then, the problem states that <math> 2x+1<100</math>. <math> (x+1)^2-x^2=x^2+2x+1-x^2=2x+1 </math>. <math> 2x+1 </math> is obviously odd, so only answer choices C and E need to be considered. | Let the smaller of the two numbers be <math>x</math>. Then, the problem states that <math> 2x+1<100</math>. <math> (x+1)^2-x^2=x^2+2x+1-x^2=2x+1 </math>. <math> 2x+1 </math> is obviously odd, so only answer choices C and E need to be considered. | ||
| − | <math> 2x+1=131 </math> refutes the fact that <math> 2x+1<100 </math>, so the answer is <math> \mathrm{(C)} | + | <math> 2x+1=131 </math> refutes the fact that <math> 2x+1<100 </math>, so the answer is <math> \boxed{\mathrm{(C)} 79} </math> |
Revision as of 21:57, 14 November 2011
Problem
Pick two consecutive positive integers whose sum is less than 
. Square both
of those integers and then find the difference of the squares. Which of the
following could be the difference?
Solution
Let the smaller of the two numbers be 
. Then, the problem states that 
. 
. 
is obviously odd, so only answer choices C and E need to be considered.
refutes the fact that , so the answer is 
