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> (x+1)+x<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> (x+1)+x<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> | + | <math> 2x+1=131 </math> contradicts the fact that <math> 2x+1<100 </math>, so the answer is <math> \boxed{\mathrm{(C)} 79} </math> |
==Video Solution by WhyMath== | ==Video Solution by WhyMath== | ||
Revision as of 23:16, 11 December 2021
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.
contradicts the fact that 
, so the answer is 
Video Solution by WhyMath
~savannahsolver
See Also
| 2007 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
