Difference between revisions of "1954 AHSME Problems/Problem 49"
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== Solution == | == Solution == | ||
| − | Although all of the forms listed can be used to show that the difference of <math>(2a+1)^2</math> and <math>(2b+1)^2</math> is necessarily divisible by <math>8</math>, we should use <math>\boxed{\textbf{( | + | Although all of the forms listed can be used to show that the difference of <math>(2a+1)^2</math> and <math>(2b+1)^2</math> is necessarily divisible by <math>8</math>, we should use <math>\boxed{\textbf{(C)}}</math> because at least one of the second and third factors are necessarily of different parities, so that their product is necessarily even and we are done. |
== See Also == | == See Also == | ||
Latest revision as of 01:31, 18 September 2020
The difference of the squares of two odd numbers is always divisible by 
. If 
, and 
and 
are the odd numbers, to prove the given statement we put the difference of the squares in the form:
![$\textbf{(A)}\ (2a+1)^2-(2b+1)^2\\ \textbf{(B)}\ 4a^2-4b^2+4a-4b\\ \textbf{(C)}\ 4[a(a+1)-b(b+1)]\\ \textbf{(D)}\ 4(a-b)(a+b+1)\\ \textbf{(E)}\ 4(a^2+a-b^2-b)$](http://latex.artofproblemsolving.com/f/a/7/fa77929520ec7fa33eff255dfa3b46759f3d0c8a.png)
Solution
Although all of the forms listed can be used to show that the difference of 
and 
is necessarily divisible by , we should use 
because at least one of the second and third factors are necessarily of different parities, so that their product is necessarily even and we are done.
See Also
| 1954 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 48 |
Followed by Problem 50 | |
| 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
