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{(D)}}</math> because then the second and third factors are necessarily of different parities, so that their product is necessarily even and we are done.
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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 $8$. If $a>b$, and $2a+1$ and $2b+1$ are the odd numbers, to prove the given statement we put the difference of the squares in the form:

$\textbf{(A)}\ (2a+1)^2-(2b+1)^2\\ \textbf{(B)}\ 4a^2-4b^2+4a-4b\\ \textbf{(C)}\ 4[a(a+1)-b(b+1)]\\ \textbf{(D)}\ 4(a-b)(a+b+1)\\ \textbf{(E)}\ 4(a^2+a-b^2-b)$

Solution

Although all of the forms listed can be used to show that the difference of $(2a+1)^2$ and $(2b+1)^2$ is necessarily divisible by $8$, we should use $\boxed{\textbf{(C)}}$ 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 48
Followed by
Problem 50
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All AHSME Problems and Solutions


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