Question

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:

$(2a+1)^2-(2b+1)^2$

$4a^2-4b^2+4a-4b$

$4[a(a+1)-b(b+1)]$

$4(a-b)(a+b+1)$

$4(a^2+a-b^2-b)$

Answer: C

Concepts:divisibility parity factoring

Difficulty rating: 1590

Small Hint:

Expand the two squares and group each variable with its successor

Big Hint:

Each product $a(a+1)$ and $b(b+1)$ is even

Solution:

Expanding and regrouping, $\begin{aligned} &(2a+1)^2-(2b+1)^2\\ &\qquad=4[a(a+1)-b(b+1)]. \end{aligned}$ Each of $a(a+1)$ and $b(b+1)$ is even, so their difference is even. The displayed expression is consequently divisible by $4\cdot2=8.$

Thus, the correct answer is C.

1954 AMC 12 Problem 49

Answer: C

Difficulty rating: 1590

Small Hint:

Big Hint:

Solution:

Problem 49 in Other Years