Question

The difference between the squares of two consecutive odd integers is always divisible which which one digit number?

I'd be thankful for explanation as well.

Answer 1

Say, consecutive odd numbers are of the form (2n - 1) and (2n + 1), with n an integer

So the difference: (2x-1)^2 - (2x+1)^2
                         = (4x^2-4x+1)-(4x^2+4x+1)
                         = - 8x

The difference between the squares of two consecutive odd integers is always divisible by 8.

Answer 2

Given :

• The difference between the squares of two consecutive odd integers.

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Find :

• Divisible by which one digit number?

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Calculations :

• It's ones digit.
• The number which has only one digit between 1-10 in the term of postive integers.

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We know that ,

• Consecutive of odd integers = (Pair of any consecutive odd integers).

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• Let "2 + 1" "2 + 3" be the consecutive pairs of two odd integers.

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• We also know that, squares of those pairs to be added in our equation.

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→ (2 + 1)² - (2 + 3)³

• Multiply their sum, deduct with each of their pair.

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→ (2 + 1 + 2 + 3) (2 + 1 - 2 - 3)

• Now, multiply their result with each other pair.

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→ (4 + 4) (2)

→ 8 (x + 1)

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Therefore, 8 is the required answer which is divisible.