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# If a and b are two odd positive integers

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Manager
Joined: 26 Dec 2011
Posts: 114
Schools: HBS '18, IIMA
If a and b are two odd positive integers  [#permalink]

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Updated on: 28 May 2015, 18:22
1
8
00:00

Difficulty:

25% (medium)

Question Stats:

76% (01:49) correct 24% (01:58) wrong based on 263 sessions

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If a and b are two odd positive integers, by which of the following integers is (a^4 - b^4) is always divisible?

(A) 3
(B) 5
(C) 6
(D) 8
(E) 12

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Originally posted by balamoon on 28 May 2015, 12:16.
Last edited by balamoon on 28 May 2015, 18:22, edited 1 time in total.
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Re: If a and b are two odd positive integers  [#permalink]

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28 May 2015, 13:19
3
2
Using the difference of squares factorization (twice), we have:

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

We've now written our number as a product of three things. But each of those three things is even, if a and b are both odd. If each of those three factors is divisible by 2, their product is divisible by 2^3 = 8.

So our number is divisible by 8. But any number divisible by 8 is also divisible by 4, so there are two correct answers: 4 and 8. The question is flawed.

Even if they meant to ask something like "what is the largest integer you can be certain is a factor of a^4 - b^4", the question is still flawed, because the answer to that question is 16. It's probably easiest to see why that's true using remainder arithmetic, but we can also see why algebraically. We know (a^2 + b^2) is divisible by 2. It turns out that (a^2 - b^2) = (a+b)(a-b) is not only divisible by 4 when a and b are odd - it actually must be divisible by 8. If a and b are odd, then for some integers s and t, we know:

a = 2s + 1
b = 2t + 1

so (a + b)(a - b) = (2s + 2t + 2)(2s - 2t) = 2*2(s + t + 1)(s - t)

Now, because addition and subtraction follow the same even/odd rules, then s+t and s-t are either both even, or both odd. So exactly one of the factors s+t+1 and s-t is even, and the other is odd, so we have another 2 in our factorization somewhere, and a^2 - b^2 is divisible by 8.
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##### General Discussion
Manager
Joined: 26 Dec 2011
Posts: 114
Schools: HBS '18, IIMA
Re: If a and b are two odd positive integers  [#permalink]

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28 May 2015, 18:23
Corrected the choices....
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Director
Joined: 23 Jan 2013
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Schools: Cambridge'16
If a and b are two odd positive integers  [#permalink]

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07 Oct 2015, 02:50
1
take 1 and 3,
1^4-3^4=1-81=-80, divisible only by 5 and 8. A,C,E out

take 3 and 5
3^4-5^4=81-625=544, not divisible by 5

D
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Re: If a and b are two odd positive integers  [#permalink]

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29 Nov 2015, 15:05
I chose 3 odd numbers to find the correct choice.

Actually it took me 1:30 minutes.

Is it ok or there is a faster way?
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Re: If a and b are two odd positive integers  [#permalink]

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02 Aug 2017, 13:24
1
one thig to consider
odd - odd = even
odd^4 - odd^4 most likely will be divisible by an even only...8=2^3, so divisible.
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Re: If a and b are two odd positive integers  [#permalink]

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02 Aug 2017, 14:19
1
alternatively break it like this
a^4 - b^4 = (a^2 - b^2) ( a^2 + b^2)
(a^2 - b^2) - even
(a^2 + b^2) - even

so you already have 2 evens

now, break a^2 - b^2 into (a+b)(a-b)
so you have the 3rd even.
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Re: If a and b are two odd positive integers  [#permalink]

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03 Aug 2017, 03:47
balamoon wrote:
If a and b are two odd positive integers, by which of the following integers is (a^4 - b^4) is always divisible?

(A) 3
(B) 5
(C) 6
(D) 8
(E) 12
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No matter what numbers you choose as a and b, the result is always going to be even (odd-odd=even). This automatically cancels out A and B, because odd numbers will sometimes be able to divide even numbers, but certainly not always. So you're left with 6, 8, and 12. You could pick numbers a=1, b=3, and then a=3, b=5 or vice versa; you'll get a pattern where all your results will be divisible by only 8.
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Re: If a and b are two odd positive integers  [#permalink]

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09 Aug 2017, 11:52
balamoon wrote:
If a and b are two odd positive integers, by which of the following integers is (a^4 - b^4) is always divisible?

(A) 3
(B) 5
(C) 6
(D) 8
(E) 12

Simplifying the given expression, we have:

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

Since a and b are both odd, we see that a - b = odd - odd = even. Similarly, a + b = odd + odd = even, and finally, a^2 + b^2 = odd^2 + odd^2 = odd + odd = even. Thus, we see that the expression is a product of three even numbers, and since each even number is divisible by 2, the expression must always be divisible by 2 x 2 x 2 = 8.

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Re: If a and b are two odd positive integers  [#permalink]

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13 Aug 2018, 11:08
If a and b are two odd positive integers, by which of the following integers is (a^4 - b^4) is always divisible?

(A) 3
(B) 5
(C) 6
(D) 8
(E) 12

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

Given That a and b both are positive odd integers so
Odd+odd =even (divisible by 2)
Hence even*even*even divisible by 8
Option D

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Re: If a and b are two odd positive integers &nbs [#permalink] 13 Aug 2018, 11:08
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