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If a and b are positive integers, is a^4-b^4 divisible by 4?

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If a and b are positive integers, is a^4-b^4 divisible by 4?  [#permalink]

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If a and b are positive integers, is \(a^4-b^4\) divisible by 4?

1) \(a+b\) is divisible by 4
2) The remainder is 2 when \(a^2+b^2\) is divided by 4


Weekly Quant Quiz #3 Question No 6


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Originally posted by gmatbusters on 06 Oct 2018, 09:25.
Last edited by gmatbusters on 07 Oct 2018, 05:04, edited 2 times in total.
Renamed the topic and edited the question.
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Re: If a and b are positive integers, is a^4-b^4 divisible by 4?  [#permalink]

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New post 06 Oct 2018, 09:25
1

Official Explanation:


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

Statement 1 :

a+b is divisible by 4, a+b = 4k
Since, \(a^4-b^4=(a-b)(a+b)(a^2+b^2)\),
we get, \(a^4-b^4=4k(a-b)(a^2+b^2)\)
Hence it is divisible by 4,
SUFFICIENT


Statement 2 :

\(a^2+b^2\) , when divided by 4 gives 2 as remainder, Hence \(a^2+b^2\) is Even.
It means either both a and b are EVEN or ODD.
When both are Even, a = 2m, b = 2n: \(a^4-b^4 = 4(m^4-n^4)\), divisible by 4
When both are oddm a = 2m +1, b = 2n+1:
\(a^4-b^4=(a-b)(a+b)(a^2+b^2)\)
=\(4(m-n)(m+n+1)(a^2+b^2)\), hence divisible by 4

SUFFICIENT

Answer D
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Re: If a and b are positive integers, is a^4-b^4 divisible by 4?  [#permalink]

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New post 06 Oct 2018, 09:29
a4−b4 = (a+b)*(a-b) (a2+b2)

Option A: a+b is divisible by 4
since a and b are integers, it definitely answers that a4-b4 will be divisible by 4

2) The remainder is 2 when a2+b2 is divisible by 4
which means a2+b2=. 4k +2

so it becomes (a+b)(a-b) (4k+2)
can't say that it is divisible by 4


A is the answer
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Re: If a and b are positive integers, is a^4-b^4 divisible by 4?  [#permalink]

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Re: If a and b are positive integers, is a^4-b^4 divisible by 4?  [#permalink]

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New post 06 Oct 2018, 09:35
a^4 - b^4 = (a-b)(a+b)(a^2-b^2)

St 1 says a+4 div by 4 - sufficient

St 2 says a^2+b^2mod 4 = 2 (remainder 2)
which means a and b are either both odd, or both even
Both cases imply a+b is even Hence product of two even numbers is div by 4

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Re: If a and b are positive integers, is a^4-b^4 divisible by 4?  [#permalink]

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New post 06 Oct 2018, 09:36
a^4-b^4 can be factorized as (a^2-b^2)*(a^2+b^2)... (a-b)*(a+b)*(a^2+b^2)

St.1 says a+b is divisible by 4. Hence sufficient.

St.2 says remainder is 2 when a^2+b^2 is divided by 4.. hence it is divisible by 2. It could have both even or both odd parts adding up to an even total... Hence either ways a+b and a-b will be even. Therefore the whole will be divisible by 4.

St2. Is also sufficient.

Hence option (d) is correct.

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Re: If a and b are positive integers, is a^4-b^4 divisible by 4?  [#permalink]

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New post 06 Oct 2018, 09:38
answer is D:
a and b
statement 1: (a^4-b^4) = (a^2+b^2) (a+b)(a-b)
if a+ b is divisible by 4 then the statement is divisible by 4

statment 2:plug in numbers. 3^2+3^2 = 18 (which has a remainder of 2 when divided by 4) and 81-81 = 0 which is divisible by 4
5^2 + 1^2 = 26
and 525-1 is divisible by 4
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Re: If a and b are positive integers, is a^4-b^4 divisible by 4?  [#permalink]

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New post 06 Oct 2018, 09:53
Answer D
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Re: If a and b are positive integers, is a^4-b^4 divisible by 4?  [#permalink]

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New post 06 Oct 2018, 09:59
a^4-b^4=(a^2+b^2)(a+b)(a-b)

1) suff
2)Suff
(a^2+b^2) div 4..... rem 2 .....===> (a^2+b^2) is divisible by 2 not by 4
case 1 : so a, b both even .... then (a+b) & (a-b) both div by 2
case 2 : so a, b both odd .... then (a+b) & (a-b) both div by 2

Hence ans D
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Re: If a and b are positive integers, is a^4-b^4 divisible by 4? &nbs [#permalink] 06 Oct 2018, 09:59
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