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Find the last 2 digits of the expansion 2^(12n) - 6^(4n)

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Find the last 2 digits of the expansion 2^(12n) - 6^(4n)  [#permalink]

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New post 18 Mar 2005, 14:23
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Find the last 2 digits of the expansion 2^(12n) - 6^(4n) when n is any positive integer

(a) 10
(b) 00
(c) 04
(d) 50
(c) 64

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Re: PS Find the last 2 digits  [#permalink]

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New post 18 Mar 2005, 15:01
anirban16 wrote:
Find the last 2 digits of the expansion 2^(12n) - 6^(4n) when n is any positive integer

(a) 10
(b) 00
(c) 04
(d) 50
(c) 64


(B) 00.

Substitute n = 1.

2^ 12 - 6^4 = 4096 - 36^2 = 4096 - 1296 = 2800.
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New post 18 Mar 2005, 18:32
There is a much simpler way, you don't need to calculate that much.
I'd wait for some more answers.
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New post Updated on: 19 Mar 2005, 09:54
"B"

2^(12n) - 6^(4n)

(2^6n)^2 - (6^2n)^2 ===> (64^n + 36^n)(64^n-36^n)

===> first term will lead to last 2 digits of 00 for all odd n, and second will be 00 for all even n, so ans is "B"

Originally posted by banerjeea_98 on 18 Mar 2005, 18:58.
Last edited by banerjeea_98 on 19 Mar 2005, 09:54, edited 1 time in total.
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Re: PS Find the last 2 digits  [#permalink]

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New post 18 Mar 2005, 22:21
Both, kapslock's and baner's, approaches are nice and short cut.

kapslock's approach is good because n must satisfy the eq with any +ve integers. this could be simple.
baner's approach is obviously excellent but little complex.
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New post 21 Mar 2005, 18:46
1
Well I'd do it like this

2^(12n) - 6^(4n) = (2^6)^2n - (6^2)^2n

As 2n is always even the expression must have a factor (2^6 + 6^2) = 64+36 = 100.

Since a^n - b^n is divisible by a + b iff n is even.

Hope that helps

Other corrolaries are

a^n + b^n is divisible by a+b if n is odd
and not divisible by a+b if n is even

a^n-b^n is divisible by a-b whether n is odd or even.
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New post 22 Mar 2005, 07:10
i got 00 too but using no shortcuts, thanks for pointing em out
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Re: Find the last 2 digits of the expansion 2^(12n) - 6^(4n)  [#permalink]

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New post 19 Feb 2018, 19:27
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Re: Find the last 2 digits of the expansion 2^(12n) - 6^(4n) &nbs [#permalink] 19 Feb 2018, 19:27
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