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20^2004 + 16^2004 – 3^2004 − 1 is divisible by:

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Bunuel
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20^2004 + 16^2004 – 3^2004 − 1 is divisible by: [#permalink] New post   24 Feb 2020, 05:21
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33% (01:50) correct 67% (02:21) wrong based on 57 sessions
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\(20^{2004} + 16^{2004} – 3^{2004} − 1\) is divisible by:


A. 91
B. 253
C. 317
D. 323
E. 324


Are You Up For the Challenge: 700 Level Questions

This one is 800+ QUESTION.
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lacktutor
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20^2004 + 16^2004 – 3^2004 − 1 is divisible by: [#permalink] New post   24 Feb 2020, 05:43
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\(20^{2004}+16^{2004}–3^{2004}−1\) is divisible by:

--> \((17+3)^{2004}–3^{2004}+(17-1)^{2004}−1\)

\(= 17m+...+3^{2004} -3^{2004} + 17n+....+ 1^{2004} -1= 17m+ 17n= 17*(m +n) \)

One of the answer choices must be divided by 17.

The answer choice D (\(323= 17*19\)) is correct.
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Re: 20^2004 + 16^2004 – 3^2004 − 1 is divisible by: [#permalink] New post   25 Feb 2020, 01:35
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\(a^n - b^n\) is divisible by (a-b) and (a+b), if n is even

\([20^{2004} - 3^{2004}]+[16^{2004}-1^{2004}]= 17x+17y\)

\([20^{2004} - 3^{2004}]+[16^{2004}-1^{2004}]\) must be divisible by 17.


Also, \([20^{2004} - 1^{2004}]+[16^{2004}-3^{2004}]= 19a + 19b\)

\([20^{2004} - 3^{2004}]+[16^{2004}-1^{2004}]\) must be divisible by 19


\(20^{2004} + 16^{2004} – 3^{2004} − 1\) is divisible by 17*19=323

Bunuel wrote:
\(20^{2004} + 16^{2004} – 3^{2004} − 1\) is divisible by:


A. 91
B. 253
C. 317
D. 323
E. 324


Are You Up For the Challenge: 700 Level Questions

This one is 800+ QUESTION.
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20^2004 + 16^2004 – 3^2004 − 1 is divisible by: [#permalink] New post   29 Mar 2020, 03:01
Bunuel wrote:
\(20^{2004} + 16^{2004} – 3^{2004} − 1\) is divisible by:


A. 91
B. 253
C. 317
D. 323
E. 324


Are You Up For the Challenge: 700 Level Questions

This one is 800+ QUESTION.


Asked: \(20^{2004} + 16^{2004} – 3^{2004} − 1\) is divisible by:

20^2004 - 1 + 16^2004 - 3^2004
= (20^1002 + 1)(20^1002 - 1) + (16^1002 + 3^1002)(16^1002 - 3^1002)

Expression is divisible by 19 since the term (20-1) = (16+3) = 19 will be common to both expressions

20^2004 - 3^2004 + 16^2004 - 1
= (20^1002 + 3^1002)(20^1002 - 3^1002) + (16^1002 + 1)(16^1002 - 1)

Expression is divisible by 17 since the term (20-3) = 17 = (16+1) will be common to both expressions

19*17 = 323

IMO D
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20^2004 + 16^2004 – 3^2004 − 1 is divisible by: [#permalink]
29 Mar 2020, 03:01
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