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24 Feb 2020, 05:21
Kudos
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
33% (01:48) correct
67%
(02:17)
wrong
based on 76
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History
Date
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Not Attempted Yet
\(20^{2004} + 16^{2004} – 3^{2004} − 1\) 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.
A. 91
B. 253
C. 317
D. 323
E. 324
Are You Up For the Challenge: 700 Level Questions
This one is 800+ QUESTION.
24 Feb 2020, 05:43
Kudos
Bookmarks
\(20^{2004}+16^{2004}–3^{2004}−1\) 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.
--> \((17+3)^{2004}–3^{2004}+(17-1)^{2004}−1\)
\(= 17m+...+3^{2004} -3^{2004} + 17n+....+ 1^{2004} -1= 17m+ 17n= 17*(m +n) \)
One of the answer choices must be divided by 17.
The answer choice D (\(323= 17*19\)) is correct.
25 Feb 2020, 01:35
Kudos
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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
\([20^{2004} - 3^{2004}]+[16^{2004}-1^{2004}]= 17x+17y\)
\([20^{2004} - 3^{2004}]+[16^{2004}-1^{2004}]\) 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
