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daviesj
Joined: 23 Aug 2012
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If n is a positive integer, is n^3 + 4n^2 - 5n divisible by 8 ?
(1) n = 4b + 1, where b is a positive integer.
(2) n^2 – n is divisible by 24.
(1) n = 4b + 1, where b is a positive integer.
(2) n^2 – n is divisible by 24.
14 Jan 2013, 06:49
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If n is a positive integer, is n^3 + 4n^2 - 5n divisible by 8 ?
\(n^3 + 4n^2 - 5n=n(n^2+4n-5)=n(n-1)(n+5)\)
(1) n = 4b + 1, where b is a positive integer --> \(n(n-1)(n+5)=(4b+1)(4b)(4b+6)=(4b+1)(8b)(2b+3)\). Sufficient.
(2) n^2 – n is divisible by 24 --> n(n-1) is divisible by 24, so its also divisible by 8. Thus n(n-1)(n+5) is also divisible by 8. Sufficient.
Answer: D.
Hope it's clear.
\(n^3 + 4n^2 - 5n=n(n^2+4n-5)=n(n-1)(n+5)\)
(1) n = 4b + 1, where b is a positive integer --> \(n(n-1)(n+5)=(4b+1)(4b)(4b+6)=(4b+1)(8b)(2b+3)\). Sufficient.
(2) n^2 – n is divisible by 24 --> n(n-1) is divisible by 24, so its also divisible by 8. Thus n(n-1)(n+5) is also divisible by 8. Sufficient.
Answer: D.
Hope it's clear.
15 Jan 2013, 11:56
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Hi Bunuel,
Can u plz xplain this step:
(4b+1)(4b)(4b+6)= (4b+1)(8b)(2b+3).Sufficient.
Thanks,
Shreeraj
Can u plz xplain this step:
(4b+1)(4b)(4b+6)= (4b+1)(8b)(2b+3).Sufficient.
Thanks,
Shreeraj
