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If x is a positive integer, is the remainder 0 when 3^x + 1

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Manager
Joined: 10 Oct 2008
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If x is a positive integer, is the remainder 0 when 3^x + 1 [#permalink]  11 Nov 2008, 12:59
If x is a positive integer, is the remainder 0 when 3^x + 1 is divided by 10?
(1) x = 4n + 2, where n is a positive integer.
(2) x > 4

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
Manager
Joined: 15 Oct 2008
Posts: 54
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Kudos [?]: 2 [0], given: 0

Re: remainder--27 [#permalink]  12 Nov 2008, 00:10
3^x +1 can be divided by 10 only when 3^x has unit digit of 9

stmt 1: I try to find x by using n=1,2,3, it shows that 3^x are all give unit digit 9 which can be divided by 10 >>> suff

stmt 2: n can be any # --> not suff

IMO A
Manager
Joined: 30 Sep 2008
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Re: remainder--27 [#permalink]  12 Nov 2008, 00:21
Jcpenny wrote:
If x is a positive integer, is the remainder 0 when 3^x + 1 is divided by 10?
(1) x = 4n + 2, where n is a positive integer.
(2) x > 4

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

(1) 3^x + 1 = 3^(4n + 2) + 1 = 9*(3^4)^n + 1 = 9 * (81)^n + 1

as 81 ^n always has the last digit = 1 (with all n) => 9 * (81)^n has the last digit = 9

=> 9 * (81)^n + 1 is divisible by 10

(1) suff
Re: remainder--27   [#permalink] 12 Nov 2008, 00:21
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