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# If x is a positive integer, is the remainder 0 when (3x +

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Senior Manager
Joined: 11 Feb 2007
Posts: 350

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If x is a positive integer, is the remainder 0 when (3x + [#permalink]

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17 Feb 2007, 03:29
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If x is a positive integer, is the remainder 0 when (3x + 1)/10?
(1) x = 3n + 2, where n is a positive integer.
(2) x > 4

I thought the answer was A because
(1) for the first three values of x when n is 1,2 3 are 5, 8, 11 and I did the calculation.

3^5 + 1 = 244 so NO

It appeared to me that they'd never end with 0.
so (1) suff. (yes, I tried only one of infinitely many substitutions but I was running out of time and couldn't afford to spend too much on this prob)

(2) 3^5 + 1 = 244 BUT
3^6 + 1 = 730 which is divisible by 10
thus (2) is insuff.

However, the answer is E, and I looked at (1) again, using a calculator and found out that 3^14 + 1 = 4782970 which IS divisible by 10.

How can I know that there is a number n such that 3^n will end with 9 such that + 1 will make it divisible by 10?

I can't afford to spend time doing 3^14 on paper and pencil...

(A good reference to study number theory perphaps?)

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Joined: 05 Jul 2006
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17 Feb 2007, 05:27
If x is a positive integer, is the remainder 0 when (3x + 1)/10?
(1) x = 3n + 2, where n is a positive integer.
(2) x > 4

3x+1 has to be both even and a multiple of 5

from one

3x+1 = 9n+7 for this to be even n has to be odd......insuff

n could be anything

from two

3N+2>4

N>2/3............STILL N COULD BE EVEN OR ODD

INSUFF

BOTH TOGETHER ARE NOW OBVIOUSLY INSUFF

Kudos [?]: 443 [0], given: 49

Senior Manager
Joined: 11 Feb 2007
Posts: 350

Kudos [?]: 186 [0], given: 0

Oops! I'm sorry it should be (3^x) + 1/10 [#permalink]

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17 Feb 2007, 19:27
Thank you for your explanation yezz.

I apologize, it should be ((3^x) + 1)/10 rather than (3x+1)/10

3 to the power of x.

Kudos [?]: 186 [0], given: 0

Oops! I'm sorry it should be (3^x) + 1/10   [#permalink] 17 Feb 2007, 19:27
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