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If x is a positive integer, is the remainder 0 when 3^(x)

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

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Hey,
I am a new member of this club.
Could anyone please help me with few problems of quant??? Here are the problems.

1.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

2.What is the sum of a certain pair of consecutive odd integers?
(1) At least one of the integers is negative.
(2) At least one of the integers is positive.


Thanks,
Bibha :)
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Re: Need some Help [#permalink]

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New post 15 Apr 2010, 03:29
bibha wrote:
Hey,
I am a new member of this club.
Could anyone please help me with few problems of quant??? Here are the problems.

1.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

2.What is the sum of a certain pair of consecutive odd integers?
(1) At least one of the integers is negative.
(2) At least one of the integers is positive.


Thanks,
Bibha :)


1. 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.

Last digit of \(3^x\) repeats in blocks of 4: {3, 9, 7, 1} - {3, 9, 7, 1} - ... So cyclicity of the last digit of 3 in power is 4. Now, \(3^{4n+2}\) will have the same last digit as \(3^2\) (remainder upon division 4n+2 upon cyclicity 4 is 2, which means that 3^{4n+2} will have the same last digit as 3^2). Last digit of \(3^2\) is \(9\). So \(3^{4n+2}+1\) will have the last digit \(9+1=0\). Number ending with 0 is divisible by 10 (remainder 0). Sufficient.

(2) x > 4. Clearly insufficient.

Answer: A.

Check Number Theory chapter of Math Book for more: math-number-theory-88376.html

2. What is the sum of a certain pair of consecutive odd integers?
(1) At least one of the integers is negative --> infinite pairs are possible: ... (-3,-1); (-17,-15); ... (-1, 1); ... Not sufficient.
(2) At least one of the integers is positive --> infinite pairs are possible: ... (3,5); (19,21); ... (-1, 1); ... Not sufficient.

(1)+(2) one odd integer must be positive and another negative. As they are consecutive odd integers, there is only one pair possible (-1, 1) --> -1+1=0. Sufficient.

Answer: C.
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Re: Need some Help [#permalink]

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New post 15 Apr 2010, 08:52
heyy,
Thank you all so much for helping. I got tricked by the "sum of consecutive pair of integers" question. I totally neglected the "pairs" part....hehe
Bibha :-D
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Re: Need some Help [#permalink]

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New post 15 Apr 2010, 11:12
Need not explain more. Very clear.....
Re: Need some Help   [#permalink] 15 Apr 2010, 11:12
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