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What is the value of x if x is the remainder obtained when

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What is the value of x if x is the remainder obtained when [#permalink] New post 25 Mar 2011, 19:19
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A
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C
D
E

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(N/A)

Question Stats:

36% (03:01) correct 64% (01:32) wrong based on 14 sessions
What is the value of x if x is the remainder obtained when \(2^{8p+2} + Z\)is divided by 5 and p is a positive integer?

1) Z= 6
2) Z is even

Last edited by dreambeliever on 25 Mar 2011, 22:36, edited 1 time in total.
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Re: Divided by 5 [#permalink] New post 25 Mar 2011, 20:16
2(8p+2)+z
=16p+(z+4)

given:

16p + (z+4) = 5k + x

where k is a whole number

Statement 1:
Given: z = 6

If p = 1

16 + 10 = 26

26/5 leaves remainder 1

If p = 2
32 + 10 = 42
42/5 leaves remainder 2

Thus, insufficient!

Statement 2:
Z is even. So let z = 6
We already know that for z = 6, the statement is insufficient!
Therefore, statement 2 is also not sufficient.

Statement 1 & 2:
z = 6 obeys both the constraints imposed by Statement 1 & 2
But z = 6 does not yield a unique value of x

Therefore, insufficient!

ANS - 'E'
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Re: Divided by 5 [#permalink] New post 25 Mar 2011, 22:35
dreambeliever wrote:
What is the value of x if x is the remainder obtained when \(2^(8p+2)\)+ Z is divided by 5 and p is a positive integer?

1) Z= 6
2) Z is even


Sorry corrected question stem.
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Re: Divided by 5 [#permalink] New post 25 Mar 2011, 23:24
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dreambeliever wrote:
What is the value of x if x is the remainder obtained when \(2^{8p+2} + Z\)is divided by 5 and p is a positive integer?

1) Z= 6
2) Z is even


\(2^{8p+2}\)

The unit digit of the above expression will always be 4.

Here's how;
\(2^n\) has a cyclicity of 4. And the unit digits appear as 2,4,8,6 with every increment in n.

When "8p+2" is divided by 4, it always render a remainder of 2.
8p is divisible by 4 leaving the remainder as 0.
When 2 is divided by 4, it will leave a remainder of 2.

Whenever; in expression \(2^n\), n divides by 4 and leaves a remainder of 2, the unit digit of the expression will be 4.

1. Z=6

4+6=0

The expression \(2^{8p+2} + Z\) will always be divisible by 5 as the unit digit will always be 0. x=0;

Sufficient.

2. Z=even;

Z=2; The expression's units digit will be 6 and x=1;
Z=6; The expression's units digit will be 0 and x=0;

Not Sufficient.

Ans: "A"
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Re: Divided by 5 [#permalink] New post 26 Mar 2011, 06:36
(2)^(8p+2) + Z

= (4)^(4p+1) + Z

4 raised to odd power ends with 4

So (4 +Z)/5 = x, x = ?

1) Z = 6, then x = 0

2) Insufficient as (4+2)/5 = Rem 1

Answer - A
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Re: Divided by 5 [#permalink] New post 01 Apr 2011, 10:02
1. Z=6
8(p+2) = 2(4p+1)
hence 2 ^(8p+2) become 4^(4P+1)
any odd power to 4 gives 4 as the last digit
2 ^(8p+2) +Z = 4^(4P+1) +6 = <no with last digit as 0>
is divisible by 5. therefore remainder = 0

A or D

2. Z=even
<no with last digit as 4> +even will give different no.s for each value of Z
hence no fixed remainder

Therefore ans= A
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Re: Divided by 5 [#permalink] New post 01 Apr 2011, 19:49
The answer should be A. Whatz the OA ?

obidan wrote:
1. Z=6
8(p+2) = 2(4p+1)
hence 2 ^(8p+2) become 4^(4P+1)
any odd power to 4 gives 4 as the last digit
2 ^(8p+2) +Z = 4^(4P+1) +6 = <no with last digit as 0>
is divisible by 5. therefore remainder = 0

A or D

2. Z=even
<no with last digit as 4> +even will give different no.s for each value of Z
hence no fixed remainder

Therefore ans= A
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Re: Divided by 5 [#permalink] New post 17 Oct 2011, 13:20
Before the correction i thought its E..But yes IMO A
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Re: Divided by 5   [#permalink] 17 Oct 2011, 13:20
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