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If x, y are positive integers, what is the unit digit of 2^(4x+2)+y?

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If x, y are positive integers, what is the unit digit of 2^(4x+2)+y?  [#permalink]

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New post 06 Jul 2016, 16:01
2
4
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

66% (01:28) correct 34% (01:19) wrong based on 185 sessions

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If x, y are positive integers, what is the unit digit of 2^(4x+2)+y?
1) x=1
2) y=2

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Re: If x, y are positive integers, what is the unit digit of 2^(4x+2)+y?  [#permalink]

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New post 06 Jul 2016, 22:23
B.
If y is 2 then equation becomes :
2^ (4*(x+1)) which will end in 6 for all x which is a positive integer.

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Re: If x, y are positive integers, what is the unit digit of 2^(4x+2)+y?  [#permalink]

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New post 06 Jul 2016, 22:35
1
the question can be rewritten as

4^(2x+1) + y

4^ odd= gives unit digit 4

therefore the unit digits entirely depnds on y

statement 1
x=1

insufficient

the value of x does not matter

statement 2

y= 2

therefore unit digit = 4+2=6

sufficient

correct answer B


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If x, y are positive integers, what is the unit digit of 2^(4x+2)+y?  [#permalink]

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New post 10 Jul 2016, 23:41
In the original condition, there are 2 variables (x and y). Hence, there is a high chance that C is the correct answer. Using 1) & 2), C is the correct answer. However, since this is an integer question, we can apply the common mistake type 4(A). From con 2), the unit digit of 2^4x+2 is always 4. Hence, we only have to know y, and the condition is sufficient. The correct answer, thus, is B.

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Re: If x, y are positive integers, what is the unit digit of 2^(4x+2)+y?  [#permalink]

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New post 24 Nov 2016, 04:01
Here we will use the principle of cyclicity =>
Cyclicity of 2 is four
2=> 4m+1
4=>4m+2
8=>4m+3
6=>4m

Hence the units digit of 2^4x+2 is always 4
so we need y

Hence statement 2 is sufficient itself and statement 1 is not
Hence B
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Re: If x, y are positive integers, what is the unit digit of 2^(4x+2)+y?  [#permalink]

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New post 22 Jan 2018, 10:46
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Re: If x, y are positive integers, what is the unit digit of 2^(4x+2)+y? &nbs [#permalink] 22 Jan 2018, 10:46
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