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# If x and y are distinct positive integers. . .

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e-GMAT Representative
Joined: 04 Jan 2015
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If x and y are distinct positive integers. . . [#permalink]

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11 Nov 2016, 05:50
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Difficulty:

75% (hard)

Question Stats:

52% (01:47) correct 48% (01:42) wrong based on 130 sessions

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If $$x$$ and $$y$$ are distinct positive integers and $$x+y$$ is even, what is the remainder when $$(x+y)^a$$ is divided by $$10$$, where $$a$$ is a positive integer?

(1) Units digit of $$y$$ is $$6$$
(2) $$(xy)^a$$ is divisible by $$10$$.

Take a stab at this fresh question from e-GMAT. Post your analysis below.

Official Solution to be provided after receiving some good analyses.

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Joined: 31 May 2015
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If x and y are distinct positive integers. . . [#permalink]

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11 Nov 2016, 08:46
1
I will try:

So basically, divisibility by 10 will require us to check the last digit so need to know last digit of x+y and how a is divisible by the corresponding cycle number.
(1) is not enough because we know nothing about x-last digit-insufficient
(2) (x.y)^a divisible by 10 so we have either x and y multiple of 2 and 5 or 10 and whatever the number
Because x+y even so we get rid of 2 and 5 and go with 10 and whatever even number. However, we don't know the last digit of that whatever number-insufficient

(1)+(2) so y must have last digit 6 and x must have last digit 0 --> (x+y) last digit 6^a
we know that cycle of 6 doesn't matter because the last digit is always 6, no matter a is--> sufficient (C)
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Joined: 26 Jan 2016
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Re: If x and y are distinct positive integers. . . [#permalink]

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11 Nov 2016, 12:58
1. Units digit of y is 6

from here all we know is that y is even and that x is even too x+y=even (given in prompt). There are a lot of different options for this. Insuff

2. (xy)^a/10 so from here we know that xy needs to be a multiple of 10. There are so many ways to do that. Insuff.

1&2. We know that the units digit of y is 6 so then X will have to involve a 0 as x+y=even. lets try 6 and 10. 6+10=16

16²=256/10=25 r =6

now lets try 6 and 20. 6+20=26
26²=676/10=67 r=6

C
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If x and y are distinct positive integers. . . [#permalink]

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Updated on: 14 Dec 2016, 20:54
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2
Let's look at the detailed solution of the above problem

Steps 1 & 2: Understand Question and Draw Inferences

• x, y are distinct integers > 0 such that x + y = even
• Hence we can have two possibilities
o both x and y are even
OR
o both x and y are odd
• a is an integer > 0

To Find:
The value of r in $$(x+y)^a=10k+r$$, where k is the quotient obtained when $$(x+y)^a$$ is divided by 10 and r is the remainder; so, $$0 ≤ r < 10$$
o Now, when a number is divided by 10, the remainder is equal to the units digit of that number.
o So, r = units digit of $$(x+y)^a$$

Step 3: Analyze Statement 1 independently

Units digit of y is 6

• It does not tell us anything about the units digit of x as well as about the value of a.

So statement 1 is not sufficient to arrive at a unique answer.

Step 4: Analyze Statement 2 independently

$$(xy)^a$$ is divisible by 10.
• As $$(xy)^a$$ is divisible by 10, the units digit of $$(xy)^a$$ = 0
• So, the units digit of xy = 0. Two cases are possible:
o Units digit of (x, y) = { 5, even number) in any order. However in this case the number with 5 as its units digit will be odd and the other number will be even. However, we’ve deduced in Steps 1 and 2 that x and y have the same even-odd nature. So, this case is not possible as it contradicts the given information (that the sum x + y is even).
o Units digit of (x, y) = (0, even number) in any order. In this case x and y are both even. So, this case is possible.

However since we do not have a unique value of units digit of both x and y and we do not know the value of a, we cannot find a unique value of the units digit of $$(x+y)^a$$

Therefore, statement 2 is sufficient to arrive at a unique answer.

Step 5: Analyze Both Statements Together (if needed)

1. From Statement 1, we know that Units digit of y = 6
2. From Statement 2, we inferred that Units digit of (x, y) = (0, even number) in any order

Combining both the statements, we can say that units digit (x) = 0 and units digit(y) = 6
So, units digit of (x+y) = 6. Now do we need the value of a to find out the units digit of $$(x+y)^a$$?

We know a number with units digit of 6 raised to any power always results in units digit of 6.

So, Units Digit of $$6^a = 6.$$
Thus r = Units Digit of $$6^a = 6$$.

Hence the correct Answer is C

Thanks,
Saquib
Quant Expert
e-GMAT
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Originally posted by EgmatQuantExpert on 14 Dec 2016, 07:36.
Last edited by EgmatQuantExpert on 14 Dec 2016, 20:54, edited 2 times in total.
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Re: If x and y are distinct positive integers. . . [#permalink]

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14 Dec 2016, 07:41
hanminhee wrote:
I will try:

So basically, divisibility by 10 will require us to check the last digit so need to know last digit of x+y and how a is divisible by the corresponding cycle number.
(1) is not enough because we know nothing about x-last digit-insufficient
(2) (x.y)^a divisible by 10 so we have either x and y multiple of 2 and 5 or 10 and whatever the number
Because x+y even so we get rid of 2 and 5 and go with 10 and whatever even number. However, we don't know the last digit of that whatever number-insufficient

(1)+(2) so y must have last digit 6 and x must have last digit 0 --> (x+y) last digit 6^a
we know that cycle of 6 doesn't matter because the last digit is always 6, no matter a is--> sufficient (C)

Hey Hanminhee,

The analysis presented by you is absolutely correct!

You have solved the question in a very methodical way, and that is how we encourage students to solve any question at e-GMAT.

Thanks
Saquib
Quant Expert
e-GMAT
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Re: If x and y are distinct positive integers. . . [#permalink]

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14 Dec 2016, 19:17
EgmatQuantExpert wrote:
So, units digit of (x+y) = 6. Now do we need the value of a to find out the units digit of (x+y)?

A small typo in the above line: Now do we need the value of a to find out the units digit of $$(x+y)^a$$?

Thanks for the detailed explanation!
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If x and y are distinct positive integers. . . [#permalink]

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16 Feb 2017, 04:32
(1) Implies that X is even. Not Suff.
(2) Implies that XY should be a multiple of 10, and have at least 2 & 5 in its base. Different options are possible. Not Suff.
(1) + (2)
Since X is even, it cannot be 5, so XY to be a multiple of 10, X must be equal to 10 at least.
That gives us understanding that any sum of X and Y will yield in 6 as units digit, Suff.
If x and y are distinct positive integers. . .   [#permalink] 16 Feb 2017, 04:32
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# If x and y are distinct positive integers. . .

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