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If x and y are distinct positive integers. . . [#permalink]
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11 Nov 2016, 05:50
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 \(10\).
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If x and y are distinct positive integers. . . [#permalink]
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11 Nov 2016, 08:46
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 xlast digitinsufficient (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 numberinsufficient
(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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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
Let's look at the detailed solution of the above problemSteps 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 independentlyUnits 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 evenodd 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 \(6^a = 6\).Sufficient to answer. Hence the correct Answer is CThanks, Saquib Quant Expert eGMAT
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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 xlast digitinsufficient (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 numberinsufficient
(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 eGMAT. Thanks Saquib Quant Expert eGMAT
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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. Answer C.




If x and y are distinct positive integers. . .
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16 Feb 2017, 04:32






