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11 Nov 2016, 05:50
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C
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Difficulty:
65%
(hard)
Question Stats:
60% (02:35) correct
40%
(02:28)
wrong
based on 312
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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?
Take a stab at this fresh question from e-GMAT. Post your analysis below.
Official Solution to be provided after receiving some good analyses.
- (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.
11 Nov 2016, 08:46
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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)
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)
11 Nov 2016, 12:58
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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
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
