If x and y are distinct positive integers and x + y is even, what is t

Question

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.

IanStewart · Accepted Answer

Statement 1 is not sufficient, since if x = 4, then x+y will end in 0, and (x+y)^a will be divisible by 10, but if x = 2, say, x+y will not be divisible by 10, and we'll get a nonzero remainder when we divide (x+y)^a by 10.

Statement 2 tells us xy is divisible by 10, so perhaps x = 10 and y = 20, in which case x+y is divisible by 10, or perhaps x = 10 and y = 2, in which case x+y is not divisible by 10, and we'll get a nonzero remainder when we divide (x+y)^a by 10.

Using both Statements, if the units digit of y is 6, then y cannot be divisible by 5. For Statement 2 to be true, so for (xy)^a to be divisible by 10, we need a "5" somewhere in (xy)^a, and if y is not divisible by 5, then x must be. But if x is divisible by 5 and x is even, then the units digit of x must be 0. So the units digit of x+y must be 0 + 6 = 6. So when we calculate (x+y)^a, we'll be raising a number with a units digit of 6 to the positive integer power a, and that will always give us a number ending in 6. Since the question asks for the remainder when we divide (x+y)^a by 10, which is just the units digit of (x+y)^a, the two Statements together are sufficient and the answer is C.

CrackverbalGMAT · Answer

This is a question based on the divisibility rule for 10, which depends on the units digit of the dividend. Since there is a power involved in the expression, you also need to know the unit digit cycles to be able to solve this question.

It is given that x+y is even. Therefore,  (x+y)^a  will also be even, regardless of the value of a. This also means that the unit digit of  (x+y)^a  can be either 2 or 4 or 6 or 8 or 0, depending on the unit digit of (x+y) and the value of a.
Therefore, any data that gives us concrete information about all these 3 variables will be sufficient data.

From statement I, we get the value of y to be 6, but we neither have the value of x nor a. So, how does this affect the answer?

If y = 16 and x = 14, (x+y) = 30; the power 'a' does not matter in this case since 30 raised to the power of any number will give us the unit digit as 0. As such, the remainder when  (x+y)^a  is divided by 10, will be 0.

But, if y = 16 and x = 12, (x+y) = 28. In this case, the unit digit of  (x+y)^a  depends on the value of a, which is unknown.

Hence, statement I is insufficient. Answer options A and D can be eliminated. Possible answer options are B, C or E.

From statement II alone,  (xy)^a  is divisible by 10. This means that the product xy is a number ending with 0. However, we do not know what the numbers x and y themselves are.

x = 6 and y = 10 is a possible set of values. For this set,  (x+y)^a  will leave a remainder when divided by 10.

x = 10 and y = 20 is another possible set. For this set,  (x+y)^a  will not leave any remainder when divided by 10.

Statement II is insufficient. Answer option B can be eliminated. Possible answer options are C or E.

From statements I and II together, we know that unit digit of y is 6 and also product of x and y ends with a 0. Therefore, unit digit of x should be 0, which in turn means that the unit digit of (x+y) will be 6.

When the units digit of a number is 6, it will give the units digit as 6 whatever power we may raise it to. Therefore, units digit of  (x+y)^a  = 6. When this number is divided by 10, the remainder will be 6, since, for divisibility by 10, we check the units digit of the number.

A point to note here is that, when analyzing the second statement alone, it’s very easy to fall into the trap of assuming that x and y are single digit numbers. That’s not mentioned anywhere in the question, so assuming that would be false.

Hope this helps!

vaibhav1221 · Answer

Okay, this is my take on the question.

Statement 1: Units digit of y is 6

A possible list of units digit of X and Y are below, given that x+y is even.

TABLE 1  
 X | Y  |  Units digit of (X + Y) 
0 | 6.....................6
2 | 6.....................8
4 | 6.....................0
6 | 6.....................2
8 | 6.....................4

Units digit of 6,8,2, and 4 follow these patterns (for units digit)

6 =>  6^1  =  6 ;  6^2  = 3 6   (Pattern 6,6,6,6,6,6,6,6...) 
8 =>  8^1  =  8 ;  8^2  = 6 4 ;  8^3  = 51 2 ;  8^4  = 409 6 ;  8^5  = 3276 8   (Pattern 8,4,2,6,8,4,2,6...) 
2 =>  2^1  =  2 ;  2^2  =  4 ;  2^3  =  2 ;  2^4  = 1 6 ;  2^5  = 3 2   (Pattern 2,4,8,6,2,4,8,6...) 
4 =>  4^1  =  4 ;  4^2  = 1 6 ;  4^3  = 6 4 ;  4^4  = 25 6   (Pattern 4,6,4,6,4,6,4,6...) 
For 0, the units digit will always be 0.

For  (X+Y)^a  to be perfectly divisible by 10, X should have 4 in its units digit (Refer  TABLE 1 ). For every other number in X's units digit will yield a different remainder. For example, the remainder when units digit of (x+y) is 6 will be different from the remainder when the units digit of (x+y) is 8, or 0, or 2, or 4.

Hence, statement 1 alone is insufficient to answer the question.

Statement 2:  (xy)^a  is divisible by 10.

Clearly, insufficient to answer the question alone.

Statement 1 & 2 together

Now we know that units digit of xy is 0, and units digit of y is 6.
For xy to be divisible by 10, units digit of x should be 0, given units digit of y = 6. Only then would the units digit of xy be 0. (units digit of X*units digit of y = 0*6 = 0)

Now that we know the units digit of X and Y, which are 0 and 6 respectively, the units digit of (X+Y) would be 6. And as pointed out earlier in the table, the units digit of 6^x where X is a positive integer always has 6 in the units place. Hence the units digit of (X+Y)^a, given and positive integer value of a, except zero, will always have 6 in the units place.
Therefore when (X+Y)^a is divided by 10, the remainder will always be 6.

I really hope this helps and I tried my best to make it as simple as possible!

Ahmed9955 · Answer

Hello  CrackVerbal ,
As per the highlighted statement- 
1.y=6
2. Product of xy should end with Zero
=> (x+y) should have unit digit of 6

My doubt is why x cannot be 5?
•if x= 5 and y=6 both the conditions are satisfied and remainder =1
 Also x= 10 and y=6 both the conditions are satisfied remainder = 6
 hence E.

Can you please clear my doubt.

IanStewart · Answer

The question also tells you that x+y is even, and if x = 5 and y = 6, then x + y is odd, so those cannot be the values of x and y here.

CEdward · Answer

So the first thing I noticed is that x + y = even which implies that x and y are both even or they are both odd.

Statement 1: Unit digit of y is 6 
This statement implies that y is even and since x + y is even, it also implies that x is even. To figure out the remainder when (x + y)^a is divided by 10, we must figure out what the units digit of (x + y)^a is. The remainder will be the units digit of that integer.

Plug and play with some numbers.
e.g. x = 1, y = 6 <-- (1 + 6)^2 = 49; 49/10 = 4 R9
e.g. x = 2, y = 6 <-- (2 + 6)^2 = 64; 64/10 = 6 R4

Therefore, statement 1 is insufficient.

Statement 2: (xy)^a/10 is an integer <-- implies that (x^a) x (y^a) = 10b (i.e. a multiple of 10).
Insufficient b/c as long as x or y is a multiple of 10 then (xy)^a/10 will equal some integer with no remainder.

Combined: Sufficient since we know that the units digit of y is 6. It doesn't matter what x is b/c that has to be a multiple of 10 to satisfy statement 2.

anshuman234 · Answer

what if instead of "0", "5" is at the unit's place of integer x?

IanStewart · Answer

I'm not sure if you're asking whether the solutions above ought to have considered that possibility, or if you're asking what the answer to the question would become if the question were different, and if we learned that x had a units digit of 5. It's impossible, in this question, for the units digit of x to be 5, because that would make x odd, and from Statement 1, we learn that y is even (since it has an even units digit). So if the units digit of x were 5, x+y would be odd, but the question tells us x+y is even.

If instead we learned, say, in the question stem that the units digit of x was 5, and the question did not tell us "x + y is even", then Statement 1 alone would be sufficient, since then the units digit of x+y would be 1, and (x+y)^a would have a units digit of 1 no matter what a was equal to. Statement 2 would not be sufficient, because y could have any even units digit, and we could get different answers to the question.

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If x and y are distinct positive integers and x + y is even, what is t

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If x and y are distinct positive integers and x + y is even, what is t

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