Is the units digit of integer x^2 - y^2 a zero?

Hello Bunuel,

Thanks very much for your explanation to this question, which by the way I came across while working through your excellent "700+ GMAT Data Sufficiency Questions With Explanations" document (wow -- they're tough!)

I had a quick question on this particular question. Firstly, I believe it is the case that x^n - y^n is always divisible by (x-y), and is divisible by (x+y) when the powers of x and y are even.

If I have not mis-stated, then would you mind pointing out the flaw in the following reasoning?
(1) "x-y is an integer divisible by 30" >> adding in the rule that "if a is a factor of b and b is a factor of c, then a is a factor of c", then: x^n - y^n is an integer divisible by (x-y), and (x-y) is an integer divisible by 30, then does it not follow that x^n - y^n is also an integer divisible by 30? and this is sufficient to say that x^n - y^n must end in zero.
(2) similar logic applies since the power of x and y are indeed even and therefore divisible by (x+y).

Would be very grateful if you could take 2 mins to point out the error in that alternative method.

Thanks very much for your contributions and for taking the time to help.

-- Well.. i see there is a rule which says...
"RULE: for x^n-y^n:
is ALWAYS divisible by (x-y) .
is divisible by (x+y) when n is even. "

Now, in this case...
1) x^2 - y^2 is divisible by (x-y), which in turn is divisible by 30. so if the property holds true, x^2 - y^2 should be divisible by 30, or in 10*3. Which would imply it would end with a 0.
2) Again, going by the above rule, x^2 - y^2 should be divisible by 70. which would imply it is divisible by 10. and thus ends with 0.

I am probably misinterpreting the property , or have got it wrong. Could someone please explain?

The property holds true if x and y are integers but we are not given that.

IS the last digit of INTEGER (x^2 - y^2) a zero?

1. x- y is an integer divisible by 30. Lets call it 30n where n is a positive integer.

Thus x^2 - y^2 = (x + y)*(x - y) = (x + y) * 30n

Now if (x + y) is an integer, it will lead to (x^2 - y^2) having units digit of zero.

eg. x = 60, y = 30; x - y = 30; x + y = 90 => x^2 - y^2 = 2700

However if (x+y) is non-integer, this would lead to (x^2 - y^2) possibly having a non-zero units digit.

eg. x = 60.3, y = 30.3 => x - y = 30 but x + y = 90.3 => x^2 - y^2 = 2709

INSUFFICIENT

2. x + y is an integer divisible by 70. Let’s call it 70m.

This is exactly analogous to statement 1, and we will still be able to choose non-integer values of x and y that lead to non-zero units digits in x^2 - y^2

INSUFFICIENT

Both 1 and 2:

x - y = 30n and x + y = 70m

Thus x^2 - y^2 = 2100mn

m and n are positive integers, so we can be sure that x^2 - y^2 has a units digit of zero.

SUFFICIENT

Correct C.

Class GMAT MATH Claudio Hurtado, GMAT CHILE

Question: Is the last digit of integer x^2 - y^2 a zero?

Usual line of thinking is
x^2 - y^2 = (x+y)(x-y)
So if any one of (x+y) and (x-y) is a Multiple of 10 then the the entire Product will be a multiple of 10
BUT THAT WILL BE WRONG because
(x+y) may be a multiple of 10 even if x and y are non-integers

Statement 1: x - y is an integer divisible by 30

No mention of whether x and y are integers or not hence

NOT SUFFICIENT

Statement 2: x + y is an integer divisible by 70

No mention of whether x and y are integers or not hence

NOT SUFFICIENT

Combining the two statements
x^2 - y^2 = (x+y)(x-y)
and (x+y) and (x-y) are both multiples of 10 hence result has to be a multiple of 100

SUFFICIENT

Answer: option C

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