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I would like to know, if the Q would have been that x and y are integers, then each statement would be sufficient to ans the Q.

Can somebody please explain why?

The question doesn't need to say that x and y are integers...

I don't understand the OA. The question says "integer x^2-y^2", which means it is already given that x^2-y^2 is an integer.

I. x^2-y^2=integer=(x+y)(x-y) Thus, if x-y is an integer divisible by 30, then x^2-y^2=30n*(x+y), which always ends in a zero as long as x+y is an integer. And x+y has to be an integer in order for x^2-y^2 to be an integer.

I would like to know, if the Q would have been that x and y are integers, then each statement would be sufficient to ans the Q.

Can somebody please explain why?

Given: \(x^2-y^2=(x-y)(x+y)=integer\). Question is the units digit of this integer zero, which can be translated as is \(x^2-y^2\) divisible by 10.

(1) \(x-y=30m\) (where \(m\) is an integer) --> if \(x\) and \(y\) are integers, then \(x+y=integer\) and \(x^2-y^2=30m*(x+y)=30m*integer\), which is divisible by 10 BUT if \(x=30.75\) and \(y=0.75\), then \(x^2-y^2=(x-y)(x+y)=30*31.5=945=integer\), which is not divisible by 10. Not sufficient.

(2) \(x+y=70n\) (where \(n\) is an integer) --> if \(x\) and \(y\) are integers, then \(x-y=integer\) and \(x^2-y^2=(x-y)*70n=integer*70n\), which is divisible by 10 BUT if \(x=69.75\) and \(y=0.25\), then \(x^2-y^2=(x-y)(x+y)=69,5*70=4865=integer\), which is not divisible by 10. Not sufficient.

(1)+(2) \(x^2-y^2=30m*70n=integer\), Which is divisible by 10. Sufficient.

I got this question correct, but I would just like to understand the problem in more depth. The explanation in the test review was rather brief. Can someone please elaborate. Thanks!

Is the last digit of integer \(x^2 - y^2\) a zero?

Re: Is the last digit of integer x^2 - y^2 a zero? 1. x - y is [#permalink]

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22 Nov 2011, 15:18

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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.

Re: Is the last digit of integer x^2 - y^2 a zero? 1. x - y is [#permalink]

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04 Jan 2012, 08:27

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I am not sure if this is the best approach to handle such problems but the following is the way I dealt with this problem.

Question: Is the last digit of (x + y)(x - y) zero?

Statement 1: x - y is an integer divisible by 30 Thus, the least possible value for (x - y) is 30. However, this does not tell us anything about (x + y). If (x + y) = 0.1, then the last digit of the product (x + y)(x - y) is not 0. Thus INSUFFICIENT.

Statement 2: x + y is an integer divisible by 70 Thus, the least possible value for (x + y) is 70. However, this does not tell us anything about (x - y). Using a similar logic as above, if (x - y) = 0.1, then the last digit of their product (x + y)(x - y) is not 0. Thus INSUFFICIENT.

Combining the two statements, we have the least possible value of (x + y) as 70 and the least possible value of (x -y) as 30. Thus, the least possible value of their product will always be divisible by 10 and thus their product will always have zero as the last digit. SUFFICIENT.

Re: Is the last digit of integer x^2 - y^2 a zero? 1. x - y is [#permalink]

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04 Jan 2012, 14:21

Very nice problem. No info on x, y so they could be frac, int

Rephrase: (x+y)(x-y) = ..htu, is u = 0?

1. x-y = 30, so x+y could be (53.3+23.3) = 76.6. Therefore, (x+y)(x-y) will clearly not have u = 0. On the other hand, (53+23) = 76. This will have u = 0. Insufficient. 2. x+y = 70. Same deal as 1.

Together, both x+y and x-y are explicitly mentioned. Sufficient - C.
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I would like to know, if the Q would have been that x and y are integers, then each statement would be sufficient to ans the Q.

Can somebody please explain why?

Given: \(x^2-y^2=(x-y)(x+y)=integer\). Question is the units digit of this integer zero, which can be translated as is \(x^2-y^2\) divisible by 10.

(1) \(x-y=30m\) (where \(m\) is an integer) --> if \(x\) and \(y\) are integers, then \(x+y=integer\) and \(x^2-y^2=30m*(x+y)=30m*integer\), which is divisible by 10 BUT if \(x=30.75\) and \(y=0.75\), then \(x^2-y^2=(x-y)(x+y)=30*31.5=945=integer\), which is not divisible by 10. Not sufficient.

(2) \(x+y=70n\) (where \(n\) is an integer) --> if \(x\) and \(y\) are integers, then \(x-y=integer\) and \(x^2-y^2=(x-y)*70n=integer*70n\), which is divisible by 10 BUT if \(x=69.75\) and \(y=0.25\), then \(x^2-y^2=(x-y)(x+y)=69,5*70=4865=integer\), which is not divisible by 10. Not sufficient.

(1)+(2) \(x^2-y^2=30m*70n=integer\), Which is divisible by 10. Sufficient.

Answer: C.

Hope it's clear.

Quote:

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.

Hi Bunuel,

I also have the same doubt. Please suggest how is this approach incorrect.

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.

So your logic is:

x-y is an integer divisible by 30 a = 30, b = x-y, c = (x-y)(x+y) a is a factor of b, b is a factor of c, therefore a is a factor of c

But you have made an assumption here that (x+y) is an integer (i.e. that c / a = (x+y) = integer), which is not necessarily true.

Re: Is the last digit of integer x^2 - y^2 a zero? [#permalink]

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21 Sep 2013, 14:13

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Re: Is the last digit of integer x^2 - y^2 a zero? [#permalink]

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19 Oct 2013, 02:03

gmatcracker2010 wrote:

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

(1) x - y is an integer divisible by 30 (2) x + y is an integer divisible by 70

This is how I approached the question:

Statement 1: \(x - y =30\) \(x = y + 30\)

Now, If any we take square any value of \(X\) and\(Y + 30\)the last digit of their difference will be O.

Let Y = -2 => then X = 28 and \((28)^2 - (-2)^2 = 784 - 4 = 780\) Let Y = -1 => then X = 29 and \((29)^2 - (1)^2 = 841 - 1 = 840\) Let Y = 0 => then X = 30 and \((30)^2 - (0)^2 = 900 - 0 = 900 0\) Let Y = 1 => then X = 31 and \((31)^2 - (1)^2 = 961 - 1 = 960\) Let Y = 2 => then X = 32 and \((32)^2 - (2)^2 = 1024 - 4 = 1020\) Let Y = 3 => then X = 33 and \((33)^2 - (3)^2 = 1089 - 9 = 1080\) Let Y = 4 => then X = 34 and \((34)^2 - (4)^2 = 1156 - 16 = 1140\) .... Let Y = 25 => then X = 55 and \((55)^2 - (25)^2 = 3025 - 625 = 2400\) .... Let Y = 107 => then X = 137 and \((137)^2 - (107)^2 = 18769 - 11449 = 7320\)

Therefore, Sufficient!

Statement 2 \(X + Y = 70\) \(Y = 70 - X\) Let X = -2 => then Y = 72 and \((-2)^2 - (72)^2 = 4 - 5184 = - 5180\) Let X = -0 => then Y = 70 and \((0)^2 - (70)^2 = 0 - 4900 = - 4900\) Let X = -1 => then Y = 69 and \((-2)^2 - (72)^2 = 4 - 5184 = - 5180\)

Sufficient

Therefore, option D

Now can anyone kindly point out the problem in this execution?

Re: Is the last digit of integer x^2 - y^2 a zero? [#permalink]

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08 Dec 2013, 23:44

gmatcracker2010 wrote:

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

(1) x - y is an integer divisible by 30 (2) x + y is an integer divisible by 70

I would like to know, if the Q would have been that x and y are integers, then each statement would be sufficient to ans the Q.

Can somebody please explain why?

-- 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?

(1) x - y is an integer divisible by 30 (2) x + y is an integer divisible by 70

I would like to know, if the Q would have been that x and y are integers, then each statement would be sufficient to ans the Q.

Can somebody please explain why?

-- 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.
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Re: Is the last digit of integer x^2 - y^2 a zero? [#permalink]

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10 Dec 2013, 11:16

since the first statement says that x-y is an integer divisible by 30 it means x-y can be any multiple of 30 so (x+y)*(x-y)=end with 0 we can pluggin some numbers and see lets assume (x+y) as any integer for e.g= 2 so 2*30 = ends with 0. now lets take a non integer 1/5*30 = 6 so doesnot end with 0 since it is a yes for one example and no for other the first statement is insufficient.

second statement says that (x+y) is divisible by 70 so (x+y)*(x-y)=end with 0 we can pluggin some numbers and see lets assume (x-y) as any integer for e.g= 2 so 70*2 = ends with 0. now lets take a non integer 140*1/5 = 28 so doesnot end with 0 since it is a yes for one example and no for other the second statement is insufficient.

both statements individually do not satisfy the condition. when we combine them it is sufficient hence the answer is C

Re: Is the last digit of integer x^2 - y^2 a zero? [#permalink]

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30 Nov 2014, 15:30

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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.

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