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Is the last digit of integer x^2  y^2 a zero?
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13 Jun 2010, 09:06
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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 (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?
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Re: Gmat club test 24
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13 Jun 2010, 11:22
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 Oa is 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: \(x^2y^2=(xy)(x+y)=integer\). Question is the units digit of this integer zero, which can be translated as is \(x^2y^2\) divisible by 10. (1) \(xy=30m\) (where \(m\) is an integer) > if \(x\) and \(y\) are integers, then \(x+y=integer\) and \(x^2y^2=30m*(x+y)=30m*integer\), which is divisible by 10 BUT if \(x=30.75\) and \(y=0.75\), then \(x^2y^2=(xy)(x+y)=30*31.5=945=integer\), which is not divisible by 10. Not sufficient. (2) \(x+y=70n\) (where \(n\) is an integer) > if \(x\) and \(y\) are integers, then \(xy=integer\) and \(x^2y^2=(xy)*70n=integer*70n\), which is divisible by 10 BUT if \(x=69.75\) and \(y=0.25\), then \(x^2y^2=(xy)(x+y)=69,5*70=4865=integer\), which is not divisible by 10. Not sufficient. (1)+(2) \(x^2y^2=30m*70n=integer\), Which is divisible by 10. Sufficient. Answer: C. Hope it's clear.
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Re: Gmat club test 24
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13 Jun 2010, 10:52
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 Oa is 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^2y^2", which means it is already given that x^2y^2 is an integer. I. x^2y^2=integer=(x+y)(xy) Thus, if xy is an integer divisible by 30, then x^2y^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^2y^2 to be an integer. Same reasoning for II  where am I wrong?



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Re: Is the last digit of integer x^2  y^2 a zero? 1. x  y is
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22 Nov 2011, 14:18
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 (xy), and is divisible by (x+y) when the powers of x and y are even.
If I have not misstated, then would you mind pointing out the flaw in the following reasoning? (1) "xy 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 (xy), and (xy) 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.



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Re: Is the last digit of integer x^2  y^2 a zero? 1. x  y is
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04 Jan 2012, 07:27
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. Answer: C
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Re: Is the last digit of integer x^2  y^2 a zero? 1. x  y is
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04 Jan 2012, 13:21
Very nice problem. No info on x, y so they could be frac, int Rephrase: (x+y)(xy) = ..htu, is u = 0? 1. xy = 30, so x+y could be (53.3+23.3) = 76.6. Therefore, (x+y)(xy) 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 xy are explicitly mentioned. Sufficient  C.
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Re: Gmat club test 24
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28 May 2012, 00:23
Bunuel wrote: 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 Oa is 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: \(x^2y^2=(xy)(x+y)=integer\). Question is the units digit of this integer zero, which can be translated as is \(x^2y^2\) divisible by 10. (1) \(xy=30m\) (where \(m\) is an integer) > if \(x\) and \(y\) are integers, then \(x+y=integer\) and \(x^2y^2=30m*(x+y)=30m*integer\), which is divisible by 10 BUT if \(x=30.75\) and \(y=0.75\), then \(x^2y^2=(xy)(x+y)=30*31.5=945=integer\), which is not divisible by 10. Not sufficient. (2) \(x+y=70n\) (where \(n\) is an integer) > if \(x\) and \(y\) are integers, then \(xy=integer\) and \(x^2y^2=(xy)*70n=integer*70n\), which is divisible by 10 BUT if \(x=69.75\) and \(y=0.25\), then \(x^2y^2=(xy)(x+y)=69,5*70=4865=integer\), which is not divisible by 10. Not sufficient. (1)+(2) \(x^2y^2=30m*70n=integer\), 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 (xy), and is divisible by (x+y) when the powers of x and y are even.
If I have not misstated, then would you mind pointing out the flaw in the following reasoning? (1) "xy 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 (xy), and (xy) 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. Thanks



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Re: Gmat club test 24
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28 May 2012, 02:46
kunalbh19 wrote: If I have not misstated, then would you mind pointing out the flaw in the following reasoning? (1) "xy 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 (xy), and (xy) 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: xy is an integer divisible by 30 a = 30, b = xy, c = (xy)(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.



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Re: Is the last digit of integer x^2  y^2 a zero? 1. x  y is
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28 May 2012, 04:00
Ohhh.. ok.. i think i got it now.. thanks



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Re: Is the last digit of integer x^2  y^2 a zero? 1. x  y is
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31 May 2012, 20:00
A tricky question.... Overlooked x+y or xy could be in Decimals.... Must not skip this
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Re: Is the last digit of integer x^2  y^2 a zero?
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01 Jun 2012, 01:01
Amazing question... Thanks Posted from my mobile device
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Re: Is the last digit of integer x^2  y^2 a zero?
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19 Oct 2013, 01:03
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 \((28)^2  (2)^2 = 784  4 = 78 0\) Let Y = 1 => then X = 29 and \((29)^2  (1)^2 = 841  1 = 84 0\) Let Y = 0 => then X = 30 and \((30)^2  (0)^2 = 900  0 = 900 0\) Let Y = 1 => then X = 31 and \((31)^2  (1)^2 = 961  1 = 96 0\) Let Y = 2 => then X = 32 and \((32)^2  (2)^2 = 1024  4 = 102 0\) Let Y = 3 => then X = 33 and \((33)^2  (3)^2 = 1089  9 = 108 0\) Let Y = 4 => then X = 34 and \((34)^2  (4)^2 = 1156  16 = 114 0\) .... Let Y = 25 => then X = 55 and \((55)^2  (25)^2 = 3025  625 = 240 0\) .... Let Y = 107 => then X = 137 and \((137)^2  (107)^2 = 18769  11449 = 732 0\) Therefore, Sufficient!Statement 2 \(X + Y = 70\) \(Y = 70  X\) Let X = 2 => then Y = 72 and \((2)^2  (72)^2 = 4  5184 =  518 0\) Let X = 0 => then Y = 70 and \((0)^2  (70)^2 = 0  4900 =  490 0\) Let X = 1 => then Y = 69 and \((2)^2  (72)^2 = 4  5184 =  518 0\) SufficientTherefore, option D Now can anyone kindly point out the problem in this execution?



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Re: Is the last digit of integer x^2  y^2 a zero?
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19 Oct 2013, 02:19
suk1234 wrote: Statement 1: \(x  y =30\) \(x = y + 30\)
Now can anyone kindly point out the problem in this execution?
Firstly, if it is said that (xy) is divisible by 30, it doesn't mean xy = 30. It just means (xy) is a multiple of 30. Again, you have taken only integral values for x,y. The fact that x,y can be nonintegral values also makes your solution incorrect. Please refer to the post above by Bunuel to get the correct solution. Hope this helps.
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Re: Is the last digit of integer x^2  y^2 a zero?
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08 Dec 2013, 22:44
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^ny^n: is ALWAYS divisible by (xy) . is divisible by (x+y) when n is even. " Now, in this case... 1) x^2  y^2 is divisible by (xy), 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?



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Re: Is the last digit of integer x^2  y^2 a zero?
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08 Dec 2013, 23:24
gmarchanda wrote: 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^ny^n: is ALWAYS divisible by (xy) . is divisible by (x+y) when n is even. " Now, in this case... 1) x^2  y^2 is divisible by (xy), 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?
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10 Dec 2013, 10:16
since the first statement says that xy is an integer divisible by 30 it means xy can be any multiple of 30 so (x+y)*(xy)=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)*(xy)=end with 0 we can pluggin some numbers and see lets assume (xy) 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



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Re: Is the last digit of integer x^2  y^2 a zero?
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30 Nov 2014, 14:30
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 noninteger, this would lead to (x^2  y^2) possibly having a nonzero 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 noninteger values of x and y that lead to nonzero 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
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Re: Is the last digit of integer x^2  y^2 a zero?
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24 Feb 2018, 23:56
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? Question: Is the last digit of integer x^2  y^2 a zero?Usual line of thinking is x^2  y^2 = (x+y)(xy) So if any one of (x+y) and (xy) 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 nonintegersStatement 1: x  y is an integer divisible by 30No mention of whether x and y are integers or not hence NOT SUFFICIENT Statement 2: x + y is an integer divisible by 70No mention of whether x and y are integers or not hence NOT SUFFICIENT Combining the two statementsx^2  y^2 = (x+y)(xy) and (x+y) and (xy) are both multiples of 10 hence result has to be a multiple of 100 SUFFICIENT Answer: option C
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