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Is the units 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
(1) \(x - y\) is an integer divisible by 30
(2) \(x + y\) is an integer divisible by 70
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13 Jun 2010, 12:22
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Is the units digit of integer \(x^2 - y^2\) a zero?
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) \(x - y\) is an integer divisible by 30. So, \(x-y=30m\) (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) \(x + y\) is an integer divisible by 70. So, \(x+y=70n\) (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) \(x^2-y^2=(x-y)(x+y)=30m*70n=integer\), which is divisible by 10. Sufficient.
Answer: C
Hope it's clear.
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) \(x - y\) is an integer divisible by 30. So, \(x-y=30m\) (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) \(x + y\) is an integer divisible by 70. So, \(x+y=70n\) (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) \(x^2-y^2=(x-y)(x+y)=30m*70n=integer\), which is divisible by 10. Sufficient.
Answer: C
Hope it's clear.
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.
Answer: C
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
, or have got it wrong. Could someone please explain? 
