M24-11

Official Solution:


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

First, let's note that we are told \(x^2 - y^2\), which can be expressed as \((x-y)(x+y)\), is an integer.

The question asks whether the units digit of this integer zero. This can be translated as: Is \(x^2-y^2\) divisible by 10?

(1) \(x - y\) is an integer divisible by 30.

The above can be expressed as \(x-y=30m\) (where \(m\) is an integer). If \(x\) and \(y\) are integers, then \(x+y\) is also an 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 \(x^2-y^2=(x-y)(x+y)=30* 31.5=945\), which is an integer but not divisible by 10. Not sufficient.

(2) \(x + y\) is an integer divisible by 70.

The above can be expressed as \(x+y=70n\) (where \(n\) is an integer). If \(x\) and \(y\) are integers, then \(x-y\) is also an 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 \(x^2-y^2=(x-y)(x+y)=69.5*70=4865\), which is an integer but not divisible by 10. Not sufficient.

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


Answer: C