Mahmoudfawzy83 wrote:
If \(z\) is a positive two-digit even number, what is the ones digit of \(z\)?
1) both digits of \(z\) are the same
2) The ones digit of \(z\) is the same as the ones digit of \(z^2\)
To find: One's digit of the two-digit even number.
As it is a two digit even number, the one's digit has to either 0,2,4,6,8.
Statement 1) The number can be 22,44,66,88. Clearly, Insufficient.
Statement 2) The square of the number will have one's digit of the square of one's digit of the number. e.g. Square of 42 will have one's digit of square of 2 i.e. 4 because any two digit number can be expressed as 10x+y, Squaring (10x+y) equals 100x^2+y^2+20xy, as 100x^2 and 20xy will only impact tens and hundreds digit. One's digit will be y^2. So, One's digit of a square of a two digit number would be the square of one's digit. Following that logic, only 6 and 0 follow. The could have 6 or 0 as its one's digit. Insufficient.
(1)+(2), The one's digit has to be 6, since from (2), 00 can't be a two digit number. Hence, one's digit would be 6. Sufficient.
IMO, Option C.
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35% (00:58) correct 
