Bunuel wrote:
12 Days of Christmas GMAT Competition with Lots of FunIf x, y, and z are positive integers, what is the units’ digit of \(34^{(2x)}*19^{(y+3)} + 17^{(2z)} + 15^{(x-y)}\) ?
(1) \(y^2*z – z\) is an odd integer
(2) \(y*z = x\)
Solution:
Analysing the question statement: 34^{(2x)}*19^{(y+3)} + 17^{(2z)} + 15^{(x-y)}
Here, 4,9,7 and 5 raised to their respective powers will decide the units digit of the entire expression.
Cyclicity of 4 is 2. Here, 4 raised to even power will always give the units digit as 6.
Cyclicity of 9 is 2. If y+3 is odd (i.e. y is even) then units digit is 9 otherwise if y+3 is even (i.e. y is odd) the. units digit is 1.
Cyclicity of 7 is 4. But 7 is raised to an even power so the units digit will either be 9 (if z is not a multiple of 4) or 1 (if z is a multiple of 4).
Cyclicity of 5 is 1. But it is raised to power (x-y). If x=y then 5^0 will give units digit as 1 and if x is not equal to y then units digit is 5.
So, things we need to know to solve the question is:
A) Whether y is even or odd.
B) Whether z is a multiple of 4 or not.
C) Whether x=y or not.
Statement 1: y^2*z – z is an odd integer.
On solving,
z(y+1)(y-1) = odd integer.
This implies that z, y+1, y-1 are odd.
So, we get z as odd and y as even. We got answer to A and B but not C.
So statement 1 is insufficient.
Statement 2: y*z = x.
Case 1: If z=1 then x=y.
Case 2: If z is not equal to 1 then x is not equal to y.
We do not get a clear answer for A, B or C.
So statement 2 is insufficient.
Combining statement 1 and 2, we get: z as odd and y as even. We have answers for A and B but not for C as again z could be or could be not equal to 1 which will determine C i.e. x=y or not. Hence, clearly insufficient.
ANSWER E.Posted from my mobile device