Bunuel wrote:
If a, b, c, and z are nonnegative integers, what is the remainder when 3^(az)×3^(bz)×3^(cz) is divided by 4?
(1) z is even.
(2) The sum of a, b, and c is odd.
Hi,
3^(az)×3^(bz)×3^(cz) =\(3^{az+bz+cz} = 3^{(a+b+c)z}\)..
Two methods --
I. check for pattern -3^1 leaves a remainder 3
and 3^2 leaves a remainder 1
3^3 leaves a remainder 3
and 3^4 leaves a remainder 1..
so ODD power leaves remainder 3 and EVEN power leaves a remainder 1..
SO if we know whether POWER is ODD or EVEN, we can answer..POWER is product of a+b+c and z
lets see the statements--(1) z is even.Hence POWER is EVEN irrespective of what a+b+c is..
remainder is 1
Suff
(2) The sum of a, b, and c is odd.we do not know about z..
If z is ODD.. power is ODD and ans is 3..
But if z is Even .. power is EVEN and ans is 1
different answers possible
Insuff
ans A
Algebrically through expansion..
(1) z is even.so \(3^{(a+b+c)z} = 3^{(a+b+c)*2*y}\) as z= 2y where y is an integer, nonnegative
\((3^2)^{(a+b+c)*y} = 9^{(a+b+c)*y} = (8+1)^{(a+b+c)*y}\)..
so all terms will be div by 8 and therefore by 4 except \(1^{((a+b+c)*y}\)
so remainder = 1..
Suff
(2) The sum of a, b, and c is odd.
we cannot work further on it..
In GMAT, isnt 0 considered as an Even number?
Insuff