really tricky problem, initially i solved it wrong
the most important thing is to notice that we are not given that z is an integer
(1) 5z=even is possible if z any even integer, or z=2/5. then 5*(2/5)=2-even but z is not even integer not suff
(2) the same is true for st 2
3z is even if z is even or z is fraction with even nominator and denominator=3, say z=2/3
then 3*(2/3) =2-even but z=2/3, not suff
both
5z=2k. and 3z=2m
z=(2/5)*k, z=(2/3)*m
(2/5)*k=(2/3)*m cancel 2 amd left with
3k=5m. that is possible if k=m=0 then z=0-even
or k=5. m=3 or k=5a,m=3a where a is +ve integer
let us insert k=5a into the 5z=2k
5z=2*5a. cancel 5 and left with z=2a where a is +ve integer, regardless of the value of a z will always be even integer
the same can be proved with
with 3z=2m, and inserting m=3a
P.S pretty tricky
thanks tonebeeze for posting!!