agnee
x ,y and z are positive integers such that x+y=2z and gcd(x,y)=2. How many different values of z are possible for 1<=z<=100
Please solve this
Prima facie, it looks like answer could be 99. What’s the OA?
99 seems fine. x+y is Even, so both x & y must either be even or odd...but gcd(x.y) is 2, which means x and y are both even. So we can start with 2+2 = 4 = 2z : z = 2....then 2+4 = 6 = 2z : z=3...and this can go on till 100. So 99 possible values