The greatest common divisor of two integers is (x+8) and their least common multiple is x(x+8), where x is a positive integer. If one of the integers is \(40\), what is the smallest possible value of the other one?
i=0;a=1; b=1;d= (a + 8) * a*(a + 8) ; if(d==40*b and gcd(40,b)==(a+8) and lcm(40,b)==a*(a+8), goto5, goto7);i++;printi,"=",(a, b); a++;if(a<2000, goto3, 0);a=1;b++;if(b<2000, goto3, 0)
The smallest possible value of the 2nd integer ==1,280
GCD[40, 1280] ==40 ==[x + 8], where x==32
LCM[40, 1280]==1280 ==x[x + 8], where x==32
