What is the smallest positive integer value n, for which 2700n is a multiple of 35?
35 = 5*7
so 2700n must be divisible by both 5 and 7
it is already divisible by 5
2700n = 2*2*5*5*3*3*3*n
so we are missing the 7
n = 7
so 2700n must be divisible by both 5 and 7
it is already divisible by 5
2700n = 2*2*5*5*3*3*3*n
so we are missing the 7
n = 7