First, consider "a number is divisible by 3, 5 and 7". This number must be a multiple of 3, 5 and 7.

3×5×7 = 105

Consider the remainder of ÷ 1 1 again:

105÷ 1 1=9……6

One 105 divided by 1 1 plus 6, and two 105 divided by 1 1 plus 6× 2-1=/kloc-0.

So this number is at least: 105×2 = 2 10.