I can't check them all!

Firstly, we need to understand what effect putting parentheses has. When there are no parentheses at all, every number in the equation except the first one (i.e. 1) will end up in the denominator of the final expression.

In the given case,

$\frac{1}{2*3*4*5*6*7}$

When you add a pair of parentheses to the equation, every number within the parentheses except the first one end up in the numerator of the final expression and every number outside the parentheses ends up in the denominator of the final expression (again, except the first one of course).

For example, $1 \div 2 \div (3 \div 4 \div 5) \div 6 \div 7$ essentially becomes

$\frac{1*4*5}{2*3*6*7}$

So, in order to make sure that the final expression results in an integer, you need to make sure that all the numbers in the denominator have factors in the numerator that can cancel them. Therefore, from observation, you can say that we need to push 5 and 7 to the numerator which implies that the parentheses must span at least from 4 to 7. But even this doesn't give us an integer because 4 remains unfactored in the denominator. This means that we need to widen the parentheses to accommodate 4 in the numerator as well.

This leaves us with $1 \div 2 \div (3 \div 4 \div 5 \div 6 \div 7)$

$\frac{1*4*5*6*7}{2*3}$

Therefore, the final answer is 140.

1÷2÷(3÷4÷5÷6÷7) = 1×4×5×6×7÷[2×3]=140 This is the least possible number because we cannot cancel the factor of 5 and 7. Therefore they must come in the numerator. Then adjust parenthesis intelligently and you will get the answer.