find the smallest positive integer n such that 11 n - 1 is divisible by 105 .
the last digit of 11 n − 1 is 0 , so 5 always divides 11 n − 1 . also , checking the remainders when the powers of 11 are divided by 3 and 7 we get that 3 divides 11 n − 1 exactly when n is even , and 7 divides 11 n − 1 exactly when 3 divides n . correct answer d
divide(multiply(multiply(const_2, const_3), 105), 105)