Tirzah Tanveer on Nearpeer MDCAT Common Room

I was watching this LR class 3 on YouTube and this question was rather tricky to solve. But, its really not that difficult: First, you need to take the sum of the individual digits of the given two digit value, for example 21, which would be 2+1=3 . Now, you need to see whether the original number(21) is divisible by the resultant (3) without any remainder. In this case, 21 is divisible by 3 since 3 x 7=21, which shows that 21 is a multiple of its digits.