In my humble opinion, although this is not exactly a problem that is tested extensively on the GMAT, there’s no harm in solving questions like these to strengthen your concepts of Permutations with numbers.
If a, b and c are the digits from which we have to form 3-digit numbers such that all the digits are non-zero and distinct, then, the number of ways of doing that is given by,
(a+b+c) * (3-1)! * (111).
In general, if there are ‘n’ distinct digits using which we have to form ‘n’ digit numbers where the digits are non-zero and distinct, then the number of ways of doing this is given by,
(Sum of the n digits) *(n-1)! *(1111….. n times).
The reason for this is as follows:
When a comes in the units place, its place value will be a. There will be (n-1)! numbers where a will be the units digit. Therefore, the sum of these values will be a * (n-1)!
When a comes in the tens place, its value will be 10a. There will be (n-1)! numbers where a will be the tens digit. The sum of these values will be 10a * (n-1)!
When a comes in the hundreds place, its value will be 100a. There will be (n-1)! numbers where a will be the hundreds digit. The sum of these values will be 100a * (n-1)!.
The sum of all these values will be (n-1)! * a (100 + 10 + 1) = (n-1)! * a * 111.
Similarly, for the other digits, b, c, d and so on, the sum of the values can be worked out as shown above.
In our case, since the sum total is (a+b+c) * 2! * 111 which is nothing but (a+b+c) * 222, we can say that the sum will be definitely divisible by 222.
The correct answer option is E.
Hope this helps!
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