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Get access to the detailed solutions to the previous years questions asked in IIM IPMAT exam
When gives n distinct numbers, and asked to find the sum of the possible n distinct numbers that can be formed,
We use the formula, (10n–1 + 10n–2 + 10n–3 + .....) [(n – 1)!] (Sum of numbers)
Inserting n = 4, (1000 + 100 + 10 + 1) (3!) (a + b + c + d)
(6666) (a + b + c + d)
We are told that this value equals, 153310 + n
Since we are told that n is a single digit natural number, the total value cannot be that much greater and 6666 should perfectly divide 153310 + n
Upon dividing 153310 by 6666 we get the quotient as 22.99
Nearest value being 23, We take 6666x23 giving us the value 153318
Hence the value of n = 8 and the value of a + b + c + d = 23
Value of a + b + c + d + n = 31