CAT 2024

Slot 1

QA

Question & Solution

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, $(10^{n-1}+10^{n-2}+10^{n-3}+...) \times ((n-1)!)\times(Sum\ of\ numbers)$

Inserting n=4, $\left(1000+100+10+1\right)\left(3!\right)\left(a+b+c+d\right)$
$\left(6666\right)\left(a+b+c+d\right)$
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.