CAT 2024 Slot 1 · Quant · Question 9

Four distinct digits give 4!=24 arrangements.

By symmetry each digit occupies each of the four places equally often, that is \frac{24}{4}=6 times.

So the grand total is 6\left( a+b+c+d \right)\times 1111, which is 6666\left( a+b+c+d \right).

Writing S for the digit sum, 6666S=153310+n with n between 1 and 9.

Dividing, \frac{153310}{6666}\approx 23.0, so test S=23.

6666\times 23=153318, which exceeds 153310 by exactly 8.

So n=8, a single-digit natural number as required, and S=23.

Therefore a+b+c+d+n=23+8=31.

Shortcut. The sum of all permutations of k distinct digits is always (repetition count) \times (digit sum) \times (repunit). Here that is 6666S, so the question is just asking which multiple of 6666 lies within 9 of 153310.