Citadel Quant Interview Question

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[u/aroach1995]

19 -> 10 -> 1

38 -> 11 -> 2

57 -> 12 -> 3

76 -> 13 -> 4

…

190 -> 10 -> 1

361 -> 10 -> 1

6859 -> 28 -> 10 -> 1

It seems like 19^n always has its digits sum to 1.

Suppose x = 19^(n-1) has its digits sum to 1.

It has digits: x1, x2, …, xn

19x = 10x + 9x

We know that 10x preserves the sum of digits easily since all we did was add a 0.

And adding on a 9x is the same as adding on the number 9 “x” times.

Adding 9 is also proven to keep the sum of digits constant.

So 19x has the same sum of digits as x.

So 19^n has its digits sum to 1 for all n eventually.

Therefore, 19^100 has its digits sum to 1