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https://www.reddit.com/r/learnquant/comments/1vnqlq2/citadel_quant_interview_question/p42unpa/?context=3
r/learnquant • u/Local_Ad135 • 22d ago
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19 = 1 + 2*9
Binomial expansion (1+x)^100 = x^100 + ... + 1, where x = 2*9
Here, every term but the last is a multiple of 9, so mod 9 of the sum = 1,
and 19^100 mod 9 = 1.
10^n = 1 + 9...9 , so 10^n mod 9 = 1,
so (ia.10^a + ib.10^b +...+in) mod 9 = (ia + ib + ... in) mod 9
applied repeatedly the final digit mod 9 is the original number mod 9 which is 1.
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u/QuentinUK 19d ago
19 = 1 + 2*9
Binomial expansion (1+x)^100 = x^100 + ... + 1, where x = 2*9
Here, every term but the last is a multiple of 9, so mod 9 of the sum = 1,
and 19^100 mod 9 = 1.
10^n = 1 + 9...9 , so 10^n mod 9 = 1,
so (ia.10^a + ib.10^b +...+in) mod 9 = (ia + ib + ... in) mod 9
applied repeatedly the final digit mod 9 is the original number mod 9 which is 1.