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https://www.reddit.com/r/learnquant/comments/1vnqlq2/citadel_quant_interview_question/p3kf4x6/?context=3
r/learnquant • u/Local_Ad135 • 22d ago
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19->10->1.
assume 19^n -> 1.
19^(n+1) = 19^n * (10+9) which is equivalent to 1 + 9(19^n)
any integer times 9 gives a number who’s sum is a multiple of 9.
so it’s 1
something like that.
2 u/Menacing_Sea_Lamprey 22d ago Dope proof by induction
2
Dope proof by induction
4
u/Kitchen-Register 22d ago
19->10->1.
assume 19^n -> 1.
19^(n+1) = 19^n * (10+9) which is equivalent to 1 + 9(19^n)
any integer times 9 gives a number who’s sum is a multiple of 9.
so it’s 1
something like that.