Calvin Lin
Reputation
31,843
79/100 score
3 35 98
Impact
~285k people reached

• 84 helpful flags
• 1,402 votes cast

# 4,714 Actions

 Apr16 comment Simplifying Sum Did you try inducting on $n$ and $m$? Mar27 awarded Peer Pressure Mar24 answered Fake induction proofs Mar24 answered Need help with a determinant proof. Mar24 answered A sequence 〈a_n〉 is defined recursively ,and find infinite series Mar24 answered Does $\lambda_1^n+ \lambda_2^n+ \dots +\lambda_k^n =0$ for all $n$ imply that $\lambda_1= \lambda_2= \dots= \lambda_k = 0$? Mar24 comment INMO Problem with even function proof. @barto Isn't a "can I have hint" question very different from a "Give me the answer" question? Or does MSE consider these duplicates of each other? Mar24 comment AMC $12A$ Problem (Sequence lengths) @GeoffreyCritzer Yes, every recurrence relation will eventually repeat modulo any number $n$. Proof: If it is recurrence on $k$ terms, consider the first $n^k+k$ terms mod $n$. There must be some string of $k$ of them which are identical. From this string, by the linear recurrence, each subsequent term is identical. Note: It is much faster to apply Chinese Remainder Theorem and work on this mod 3 and mod 4 separately (then to figure out the period is 52). Mar23 comment AMC $12A$ Problem (Sequence lengths) @GeoffreyCritzer Note you we only need the answer mod 12. So, simply list it out mod 12, and find out the period. Mar22 awarded Sportsmanship Mar22 revised INMO Problem with even function proof. added 225 characters in body Mar22 comment INMO Problem with even function proof. Well, use the fact that $(n+1) - n$ = 1 to conclude that if $n < ik \leq n+1$, then we must have $ik = n+1$, or that $k$ divides $n+1$. Mar22 comment INMO Problem with even function proof. @INMO No, each integer has at least 2 divisors, namely 1 and itself. For primes $p$, we have $d(p) = 2$. But for composite numbers, $d(p) > 2$. For example, $d(4) = 3, d(6) = 4$. Mar22 comment INMO Problem with even function proof. @INMO We must have $\frac{n}{k} < i \leq \frac{n+1}{k}$ in order for the terms to be different. This implies that $n < ik \leq n+1$. And then .... (can you continue?) Mar22 revised INMO Problem with even function proof. added 225 characters in body Mar22 comment INMO Problem with even function proof. @INMO Note: You should not accept an answer that you don't really understand. The pattern is "the terms are different when $k$ divides $n+1$. In this case, the value increases by exactly 1". This tells you that the sum of terms $\lfloor \frac{n+1}{k} \rfloor$ increases by the value of the number of divisors of $n+1$, IE $d(n+1)$ in Brian's solution below. Mar22 comment INMO Problem with even function proof. @INMO Yes, they are also different at $\lfloor \frac{2}{2} \rfloor$ because this is a new term that was introduced. More generally, can you provide a classification for when this changes? (An example of a potential classification is "the number changes when $k$ has the same parity as $n+1$" or "the number changes every time") Mar22 answered AMC $12A$ Problem (Sequence lengths) Mar22 comment INMO Problem with even function proof. +1 While I agree that is the underlying idea to the problem, writing it up this way makes it appear as if "Oh, I must know this special fact to solve this problem." Mar22 answered INMO Problem with even function proof.