My First AIME Mocks!
I have been very inactive in this diary. I invested most of my time in my notes on Real Analysis and Modern Algebra. I’ll try to publish them soon if I think they are up to standard.
I participated in two different mock AIMEs today, so it’s safe to say that my brain is pretty fried. I won’t post any write-ups right now because neither are finished, but I will link the contests.
The first one I did in the morning was the IMTC contest. 15 questions with 3 hours. I answered 7 of them, and guessed on 2 more. The first few were quite easy and caused my confidence to go up until the last few questions, which killed me.
The second contest was the TTLM contest, which I started in the evening. I was more relaxed doing this, I did not even plan on using paper for anything other than the geo problems, so I stuck to my whiteboard. I answered 6, and guessed on 1 more. Again, I started to get stuck on some problems. I will also note that the time on this exam was 2 hours compared to the normal time period of 3 hours. The questions however seemed easier, so it seemed to balance out the difficulty of this exam.
LaTeX to Markdown Test
Problem How many positive integers $n \leq 242$ exist such that the remainders when $n$ is divided by $1, 2, 11, 22, 121$ and $242$, respectively, form a non-decreasing sequence?
Denote $n_k$ as taking $n \bmod{k}$. Our series becomes \(0, n_2, n_{11}, n_{22}, n_{121}, n\) We can simply compare consecutive terms, and as long as the next one is greater than or equal to the previous one, we are good.
It is easy to verify that $n_2 \geq 0, n_{22} \geq n_{11},$ and $n_{121} \geq n$ are always true. That leaves us to find examples of $n_{11}<n_2$ and $n_{121}<n_{22}$.
When $n=11$, \(n_2=1 > 0 = n_{11}.\) Past this, $n_{11}$ continues to increase, while $n_2$ stays at $0$ or $1$. Even at $n=22$, $n_2 = 0 = 0 = n_{11}$. But when $n=33$, $n_2 = 1 > 0 = n_{11}$. We conclude that when $n \equiv 11 \pmod{22}$, we have $n_{11}<n_2$. This is also easy to verify.
For $n_{121}<n_{22}$, when $n<121$, $n_22$ remains small, while $n_{121}$ continues to increase, so all of these numbers should be good. When $n=121,$ we have \(n_{22} = 11 > 0 = n_{121}.\) $n_{22}$ continues to increase at the same pace of $n_{121}$ until $n=132$, where \(n_{22} = 0 < 11 = n_{121}.\) Past this, $n_{121}$ continues to increase until $n=242$, which is where we would have stopped anyway.
| Condition | Is false… |
|---|---|
| $n_2 \geq 0$ | Never |
| $n_{11} \geq n_2$ | $n \equiv 11\bmod{22}$ |
| $n_{22} \geq n_{11}$ | Never |
| $n_{121} \geq n_{22}$ | $121 \leq n \leq 131$ |
| $n_{121} \geq n$ | Never |
There are $11$ $n$ that satisfy $n \equiv 11\bmod{22}$, and $11$ $n$ that satisfy $121 \leq n \leq 131$. However, don’t forget PIE, the number $121$ is counted twice! Therefore the $n$ that satisfy the conditions are \(242 - (11 + 11 - 1) = \boxed{221}.\)
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