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How to get good at comp math
fossasor   23
N an hour ago by fossasor
I'm a rising ninth grader who wasn't in the school math league this year, and basically put aside comp math for a year. Unfortunately, that means that now that I'm in high school and having the epiphany about how important comp math actually is, and how much it would help my chances of getting involved in other math-related programs. In addition, I do enjoy math in general, and suspect that things like the AMCs are probably going to be some of the best practice I can get. What this all means is that I'm trying to go from mediocre to orz, 2 years after I probably should have started if I wanted to be any good.

So my question is: how do I get good at comp math?

This year, my scores on AMC 10 (and these are the highest I've ever gotten) were a 73.5 and an 82.5 (AMC 8 was 21/25, but that doesn't matter much). This is not good enough to qualify for AIME, and I probably need to raise my performance on each by at least 10 points. I've been decently good in the past at Number Theory, but I need to work on Geo and Combinatorics, and I'm trying to find the best resources to do that. My biggest flaw is probably not knowing many algorithms like Stars and Bars, and the path is clear here (learn them) but I'm still not sure which ones I need to know.

I'm aware that some of this advice is going to be something like "Practice 5 hours a day and start hardgrinding" or something along those lines. Unfortunately, I have other extracurriculars I need to balance, and for me, time is a limiting resource. My parents are somewhat frowning upon me doing a lot of comp math, which limits my time as well. I have neither the time nor motivation to do more than an hour a day, and in practice, I don't think I can be doing that consistently. As such, I would need to make that time count.

I know this is a very general question, and that aops is chock-full of detailed advice for math competitions. However, I'd appreciate it if anyone here could help me out, or show me the best resources I should use to get started. What mocks are any good, or what textbooks should I use? Where do I get the best practice with the shortest time? Is there some place I can find a list of useful formulas that have appeared in math comps before?

All advice is welcome!

23 replies
fossasor
Apr 10, 2025
fossasor
an hour ago
No more topics!
generic whiteboard problem
ostriches88   19
N Nov 16, 2024 by happypi31415
Source: 2024 AMC10B p16
Jerry likes to play with numbers. One day, he wrote all the integers from $1$ to $2024$ on the whiteboard. Then he repeatedly chose four numbers on the whiteboard, erased them, and replaced them with either their sum or their product. (For example, Jerry's first step might have been to erase $1, 2, 3$, and $5$, and then write either $11$, their sum, or $30$, their product, on the whiteboard.) After repeatedly performing this operation, Jerry noticed that all the remaining numbers on the board were odd. What is the maximum possible number of integers on the board at that time?

$
\textbf{(A) }1010 \qquad
\textbf{(B) }1011 \qquad
\textbf{(C) }1012 \qquad
\textbf{(D) }1013 \qquad
\textbf{(E) }1014 \qquad
$
19 replies
ostriches88
Nov 13, 2024
happypi31415
Nov 16, 2024
generic whiteboard problem
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G H BBookmark kLocked kLocked NReply
Source: 2024 AMC10B p16
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ostriches88
1527 posts
#1
Y by
Jerry likes to play with numbers. One day, he wrote all the integers from $1$ to $2024$ on the whiteboard. Then he repeatedly chose four numbers on the whiteboard, erased them, and replaced them with either their sum or their product. (For example, Jerry's first step might have been to erase $1, 2, 3$, and $5$, and then write either $11$, their sum, or $30$, their product, on the whiteboard.) After repeatedly performing this operation, Jerry noticed that all the remaining numbers on the board were odd. What is the maximum possible number of integers on the board at that time?

$
\textbf{(A) }1010 \qquad
\textbf{(B) }1011 \qquad
\textbf{(C) }1012 \qquad
\textbf{(D) }1013 \qquad
\textbf{(E) }1014 \qquad
$
This post has been edited 1 time. Last edited by ostriches88, Nov 13, 2024, 6:08 PM
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saturnrocket
1306 posts
#2
Y by
i think the answer's 1011?
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mathMagicOPS
849 posts
#3
Y by
Did anyone else get 1010? I just analyzed each operation (ex. eeee->e)
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ostriches88
1527 posts
#4
Y by
its 1010, you must have at least one of the moves that removes two odd numbers from the board
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eg4334
629 posts
#5
Y by
Note that the total number is invariant mod 3 and the number of odds is nonincreasing. Therefore 1010.
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pingpongmerrily
3564 posts
#6
Y by
this problem fr so generic
can confirm 1010 even tho i didn't solve it
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aleyang
192 posts
#7
Y by
I did OEEE and 1 OOOE
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EaZ_Shadow
1220 posts
#8
Y by
BROOOO I CHOKED I GOT C :( it was bc that you can only lose or no gain, so I thought 1012 was the answer, but it wasn't achievable
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lucaswujc
294 posts
#9
Y by
saturnrocket wrote:
i think the answer's 1011?

was 1010 iirc
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megarnie
5576 posts
#10
Y by
The answer is $\boxed{\textbf{(A)}\ 1010}$.

Bound: Note that each move removes at most $3$ evens, and there are $1012$ evens, meaning we must make at least $338$ operations and therefore lose at least $338 \cdot 3 = 1014$ numbers.

Construction (didn't need this in test):

Claim: You can turn $4n$ evens into $n$ evens in $n$ turns.
Proof: Just group them into $n$ groups of $4$ and operate on each one. $\square$

Extending this claim, we can also turn $4n + k$ evens into $n + k$ evens in $n$ operations.

Note that the number of evens is initially $1012$. Consider $1012 \to 253 \to 64 \to 16 \to 4 \to 1$ takes $253 + 63 + 16 + 4 + 1 = 337$ operations. Now operate the final even with three random odds and we are done.
This post has been edited 1 time. Last edited by megarnie, Nov 13, 2024, 6:19 PM
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andrewcheng
525 posts
#11
Y by
notice that each operation takes 3 numbers out
to remove all evens we must remove at least 1012 entries(all the evens)
1014 is the smallest mult of 3 >1012
2024-1014=1010
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alexanderhamilton124
388 posts
#12
Y by
I got A; did (O, O, O, E) operations, followed by one last (E, O, O, O).
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elizhang101412
1203 posts
#13
Y by
i tried to cheese this by finding 2024-3x; should have went with 1010

now that I think about it 1013 isn't even realistically possible idk what I was on :(
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giratina3
492 posts
#14
Y by
I got 1009, but realized that I didn’t got the new odd numbers formed and got 1010
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KevinChen_Yay
218 posts
#15
Y by
I got 1010. Every time the total amount of integers decreases by 3 because it's 4-->1, therefor the final answer is either (A) or (D). However, since the amount of odd integers can only decrease (the only case where it may increase is all 4 even --> 1 odd but that's impossible), the answer is $\boxed{\mathbf{(A)}\ 1010}$.
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joshualiu315
2513 posts
#16
Y by
the answer choices here are so stupid bruh


Note that each operation removes three numbers from the whiteboard. It is easy to see that there are many ways to have all three removed numbers be even for each move, but we still need at least $338$ operations to remove $1012$ even numbers. Hence, the answer is $\boxed{1010}$ (some odd numbers will be removed on the last operation).


Remark: the only possible answer choices are A and D lol
This post has been edited 1 time. Last edited by joshualiu315, Nov 13, 2024, 8:51 PM
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andrewcheng
525 posts
#17
Y by
I threw this by getting that multiplying 4 es only subtracted 2 I had doubts about my ans here
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FuturePanda
235 posts
#18
Y by
Yes! Me got it right!
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Tetra_scheme
90 posts
#19
Y by
during the test this problem gave me hope the 10b was high quality. The total is invariant mod 3 and there can't be more odds so it is the next number that is equal to $2$ mod 3 or 1010.
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happypi31415
742 posts
#20
Y by
just note that it must be less then or equal to $1012$, because there are $1012$ odd numbers, and the answer is invariant $\pmod {3}$, which gives the only possible answer as $\boxed{\textbf{(A)}\ 1010}$
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