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Contests & Programs AMC and other contests, summer programs, etc.
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9 When To Start Grinding Math Again
Existing_Human1   142
N 3 minutes ago by xTimmyG
For context, I got a 103.5 on AMC 10B, so I am not sure if I made AIME. I have AMC 8 and Mathcoutns coming up as competitions
142 replies
Existing_Human1
Nov 23, 2024
xTimmyG
3 minutes ago
Counterexample
avn   35
N 4 minutes ago by xTimmyG
Source: 2019 AMC 10B #2 / 12B #2
Consider the statement, "If $n$ is not prime, then $n-2$ is prime." Which of the following values of $n$ is a counterexample to this statement?

$\textbf{(A) } 11 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 19 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 27$
35 replies
avn
Feb 14, 2019
xTimmyG
4 minutes 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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Source: 2024 AMC10B p16
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ostriches88
1451 posts
#1
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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
1305 posts
#2
Y by
i think the answer's 1011?
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mathMagicOPS
809 posts
#3
Y by
Did anyone else get 1010? I just analyzed each operation (ex. eeee->e)
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ostriches88
1451 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
413 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
2552 posts
#6
Y by
this problem fr so generic
can confirm 1010 even tho i didn't solve it
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aleyang
185 posts
#7
Y by
I did OEEE and 1 OOOE
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EaZ_Shadow
377 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
290 posts
#9
Y by
saturnrocket wrote:
i think the answer's 1011?

was 1010 iirc
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megarnie
5324 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
482 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
250 posts
#12
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I got A; did (O, O, O, E) operations, followed by one last (E, O, O, O).
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elizhang101412
1068 posts
#13
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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
247 posts
#14
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I got 1009, but realized that I didn’t got the new odd numbers formed and got 1010
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KevinChen_Yay
136 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
2431 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
482 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
143 posts
#18
Y by
Yes! Me got it right!
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Tetra_scheme
22 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
668 posts
#20
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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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