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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!
Problem 2
evt917   50
N Nov 17, 2024 by sadas123
Source: 2024 AMC 12B #2 / AMC 10B #2
What is $10! - 7! \cdot 6!$?

$
\textbf{(A) }-120 \qquad
\textbf{(B) }0 \qquad
\textbf{(C) }120 \qquad
\textbf{(D) }600 \qquad
\textbf{(E) }720 \qquad
$
50 replies
evt917
Nov 13, 2024
sadas123
Nov 17, 2024
Problem 2
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Source: 2024 AMC 12B #2 / AMC 10B #2
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evt917
1584 posts
#1
Y by
What is $10! - 7! \cdot 6!$?

$
\textbf{(A) }-120 \qquad
\textbf{(B) }0 \qquad
\textbf{(C) }120 \qquad
\textbf{(D) }600 \qquad
\textbf{(E) }720 \qquad
$
This post has been edited 1 time. Last edited by LauraZed, Nov 13, 2024, 5:47 PM
Reason: adding AMC 10 to source
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Jack_w
61 posts
#2 • 40 Y
Y by mathlearner2357, weihang, eg4334, vsamc, EpicBird08, allenshen1226, pog, Sedro, alexanderhamilton124, AbhayAttarde01, mahaler, Amkan2022, Challengees24, Firebreather14, Awesomeness_in_a_bun, cweu001, Metavaria, aidan0626, Ilikeminecraft, bot1132, forestcaller2010, alan9535, Andyluo, ap246, happypi31415, suvamkonar, studymoremath, UnearthedCyclone, NoSignOfTheta, kafuu_chino, centslordm, ihatemath123, giratina3, Jndd, goldenuni678, OronSH, evt917, KLBBC, AlexWin0806, Turtwig113
Divisible by $6!$ so $\textbf{(E) }720$
This post has been edited 1 time. Last edited by Jack_w, Nov 13, 2024, 5:13 PM
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HonestCat
950 posts
#3 • 3 Y
Y by studymoremath, alexanderhamilton124, MATHMAN9999
Jack_w wrote:
Divisible by $6!$ so $\textbf{(E) }720$

I got 0
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HumanCalculator9
6230 posts
#4 • 1 Y
Y by golue3120
Simply notice that only 0 is a multiple of 7, and as obviously $10!$ and $7!$ are both multiples of 7, the answer must be $0$

What was MAA thinking with the answer choices bro :skull:
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yambe2002
1659 posts
#5
Y by
$6!=8\cdot9\cdot10$ so $0$
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BS2012
558 posts
#6
Y by
$$6!*7(10*9*8-6!)=0$$
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megarnie
5324 posts
#7 • 1 Y
Y by Mathandski
Divisible by $7!$, so $\boxed{\textbf{(B)}\ 0}$.
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titanlord
197 posts
#8 • 2 Y
Y by Soccerstar9, X423anp1
Jack_w wrote:
Divisible by $6!$ so $\textbf{(E) }720$

bruh oh my lord this is tragic
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Squidget
380 posts
#9
Y by
I actually multiplied it out because I was too paranoid to go along with my original answer.
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paganiniana
204 posts
#10
Y by
note that $6!=720=8\cdot9\cdot10$, and $10!=7!\cdot8\cdot9\cdot10$ so 0
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alexanderhamilton124
250 posts
#11
Y by
Jack_w wrote:
Divisible by $6!$ so $\textbf{(E) }720$

Bro got cooked so hard

10! - 7! 6! = 6!(10*9*8*7 - 7!), and 10*9*8*7 = 7!, so is zero
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NoSignOfTheta
1590 posts
#12
Y by
Rip Jack_w
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pqr.
147 posts
#13
Y by
Jack_w wrote:
Divisible by $6!$ so $\textbf{(E) }720$

This guy bruh
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ChromeRaptor777
1890 posts
#14
Y by
evt917 wrote:
What is $10! - 7! \cdot 6!$?

$
\textbf{(A) }-120 \qquad
\textbf{(B) }0 \qquad
\textbf{(C) }120 \qquad
\textbf{(D) }600 \qquad
\textbf{(E) }720 \qquad
$

sol1
sol2
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MathRook7817
285 posts
#15
Y by
just factor 6! or 7! out and it become trivial
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