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Source: 2024 AMC 10A #5/2024 AMC 12A #4

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#1 h Nov 7, 2024, 4:42 PM

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What is the least value of $n$ such that $n!$ is a multiple of $2024$ ?

$\textbf{(A) }11 \qquad \textbf{(B) }21 \qquad \textbf{(C) }22 \qquad \textbf{(D) }23 \qquad \textbf{(E) }253 \qquad$

This post has been edited 1 time. Last edited by Marcus_Zhang, Nov 7, 2024, 4:44 PM

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4337 posts

#2 h Nov 7, 2024, 4:43 PM • 2 Y

Y by evanhliu2009, ranu540

(D) confirm?

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5845 posts

Sedro

#3 h Nov 7, 2024, 4:43 PM

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This is D, obviously $n\ge 23$ and $n=23$ works.

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174 posts

hsuya1

#4 h Nov 7, 2024, 4:44 PM

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yeah, 2024=2^3*11*23 and since 23 is prime n must be at least 23

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180 posts

#5 h Nov 7, 2024, 4:44 PM

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$n\geq 23,$ and $23!$ is abundant in factors of $2$ and $3$ , so we're good: $\textbf{(D)\ 23}.$

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2376 posts

evt917

#6 h Nov 7, 2024, 4:46 PM

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By prime factorizing you must have a $23$ so $D$

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4917 posts

pog

#7 h Nov 7, 2024, 4:47 PM

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Write $2024 = 45^2 - 1 = (45-1)(45+1) = 23 \cdot 2 \cdot 11 \cdot 4$ . Clearly $2,4,11,23$ appear in $23!$ , and $23$ does not divide any smaller factorial. So we get $\boxed{\textbf{(B) }23}$ .

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141 posts

#8 h Nov 7, 2024, 4:48 PM

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the answer is 23, the greatest prime factor of 2024

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636 posts

eg4334

#9 h Nov 7, 2024, 4:48 PM

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i guess knowing the prime factorization of 2024 did save some time

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alexanderhamilton124

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alexanderhamilton124

#10 h Nov 7, 2024, 4:48 PM

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2024 = 23* 11 * 2^3, so 23 is the bigest prime factor which means $n \geq 23$ , and $11 \mid 23!$ , and $8 \leq 23$ , so done.

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34 posts

#11 h Nov 7, 2024, 4:51 PM

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D. This is why you should remember prime factorizations!

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980 posts

#12 h Nov 7, 2024, 4:52 PM

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eg4334 wrote:

i guess knowing the prime factorization of 2024 did save some time

when difference of squares comes in clutch ( $45^2 - 1$ )

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pingpongmerrily

3577 posts

pingpongmerrily

#13 h Nov 7, 2024, 4:53 PM

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just realize that $2024=2025^2-1=46\cdot44$ and then finish the rest

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3377 posts

dbnl

#14 h Nov 7, 2024, 5:03 PM

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me when spend 5 mins prime factorizing 2024

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pingpongmerrily

3577 posts

pingpongmerrily

#15 h Nov 7, 2024, 5:03 PM

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dbnl wrote:

me when spend 5 mins prime factorizing 2024

one one question i started doing long division to divide 2560 instead of just realizing its 256*10 :sob:

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golden_star_123

206 posts

golden_star_123

#16 h Nov 7, 2024, 5:10 PM

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This problem was really misplaced and should have been #2. But the greatest prime in the prime factorization of $2024$ is $\boxed{23}$ .

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ChromeRaptor777

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ChromeRaptor777

#17 h Nov 7, 2024, 5:18 PM

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Marcus_Zhang wrote:

What is the least value of $n$ such that $n!$ is a multiple of $2024$ ?

$\textbf{(A) }11 \qquad \textbf{(B) }21 \qquad \textbf{(C) }22 \qquad \textbf{(D) }23 \qquad \textbf{(E) }253 \qquad$

sol

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144 posts

#18 h Nov 7, 2024, 5:19 PM

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Prime factorization of $2024 = 2^3 \cdot 11 \cdot 23$

So the answer is clearly D

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SomeonecoolLovesMaths

3209 posts

SomeonecoolLovesMaths

#19 h Nov 7, 2024, 6:08 PM

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D confirmed

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pingpongmerrily

3577 posts

pingpongmerrily

#20 h Nov 7, 2024, 6:10 PM

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answer choice study works too if you see that $253=11\cdot23$

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660 posts

#21 h Nov 7, 2024, 6:16 PM

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$23$ . Anything below does not contain a factor of $23$ and $2^3|4!|23!$ and $11|11!|23!$

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159 posts

#22 h Nov 7, 2024, 7:26 PM

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pog wrote:

Write $2024 = 45^2 - 1 = (45-1)(45+1) = 23 \cdot 2 \cdot 11 \cdot 4$ . Clearly $2,4,11,23$ appear in $23!$ , and $23$ does not divide any smaller factorial. So we get $\boxed{\textbf{(B) }23}$ .

it says 23 is D not B
But confirmed

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159 posts

#23 h Nov 7, 2024, 7:28 PM • 1 Y

Y by cirrus20

dbnl wrote:

me when spend 5 mins prime factorizing 2024

Always know the prime factorization of that year. PREPARE FOR 2025 $3^4$ × $5^2$

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223 posts

#24 h Nov 7, 2024, 7:56 PM

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Tears I thought the prime factorization for 2024 is 43 instead of 23 cuz I make rly stupid mistakes

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401 posts

#25 h Nov 7, 2024, 8:06 PM

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it payed off to memorize 2024 factorization :thumbsup:

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1059 posts

#26 h Nov 7, 2024, 11:29 PM

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when solving this i almost did a funny by choosing $11$ even though i knew $2^3\cdot11\cdot23=2024$ lol

anyway if you didn't memo the factorization at least you can get $8*253=2024$ and $253$ is obviously divisible by $11$ (bc $2+3=5$ ) so good problem

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3183 posts

#27 h Nov 10, 2024, 10:33 PM

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The answer is $\boxed{23}$ .

Claim: $n\ge 23$ . Since $23$ divides $2024$ , $n!$ has a factor of $23$ . Since $23$ is prime, the claim is proved.

Now, we prove that $23$ actually works. Note that we only have to verify that $2^3\cdot 11\cdot 23$ divides $23!$ . Clearly, $23!$ has factors of the two primes $11$ and $23$ . Also, $8$ is a factor of $23!$ . This finishes the problem.

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5 posts

#28 h Mar 25, 2025, 9:23 PM

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obviously D
2024=2*2*2*11*23
largest prime is 23,D

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26 posts

#29 h Apr 4, 2025, 3:55 PM

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2024=23×11×2×2×2
So smallest value of n will be 23(D)

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