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# Number Theory: Multiples

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In the prime factorization of $$24!$$, what is the exponent of $$3$$? (Source: Aops Staff)

Jun 13, 2020

#1
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First of all, let's find the multiples of 3 in 24!.

1*3, 2*3 ---> 8*3 = 8 multiples

But we can't forget about the multiples of 32. These have an extra 3.

9*1, 9*2 = 2 multiples

8 + 2 = 10

The exponent is 10: 310.

Jun 13, 2020
#3
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thx a lot

amazingxin777  Jun 13, 2020
#2
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it has something to do with number theory because I did it before and I forgot the answer

Jun 13, 2020
#4
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First, notice that 24! is an enormous number, and it is the product of 24 consecutive integers, so it has a lot of factors.

Also, notice that if we omit numbers that aren't divisible by 3, we have $$3\times 6 \times 9 \times \cdots \times 24$$ as a factor of 24!.

We will find the exponent of 3 in this expression because other terms are not divisible by 3.

$$3\times 6 \times 9 \times \cdots \times 24 = (3\times 1) \times (3\times 2) \times (3\times 3) \times \cdots \times (3\times 8)$$

You may notice that the product every 3 consecutive terms is divisible by $$3^4$$.

We have

$$(3^4)^2 \times 3^2 | 24!$$

Simplify to get

$$3^{10} | 24!$$

And we cannot find any other multiple of 3. So the exponent of 3 is 10.

Jun 13, 2020
#5
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nice solution thanks too!

amazingxin777  Jun 13, 2020