In the prime factorization of \(24!\), what is the exponent of \(3\)? (Source: Aops Staff)
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.
it has something to do with number theory because I did it before and I forgot the answer
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.