What is the highest power of 3 that divides 1000! (1000 factorial)
498
To find the highest power of 3 that divides 1000!, we need to count the number of multiples of 3, multiples of 9, multiples of 27, and so on up to the highest power of 3 less than or equal to 1000.
1. Multiples of 3: There are 1000/3 = 333 multiples of 3 in the range 1 to 1000.
2. Multiples of 9: There are 1000/9 = 111 multiples of 9 in the range 1 to 1000.
3. Multiples of 27: There are 1000/27 = 37 multiples of 27 in the range 1 to 1000.
4. Multiples of 81: There are 1000/81 = 12 multiples of 81 in the range 1 to 1000.
5. Multiples of 243: There are 1000/243 = 4 multiples of 243 in the range 1 to 1000.
6. Multiples of 729: There is only 1 multiple of 729 in the range 1 to 1000.
Since the highest power of 3 that divides 1000! is determined by the highest power of 3 less than or equal to 1000, we can see that the highest power is 3^333.
To find the highest power of 3 that divides 1000!, we need to determine how many multiples of 3 are there in the product 1000 x 999 x 998 x ... x 3 x 2 x 1.
To count the number of multiples of 3, we can divide 1000 by 3, which gives us 333 multiples of 3.
However, some numbers (e.g., 6, 9, 12, ...) have more than one factor of 3. To account for this, we need to consider the numbers that are divisible by 3 twice (i.e., multiples of 9) and calculate how many of these there are.
We can divide 1000 by 9 to find that there are 111 multiples of 9.
Next, we need to consider the numbers that are divisible by 3 three times (i.e., multiples of 27). We can divide 1000 by 27 to find that there are 37 multiples of 27.
We continue this process for higher powers of 3 until we reach a number larger than 1000.
Now, to find the highest power of 3 that divides 1000!, we sum up the number of multiples of 3, 9, 27, and so on until we reach a power of 3 larger than 1000.
333 + 111 + 37 + ... = 333 + 111 + 37 + 12 + 4 + 1 = 498
Therefore, the highest power of 3 that divides 1000! is 3^498.