Question

In the prime factorization of 2500 factorial what is the power of 7 factorial

Answer 1

409

Step-by-step explanation:

The largest power of 7 less than or equal to 2500 is 7^3 = 343, so we need to count the number of multiples of 7, 49, and 343 that are less than or equal to 2500.

Multiples of 7: There are 357 multiples of 7 less than or equal to 2500 (since 357 x 7 = 2499).

Multiples of 49: There are 51 multiples of 49 less than or equal to 2500 (since 51 x 49 = 2499).

Multiples of 343: There is only 1 multiple of 343 less than or equal to 2500 (since 1 x 343 = 343).

Therefore, the total number of factors of 7 in the prime factorization of 2500! is:

357 + 51 + 1 = 409

Answer 2

To determine the power of 7 in 2500 factorial, one would sum up the number of times 2500 can be divided by 7 and its subsequent powers until the quotient is less than 7.

Step-by-step explanation:

The student is asking about the power of 7 in the prime factorization of 2500 factorial (2500!). To find the power of a prime within a factorial, one must sum up the integer quotients resulting from dividing the factorial number (2500 in this case) by the prime number (7 here) and each of its powers until the quotient is less than the prime.

Here's a quick example with a smaller number to illustrate the method: For 10! the power of factor 2 would be calculated as:

• 10/2 = 5
• 10/4 = 2
• 10/8 = 1

We would then add up those quotients: 5 + 2 + 1 = 8, so 2 is raised to the 8th power in the prime factorization of 10!.

In the case of 2500! and the prime number 7, this would require a significant number of calculations as 7 raised to various powers will divide the number 2500 numerous times.