4.h Factorial Notation

Question 1

The product of any set of consecutive integers is divisible by ___________ and _____________

Answer 1

The product of any set of consecutive integers is divisible by ANY OF THE INTEGERS IN THE SET and ANY FACTOR COMBINATIONS OF THE NUMBERS

Question 2

5 X 6 X 7 X 8 = 1,680 What is this number divisible by?

Answer 2

5 X 6 X 7 X 8 = 1,680

Thus 1,680 must be divisible by 5, 6, 7, 8 as well as any factor combinations, such as 30 (5 x 6)

Question 3

n! is divisible by?

Answer 3

n! is divisible by all numbers from 1 to n, as well as any factor combinations

Question 4

What is 4! divisible by?

Answer 4

4! = 4 x 3 x 2 x 1

Therefore 4! must be divisible by 4, 3, 2, 1, as well as any factor combinations such as 12 (4 x 3)

Question 5

What is the product of any consecutive integers divisible by?

Answer 5

The product of any set of n consecutive integers must be divisible by n!

Question 6

What must 5 x 6 be divisible by?

Answer 6

5 x 6 are two consecutive integers, and thus must be divisible by 2!

Question 7

What is the largest number that must be a factor of the product of any four consecutive positive integers?

Answer 7

4 consecutive integers, so must be divisible by 4!

4! = 4 x 3 x 2 x 1 == 24

Question 8

How can you determine the largest number of a prime number x that divide into y! ?

Answer 8

divide y (not y!) by x^1, x^2, x^3….x^k

Keep track of the quotients and ignore any remainders. Stop dividing when the quotient = 0.

Add all quotients together, the sum represents the number x’s in the prime factorization of y!

Question 9

How many 3’s are there in 21! ?

Answer 9

21(NOT 21 factorial)/ 3^1 = 7
21/ 3^2 = 2 (ignore remainders)
21/ 3^3 = 0

7 + 2 = 9
So, there are 9 x’s in the prime factorization of 21!

Question 10

Answer the following:

Answer 10

99

Question 11

How would you answer this?

Answer 11

Watch out! First break 6 into prime factors.

Then divide 40 by each of the prime factors and see how many pairs there are!

Question 12

How can you determine the largest number of a non-prime number x that divides into y! ?

Answer 12

First break x into prime factors.

Then divide y by each of the prime factors and see how many there are of each.

Then determine the number of pairs of the prime factors there are (basically the lowest power), that is the final amount of x’s in y!.

Question 13

How can you solve this?

Answer 13

Break 8 into its prime factors, this would be 2^3.

Then divide 100 by 2, 2^2, 2^3…. etc until the quotient is 0.

Sum all quotients together, which is 97 in this case.

Then remember, to determine the largest integer value of n:

2^97 / 2^3n, so 97 = 3n, thus n = 32.33, so largest integer is 32.