How Can the Product of Factorials from 1 to 15 Be Expressed as \(A^2B!\)?

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  • Thread starter anemone
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    2015
In summary, a product of factorials is a mathematical expression that involves multiplying a series of consecutive whole numbers, starting from 1. To solve for A and B in a product of factorials, you can use the formula A! x B! = (A + B + 1)!. This formula is known as the "sum of factorials" formula. POTW #186 is a mathematical problem that challenges students to solve for A and B in a product of factorials and helps reinforce the concept of factorials. There are techniques such as breaking down larger factorials and using the "sum of factorials" formula that can make solving product of factorials problems easier. Real-world applications of product of factorials include statistics, computer
  • #1
anemone
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Here is this week's POTW:

-----

Find $A$ and $B$ such that the product \(\displaystyle \prod_{n=1}^{15} (n+1)!\) can be written in the form $A^2B!$, where $A,\,B$ are positive integers. -----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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  • #2
Congratulations to the following members for their correct solution:

1. kaliprasad
2. greg1313 Solution from kaliprasad:
Because $(n+1)! = n! * (n+1)$
we have
$ \prod_{n=1}^{15}(n+1)!=2! * (3!)^2 * 4 * (5!)^2 * 6 * (7!)^2 * 8 * (9!)^2 * 10 * (11!)^2 * 12 * (13!)^2 * 14 * (15!)^2 * 16$
= $2! * (3! * 5! * 7!*9!*11!*13!*15!)^2 * 4 * 6 * 8 * 10 * 12 *14 * 16$
= $2! * (3! * 5! * 7!*9!*11!*13!*15!)^2 * 2^2 * 2* 3 * 2^3 * 2* 5 * 2^2 * 3 * 2* 7 * 2^4$
= $2! * (3! * 5! * 7!*9!*11!*13!*15!)^2 * 2^{14}* 3^2 * 5 * 7$
(there is an isolated 7 and an isolated 5 on the expression so this should get absorbed in factorial)
= $3! * (3! * 5! * 7!*9!*11!*13!*15!)^2 * 2^{14}* 3 * 5 * 7$
= $4 ! * (3! * 5! * 7!*9!*11!*13!*15!)^2 * 2^{12}* 3 * 5 * 7$
= $5! * (3! * 5! * 7!*9!*11!*13!*15!)^2 * 2^{12}* 3 * 7$
= $6! * (3! * 5! * 7!*9!*11!*13!*15!)^2 * 2^{11} * 7$
= $7! * (3! * 5! * 7!*9!*11!*13!*15!)^2 * 2^{11}$

(Now we have 2^11 is not a power of 2 but 2^3 =8 so we get)

= $8! * (3! * 5! * 7!*9!*11!*13!*15!)^2 * 2^8$ solution (1)

as $3^3 = 9$ so we get also

= $9! * (2 * 5! * 7!*9!*11!*13!*15!)^2 * 2^8$ solution (2)

so there are 2 solutions
solution 1

A = 3! * 5! * 7!*9!*11!*13!*15! * 2^4 and B = 8

solution 2

A = 32 * 5! * 7!*9!*11!*13!*15! and B = 9
 

FAQ: How Can the Product of Factorials from 1 to 15 Be Expressed as \(A^2B!\)?

What is a product of factorials?

A product of factorials is a mathematical expression that involves multiplying a series of consecutive whole numbers, starting from 1. For example, 4! (read as "4 factorial") is equal to 1 x 2 x 3 x 4 = 24.

How do you solve for A and B in a product of factorials?

To solve for A and B in a product of factorials, you can use the formula A! x B! = (A + B + 1)!. This formula is known as the "sum of factorials" formula. You can rearrange the equation to solve for A or B, depending on which variable you are trying to find.

What is the significance of the POTW #186 problem involving product of factorials?

POTW #186 is a mathematical problem that challenges students to solve for A and B in a product of factorials. This problem helps students practice their algebraic manipulation skills and reinforces the concept of factorials. It also encourages critical thinking and problem-solving abilities.

Are there any shortcuts or techniques for solving product of factorials problems?

Yes, there are a few techniques that can make solving product of factorials problems easier. One technique is to break down the larger factorials into smaller ones, for example, if you have 7! x 8!, you can rewrite it as (4! x 3! x 2!) x (4! x 3!), which can then be simplified further. Another technique is to use the "sum of factorials" formula mentioned earlier.

What are some real-world applications of product of factorials?

Product of factorials have various applications in mathematics, computer science, and statistics. For example, they are used in calculating binomial coefficients in statistics and in counting the number of possible combinations in a set of objects. In computer science, product of factorials are used in the analysis of algorithms and in finding the complexity of certain problems. They can also be used in probability and in calculating permutations.

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