positive integer solution

(1) Total no. of positive integer ordered pairs $x,y,z$ in $x!+y!+z! = x!\cdot y! $

(2) Total no. of positive integer ordered pairs $x,y,z$ in $x!+y!+z! = x!\cdot y!\cdot z!$

(3) If $x!\cdot y! = x!+y!+2^z$ , Then no. of positive integer ordered pairs $(x,y,z$ is

(4) Total no. of positive integer ordered pairs in $x,y,z,t$ in $x!+y!+z! = 3^t$

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

For number 2, there no positive integer solutions:

The equation can be expressed as (1/y!z!) + (1/x!z!) + (1/x!y!) = 1 Let a = y!z! b = x!z! c = x!y!

Solving the equation (1/a) + (1/b) + (1/c) = 1

By case-to-case analysis: Ordered pairs are given (a, b, c)

Since the equation is symmetric, one can assume the value of a.

For a = 2, there re 3 solutions: (2, 4, 4), (2, 3, 6), (2, 6, 3)

For a = 3, there are 3 solutions: (3, 2, 6), (3, 6, 2), (3, 3, 3)

For a = 4, there are 2 solutions: (4, 2, 4), (4, 4, 2)

For a greater than or equal to 5: There are no more solutions. Except for a = 6: (6, 3, 2), (6, 2, 3)

Now substituting the variables to their given values as stated earlier. You can check that there are no solutions existing for any positive integers x, y, z.

John Ashley Capellan - 7 years, 6 months ago

For number 4, there 13 positive integer ordered pairs for the equation.

When t = 1:

(x, y, z, t): (1, 1, 1, 1)

When t = 2:

(x, y, z, t): (1, 2, 3, 2), (1, 3, 2, 2), (2, 1, 3, 2), (2, 3, 1, 2), (3, 1, 2, 2), (3, 2, 1, 2)

When t = 3:

(x, y, z): (1, 2, 4, 3), (1, 4, 2, 3), (2, 1, 4, 3), (2, 4, 1, 3), (4, 1, 2, 3), (4, 2, 1, 3)

Explanation:

Using case-to-case analysis for t:

When t = 1, trivially, the solution is when x! + y! + z! = 1 + 1 + 1 = 3

When t = 2, we try maximize one of the values in order to obtain the equation x! + y! + z! = 9. Hence, the solutions above.

When t = 3, (The same as when t = 2), hence, the solutions above.

How about when t greater than or equal 4? Some values of t, when we try to maximize one of the values and try case-to-case basis... the value does not equate to the right-hand side of the expression. (e.g. For t = 4, we try to maximize x! for x = 4, as well as for y! and z! for y = z = 4 which implies that they are not equal.)

We can also try comparing modulo 10's....

For x, y, z: When x, y, z = 1, then x!, y!, z! is congruent to 1 modulo 10. When x, y, z = 2, then x!, y!, z! is congruent to 2 modulo 10. When x, y, z = 3, then x!, y!, z! is congruent to 6 modulo 10. When x, y, z = 4, then x!, y!, z! is congruent to 4 modulo 10. When x, y, z is greater than or equal to 5, then x!, y!, z! is congruent to 0 modulo 10.

For 3^t, 3^t can be congruent to 1, 3, 7, 9 modulo 10. Some values of 3^t, when we try to express them as sum of factorials, we need to maximize the values, but comparing their modulo 10's doesn't comply to the right-hand side of the equation.

John Ashley Capellan - 7 years, 6 months ago

For number 1: There is only one solution (x, y, z): (3, 3, 4).

Explanation:

We can use case-to-case basis for x or y and z.

If z = 1, 2, 3: There are no solutions for this equation.

If z = 4, there is one solution (x, y, z): (3, 3, 4)

If z greater than or equal to 5, there are no more solutions.

John Ashley Capellan - 7 years, 6 months ago

For number 3:

This is an implication of number 1: This case let w! = 2^z.

Using the solutions that we solved in number 1: w! = 24 which is not a power of 2, so z is not integral.

Therefore, there are no integral solutions.

John Ashley Capellan - 7 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$