(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 \(x!+y!+z! = 3^t \)

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TopNewestFor 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 · 2 years, 10 months ago

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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 · 2 years, 10 months ago

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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 · 2 years, 10 months ago

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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 · 2 years, 10 months ago

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