(1) If \(\left(x,y,z∈N+\left\{0\right\}\right)\). Then the number of solutions of the equation \(x!+y!=z!\) is?

(2) If \(\left(x,y,z,t∈N+\left\{0\right\}\right)\). Then the number of solutions of the equation \(x!+y!+z!=t!\) is?

(3) If \(\left(x,y,z,t,u∈N+\left\{0\right\}\right)\). Then the number of solutions of the equation \(x!+y!+z!=t!+u!\) is?

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TopNewest1) assume x!>y! using this we get z! < 2(x!). but z is greater than x and we can prove for the sequence of factorials that : z! > 2(x!) if z>x>1 because even (n+1)! > 2(n!) for n>1. rest should be simple. – Hemang Sarkar · 3 years ago

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For number 1, the answer is 4. For (x, y, z): (0, 0, 2), (0, 1, 2), (1, 0, 2), and (1, 1, 2). Trivially.

For number 2, the answer is 2. For (x, y, z, t): (2, 2, 2, 3). Trivially.

For number 3, – John Ashley Capellan · 3 years, 7 months ago

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