Saturday, January 18, 2014

Students' corner in the Indian Express

The Indian Express newspaper keeps some space for students, preparing for exams.  Most of what I saw was not great; but there was one neat little puzzle:  find all solutions of n! = a! + b! + c!.
Posted by Arun at 12:47 PM

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CIP · 585 weeks ago

I convinced myself that there is only a single solution. Not quite sure I'd call it a proof: n has to be greater than a, b, or c. Also, if n > 3, the sum a! + b! + c! is less than n!. Which just leaves one case.
3 replies · active 583 weeks ago
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macgupta 81p · 585 weeks ago

The clue is one has to use a fact which is accessible to high school students.

Suppose n ≥ a ≥ b ≥ c.

Divide both sides of n! = a! plus b! plus c! by a!.

n!/a!, and 1 are integers so (b! plus c!)/a! has to be an integer. So b! plus c! = m a! for some integer m.

If you divide the above by b!, since 1, m(a!/b!) are integers, so c!/b! must be an integer and can only be 1, since b ≥ c.

So 2 b! = m a!; and 2/m has to be an integer. m can only be 1 or 2.

If m==1, b must be 0 or 1; if m==2 then b! = a! and 3 a! = n!.

It is easy to check that only m==2 works.
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CIP · 583 weeks ago

I think my method is simpler.

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