Factorial Product = Factorial Sum!

Number Theory Level 5

a ! ⋅ b ! = a ! + b ! + c ! \large{ a! \cdot b! = a! + b! + c! }

How many ordered triplets of positive integers ( a , b , c ) {(a,b,c)} exists for which the above equation is satisfied?

∞ \infty 3 6 0 4 8 1 2

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Satyajit Mohanty by Satyajit Mohanty , 24, India

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2 solutions

Curtis Clement
Aug 18, 2015

Firstly, lets assume without loss of generality that a ≤ b \ a \leq\ b . Testing a = 1 and a = 2 leads to no solutions, so a ≥ 3 \ a \geq\ 3 . Now dividing the whole equation by a!: b ! = 1 + b ! a ! + c ! a ! ( 1 ) \ b! = 1 + \frac{b!}{a!} + \frac{c!}{a!} \ ~~~ \ (1) From this we can see that c ≥ a \ c \geq\ a otherwise the RHS is not an integer, but the LHS is, which forms a contradiction. Looking at b and c we have 2 possibilities: c < b o r c ≥ b \ c < b \ ~~ \ or \ ~~ \ c \geq\ b . The first possibility implies that a ! b ! = a ! + b ! + c ! < 3 b ! ⇒ a ! < 3 \ a!b! = a! +b! +c! < 3b! \Rightarrow\ a! < 3 . This is a contradiction so c ≥ b \ c \geq\ b . However b = c leads to the following equation: 1 = b ! a ! ( a ! − 2 ) > 1 \ 1 = \frac{b!}{a!}(a! - 2) > 1 so c > b \ c > b . Now looking at (1): b ! a n d c ! a ! \ b! \ ~ \ and \ \frac{c!}{a!} are both even so b ! a ! \frac{b!}{a!} is odd. Testing b=a+1 in the original equation gives: a ! ( a + 1 ) ! = a ! + ( a + 1 ) ! + c ! \ a!(a+1)! = a! +(a+1)! +c! ⇒ ( a + 1 ) ! = a + 2 + c ! a ! = ( a + 1 ) + 1 + ( c ! a ! ) \Rightarrow\ (a+1)! = a+2 + \frac{c!}{a!} = (a+1) + 1+ ( \frac{c!}{a!}) so (a+1) divides every term except 1 which is not possible. This contradiction means that a =b. (i.e. 3 ≤ a = b < c \ 3 \leq\ a=b < c ). From (1) we obtain: a ! − 2 = c ! a ! \ a! - 2 = \frac{c!}{a!} of which 3 doesn't divide either side. This implies that RHS = a+1 or (a+1)(a+2). The first produces a ! − a = 3 ⇒ a ∣ 3 \ a! -a = 3 \Rightarrow\ a|3 and the 2nd produces a ! − a ( a + 3 ) = 4 ⇒ a ∣ 4 \ a! - a(a+3) = 4 \Rightarrow\ a|4 . Hence, a = 3 or 4, but testing leads to only a = 3. ∴ ( a , b , c ) = ( 3 , 3 , 4 ) i s a u n i q u e s o l u t i o n \therefore\ (a,b,c) = (3,3,4) \ is \ a \ unique \ solution

10 Helpful 0 Interesting 0 Brilliant 0 Confused

Great explanation!

Brett Davidson - 4 years, 11 months ago

S i n c e a a n d b a r e i n t e r c h a n g e a b l e , w e c a n s a f e l y a s s u m e a = b o r a > b . N o t e t h a t n ! > 0 , a n d s o m i n i m u m c ! = 1. I f b = 1 , a ! ∗ b ! < a ! + b ! , ⟹ a = b > 1 I f a = b = 2 , a ! ∗ b ! = a ! + b ! + 0 , b u t c ! ≠ 0 , ∴ a = b = 2 n o t a s o l u t i o n . 12 = 3 ! ∗ 2 ! = 3 ! + 2 ! + 4. B u t c ! ≠ 4. n o s o l u t i o n . N e x t 3 ! ∗ 3 ! = 36 = 3 ! + 3 ! + 24 s o c ! = 24 a n d c = 4. S o l u t i o n ( a , b , c ) = ( 3 , 3 , 4 ) o n l y . S i n c e a > 3 , a ! ∗ b ! = a ! + b ! + X . B u t X i s n e v e r = c ! . Since~ a~ and~ b ~are~ interchangeable,~ we~ can~ safely~ assume~ a=b~ or~ a>b.\\ Note~ that ~n!>0,~ and~so~minimum~c!=1.\\ If~b=1, a!*b!<a!+b! ,~~ \implies~ a=b>1\\ If~a=b=2,~~a!*b!=a!+b!+0,~but~c!\neq 0,~\therefore~a=b=2~not~a~solution.\\ 12=3!*2!=3!+2!+4. ~~But~c! \neq 4. ~~no~solution.\\ Next~3!*3!=36=3!+3!+24~~so~c!=24~~and~c=4.\\ Solution~(a,b,c)=(3,3,4) ~only.\\ Since~a>3, a!*b!=a!+b!+X. ~~But~X~is~never~=c!.

I missed it for I took (3,3,4),(3,4,3),(4,3,3) as solutions !!!

Niranjan Khanderia - 2 years, 8 months ago
Aaysha Babu
Aug 30, 2015

i don't know no. theory i just checked with some no.s and guessed it

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Sorry there was a slight mistake in my original solution, but it has been fixed now.

Curtis Clement - 5 years, 9 months ago

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