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[SOLVED] Factorials and Exponent Challenge

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,588
Find all positive integer solutions $(a,\,b,\,c,\,n)$ of the equation $2^n=a!+b!+c!$.
 

kaliprasad

Well-known member
Mar 31, 2013
1,281
Without loss of generalty we can have $a>=b>=c$.

Now we get a multiple of power of 2 only when we add multiples of same power of 2

So $a!$ and $b!+c!$ should be muiltiple of same power of 2. and when we add the 2
we shall get multiple of power of 2 say of the form $m2^x$. if m is power of 2 then we are
done.

Now b and c should be multiple of same power of 2 and when we add the same we get a multiple of
higher power and further this should be same as multiple of power of 2 of a.

So we have 2 cases to check

$b = c$ and $a = 2$ or 3 (for reason please see below )-

And $ b = c + 1$ then any a.

As c devides a!+b!+c! so c can not be greater than 2 as sum is power of 2

Now

Put the values b= 1, c = 1 giving a = 2 or 3 as a =4 gives a! divsible by 4 but b!+c! is not

a =2 gives c = 2
a =3 gives c = 3

c= 2, b= 3 gives b! + c! = 8 so we need to check for a = 4 and 5 only as a = 6 or above a! is
divisible by 16 so it is not possible

a = 4 gives 32 power of 2 so n = 5 so solution (4,3,2,7)
a = 5 gives 128 power of 2 so n = 7 so solution (5,3,2,7)

so solution set $(2,1,1,2), (3,1,1,3), (4,3,2,7),(4,3,2,7)$ and any permutation of 1st 3 numbers is each set
 
Last edited:

kaliprasad

Well-known member
Mar 31, 2013
1,281
Without loss of generalty we can have $a>=b>=c$.

Now we get a multiple of power of 2 only when we add multiples of same power of 2

So $a!$ and $b!+c!$ should be muiltiple of same power of 2. and when we add the 2
we shall get multiple of power of 2 say of the form $m2^x$. if m is power of 2 then we are
done.

Now b and c should be multiple of same power of 2 and when we add the same we get a multiple of
higher power and further this should be same as multiple of power of 2 of a.

So we have 2 cases to check

$b = c$ and $a = 2$ or 3 (for reason please see below )-

And $ b = c + 1$ then any a.

As c devides a!+b!+c! so c can not be greater than 2 as sum is power of 2

Now

Put the values b= 1, c = 1 giving a = 2 or 3 as a =4 gives a! divsible by 4 but b!+c! is not

a =2 gives c = 2
a =3 gives c = 3

c= 2, b= 3 gives b! + c! = 8 so we need to check for a = 4 and 5 only as a = 6 or above a! is
divisible by 16 so it is not possible

a = 4 gives 32 power of 2 so n = 5 so solution (4,3,2,7)
a = 5 gives 128 power of 2 so n = 7 so solution (5,3,2,7)

so solution set $(2,1,1,2), (3,1,1,3), (4,3,2,7),(4,3,2,7)$ and any permutation of 1st 3 numbers is each set
above one has mistake in last 3 lines

it should be

a = 4 gives 32 power of 2 so n = 5 so solution (4,3,2,5)
a = 5 gives 128 power of 2 so n = 7 so solution (5,3,2,7)

so solution set $(2,1,1,2), (3,1,1,3), (4,3,2,5),(5,3,2,7)$ and any permutation of 1st 3 numbers is each set