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- Feb 14, 2012

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Find all positive integer solutions $(a,\,b,\,c,\,n)$ of the equation $2^n=a!+b!+c!$.

- Thread starter anemone
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- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,631

Find all positive integer solutions $(a,\,b,\,c,\,n)$ of the equation $2^n=a!+b!+c!$.

- Mar 31, 2013

- 1,292

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:

- Mar 31, 2013

- 1,292

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

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