- #1

anemone

Gold Member

MHB

POTW Director

- 3,885

- 11,584

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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- MHB
- Thread starter anemone
- Start date

- #1

anemone

Gold Member

MHB

POTW Director

- 3,885

- 11,584

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

- #2

kaliprasad

Gold Member

MHB

- 1,339

- 0

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:

- #3

kaliprasad

Gold Member

MHB

- 1,339

- 0

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

Share:

- Last Post

- Replies
- 1

- Views
- 495

- Replies
- 12

- Views
- 835

- Replies
- 66

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 581

- Last Post

- Replies
- 2

- Views
- 573

- Last Post

- Replies
- 2

- Views
- 586

- Last Post

- Replies
- 0

- Views
- 436

- Last Post

- Replies
- 2

- Views
- 578

- Last Post

- Replies
- 2

- Views
- 588

- Last Post

- Replies
- 1

- Views
- 472