The three-digit number $abc$ has $3$ distinct digits. Find all possible numbers of the form $abc$ which satisfy

$abc = c (a+b+c)^2$

## Number Theory Problem

- Abdul Muntakim Rafi
**Posts:**173**Joined:**Tue Mar 29, 2011 10:07 pm**Location:**bangladesh,the earth,milkyway,local group.

### Number Theory Problem

**Man himself is the master of his fate...**

### Re: Number Theory Problem

Quite late a reply, but a reply indeed.

Prodip, you need to show how you obtained the solution. And also, you need to find the complete solution and must prove that there are none beyond what you show.

In this case, the solutions are $162, 243, 324, 405, 605$

Lets rearrange the given equation as $100a + 10b = c[(a+b+c)^2 - 1]$

For the right hand side to be divisible by $10$, $a+b+c$ has to be $\pm 1 modulo 10$. Since $a,b,c$ are digits, the possibilities are $a+b+c = 9, 11, 19, 21$

Putting these values in the earlier equation, we are left with the following linear Diophantine equations-

$ 2a + b = 8 $ when $a+b+c=9$

$ 22a +13b = 132 $ when $a+b+c=11$

$46a + 37b = 684$ when $a+b+c = 19$

$ 54a + 45b = 924 $ when $a+b+c=21$

Hoping that I did no mistake, solving these leave us with the solutions mentioned earlier

Prodip, you need to show how you obtained the solution. And also, you need to find the complete solution and must prove that there are none beyond what you show.

In this case, the solutions are $162, 243, 324, 405, 605$

Lets rearrange the given equation as $100a + 10b = c[(a+b+c)^2 - 1]$

For the right hand side to be divisible by $10$, $a+b+c$ has to be $\pm 1 modulo 10$. Since $a,b,c$ are digits, the possibilities are $a+b+c = 9, 11, 19, 21$

Putting these values in the earlier equation, we are left with the following linear Diophantine equations-

$ 2a + b = 8 $ when $a+b+c=9$

$ 22a +13b = 132 $ when $a+b+c=11$

$46a + 37b = 684$ when $a+b+c = 19$

$ 54a + 45b = 924 $ when $a+b+c=21$

Hoping that I did no mistake, solving these leave us with the solutions mentioned earlier

"Je le vois, mais je ne le crois pas!" - Georg Ferdinand Ludwig Philipp Cantor

### Re: Number Theory Problem

my solution indicates $162, 243, 324$ and $405$ only, but $605$ is also correct(in calculator), so somewhere there could be a problem or something missed.. here it is shown

the first part is, like avik da's way $a+b+c= 9,11,19,21$

if $a+b+c= 9 $then $10a+b-8c=0..........$..(line 1

$a+b+c= 11 $then $10a+b-12c=0......$.....(line 2)

$a+b+c= 19$ then $10a+b-36c=0.....$......(line 3)

$a+b+c= 21$ then $10a+b-44c=0....$.......(line 4)

since a, b, c are digits, so, $a-b, b-c,$ or $c-a$, should be an integer, now we can find from the three lines that only for line one c-a (or a-c) is an integer and c-a =1, which means $c= a+1$

so we should consider only the equations derive from line 1 now

since the number is like $100a+10b+c$, it can be written now $100a+10b+1+a= 101a+10b+1$

since c-a=1 again we can find $2a+b= 8$

or $b= 8-2a$, again entering the value of b the number becomes like this

$101a+80-20a+1=81a+81$

so ultimately the number is =$81(a+1)$

considering a>=5 we see the first digit of $81(a+1)$ is not the same as a (example a=5 then the number is 486,here a=4,contradiction

a<5 we can see the frist digit of 81(a+1)= a

so the numbers are $81(a+1), a= [1,4]$

now my question, how can the digit become 605??? , please give me help..

the first part is, like avik da's way $a+b+c= 9,11,19,21$

if $a+b+c= 9 $then $10a+b-8c=0..........$..(line 1

$a+b+c= 11 $then $10a+b-12c=0......$.....(line 2)

$a+b+c= 19$ then $10a+b-36c=0.....$......(line 3)

$a+b+c= 21$ then $10a+b-44c=0....$.......(line 4)

since a, b, c are digits, so, $a-b, b-c,$ or $c-a$, should be an integer, now we can find from the three lines that only for line one c-a (or a-c) is an integer and c-a =1, which means $c= a+1$

so we should consider only the equations derive from line 1 now

since the number is like $100a+10b+c$, it can be written now $100a+10b+1+a= 101a+10b+1$

since c-a=1 again we can find $2a+b= 8$

or $b= 8-2a$, again entering the value of b the number becomes like this

$101a+80-20a+1=81a+81$

so ultimately the number is =$81(a+1)$

considering a>=5 we see the first digit of $81(a+1)$ is not the same as a (example a=5 then the number is 486,here a=4,contradiction

a<5 we can see the frist digit of 81(a+1)= a

so the numbers are $81(a+1), a= [1,4]$

now my question, how can the digit become 605??? , please give me help..

### Re: Number Theory Problem

Not necessarily, e.g. $(a+b+c)^2-1$ can be $5\pmod{10}$ while $c$ is any even digit.Avik Roy wrote:Lets rearrange the given equation as $100a + 10b = c[(a+b+c)^2 - 1]$

For the right hand side to be divisible by $10$, $a+b+c$ has to be $\pm 1 modulo 10$...

"Everything should be made as simple as possible, but not simpler." - Albert Einstein