Number Theory Problem

For students of class 11-12 (age 16+)
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Abdul Muntakim Rafi
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Number Theory Problem

Unread post by Abdul Muntakim Rafi » Thu Jun 09, 2011 6:56 pm

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

$abc = c (a+b+c)^2$
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prodip
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Re: Number Theory Problem

Unread post by prodip » Fri Jun 10, 2011 1:27 am

$(a,b,c) =(3,2,4) $

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Avik Roy
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Re: Number Theory Problem

Unread post by Avik Roy » Tue Aug 30, 2011 11:21 pm

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
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Shifat
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Re: Number Theory Problem

Unread post by Shifat » Wed Sep 28, 2011 1:01 am

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??? :o, please give me help..

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nayel
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Re: Number Theory Problem

Unread post by nayel » Wed Sep 28, 2011 5:12 pm

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 $\pm 1 modulo 10$...
Not necessarily, e.g. $(a+b+c)^2-1$ can be $5\pmod{10}$ while $c$ is any even digit.
"Everything should be made as simple as possible, but not simpler." - Albert Einstein

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