This is Q27 from Australian Maths 2013.

$3$ different non-zero digits are used to form $6$ different $3$-digit numbers.The sum of $5$ of them is $3231$.What is the $6$ th number?

What I tried:

Let $a,b,c$ be the different digits.

$(100a+10b+c)+(100a+10c+b)+(100b+10a+c)+(100b+10c+a)+(100c+10a+b) =3231 $

From there,I can see that

$a+2b+2c =10x +1 $,where $x$ is some integer.

$2a+b+2c =10j+(3-x)$,where $j$ is some integer.

$2a+2b+c =32-j$

Using substitution to sub in the values of $j$ and $x$,

$221a +212b+122c=3231$,which leads me back to where I started from...


With three digits $a,b,c$, You should be able to get at most six different $3$-digits numbers, and they are: $abc,acb,bac,bca,cab,cba$
So when you add them up, the equation should be$$200(a+b+c)+20(a+b+c)+2(a+b+c)=222(a+b+c)=3231+n$$where $n$ is the unknown $6$th number. By quick estimation you can find that,
When $a+b+c=15, n=99$, does not qualify;
When $a+b+c=16, n=321$, does not qualify;
When $a+b+c=17, n=543$, does not qualify;
When $a+b+c=18, n=765$, qualified.

  • $\begingroup$ What do you mean by estimation?Like how do you know what are some values to estimate? $\endgroup$ – Arc Neoepi Jun 27 '16 at 15:47
  • $\begingroup$ @ArcNeopi So you want to find a number that's larger than $3231$, it is easy to notice that $200\times15$ is $3000$, so $222\times15$ will roughly be some number near $3231$ $\endgroup$ – Paul Jun 27 '16 at 15:49
  • $\begingroup$ But,what if $a+b+c$ is a very big number like(for example), $27$ ?Then do we have to do till we reach $27$ ?Because no calculator is allowed too. $\endgroup$ – Arc Neoepi Jun 27 '16 at 15:53
  • $\begingroup$ No it will never be $27$, your $n$ is a three-digit number $\endgroup$ – Paul Jun 27 '16 at 15:58
  • $\begingroup$ For example,$a=9,b=9,c=9$,then $n =999$ but that would cause the RHS to not equal to the LHS right? $\endgroup$ – Arc Neoepi Jun 27 '16 at 16:02

The sum of all six numbers is $222(a+b+c)$. Now you can check the multiples of $222$ wich exceed $3231$, to find that $222\cdot18$ does the job.

  • $\begingroup$ So,$a+b+c$=18.So then,I have to use elimination to find each value of $a,b,c$? $\endgroup$ – Arc Neoepi Jun 27 '16 at 15:49
  • 1
    $\begingroup$ No, $18\cdot 222=3996$, and $3996-3231=765$, so you know the digits are $7$, $6$ and $5$ - and you weren't even asked for the digits, just for the sixth number. $\endgroup$ – Henrik Jun 27 '16 at 15:52
  • $\begingroup$ Wow,thanks.Umm,this may be a bit off topic,but how do you recognize what to do so easily( for the question)? $\endgroup$ – Arc Neoepi Jun 27 '16 at 15:57
  • $\begingroup$ When I saw your five-term sum I immediately recognized that in the full sum $a+b+c$ would factor out. And finding divisors of a number is very likely to help you in finding it. $\endgroup$ – Aretino Jun 27 '16 at 19:23

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