3
$\begingroup$

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...

$\endgroup$
9
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.