4
$\begingroup$

A number '$Z$' contains all the digits from $1$ to $9$ exactly once. $Z$ is divisible by $99$. What will be the number on its hundreds place (i.e. its third-to-last digit)?

$99=9\cdot11$ so the addition of numbers $1,2,3,4,5,6,7,8,9$ is $45$ which is divisible by $9$.

To be divisible by $11$, the sum of the odd places of a number subtracted by the even places of a number must be a multiple of $11$.

How do I proceed?

$\endgroup$
  • 3
    $\begingroup$ As another remark, note that if it were possible then you could interchange the hundreds place with any other even place without changing divisibility. $\endgroup$ – lulu Jan 7 '17 at 17:32
  • 1
    $\begingroup$ @CaitlinZara To illustrate your point, $6+4+5+2=17$ and $1+3+7+8+9=28$. Thus $163475829$ is divisbile by $99$, indeed the quotient is $1651271$. But we can swap the $8$ and the $9$, say, to get $163475928$ to get another possible solution, this time the quotient is $1651272$. $\endgroup$ – lulu Jan 7 '17 at 17:39
  • 1
    $\begingroup$ This question is busted. $\endgroup$ – TonyK Jan 7 '17 at 17:39
  • 2
    $\begingroup$ Another viable partition has $\{7,3,6,1\}$ in the even slots... $\endgroup$ – lulu Jan 7 '17 at 17:44
  • 2
    $\begingroup$ And another has $\{9,1,5,2\}$ and yet another has $\{8,2,4,3\}$ so now I agree...we can get any digit in any slot. $\endgroup$ – lulu Jan 7 '17 at 17:45
1
$\begingroup$

You must found the digits those sum's difference is a multiple of $11$ and place them on odd and even places.

One of the set of solution is $6+4+5+2=17$ and $9+8+7+3+1=28$ since $28-17=11$ Thus $968472351$ is divisible by $99$. But we can swap the odd positioned digits with odd positioned digits and even positioned digits with even positioned digits. So the problem can have more than $(5!\times 4!=2880)$ solutions and I can't provide all the possible solutions one by one.

$\endgroup$
  • $\begingroup$ How can you say that it have more than $2880$ solution. Can you suggest at least $1$ different from the above mentioned. $\endgroup$ – user401699 Jan 8 '17 at 10:11
  • $\begingroup$ @JustinBieber Why not just interchange $6+5+3+2+1=17$ and $9+8+7+4=28$ those are also $2880$ again $\endgroup$ – Harsh Kumar Jan 8 '17 at 10:16
  • $\begingroup$ OK, Thats good.Nice answer +1 $\endgroup$ – user401699 Jan 8 '17 at 10:18

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.