# 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?

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?

• As another remark, note that if it were possible then you could interchange the hundreds place with any other even place without changing divisibility. – lulu Jan 7 '17 at 17:32
• @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 $1651272$. – lulu Jan 7 '17 at 17:39
• This question is busted. – TonyK Jan 7 '17 at 17:39
• Another viable partition has $\{7,3,6,1\}$ in the even slots... – lulu Jan 7 '17 at 17:44
• 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. – lulu Jan 7 '17 at 17:45

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