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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?

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    how can this be possible? The total sum is $45$ as you point out...but that is an odd number, hence can not be split into two equal sums. – 2017-01-07
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    @lulu The sums don't have to be equal. They may differ by 11, for example ... – 2017-01-07
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    As another remark, note that if it were possible then you could interchange the hundreds place with any other even place without changing divisibility. – 2017-01-07
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    DIVISIBILITY CONDITION FOR 9 – 2017-01-07
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    @jerry Please do not use all caps. It is annoying...looks like you are shouting. – 2017-01-07
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    OK ..........@lulu – 2017-01-07
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    @CatalinZara Good point, thanks. Did I also blunder in my second comment? – 2017-01-07
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    @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 $1651272$. – 2017-01-07
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    This question is busted. – 2017-01-07
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    @lulu Your second comment is correct and leads to the conclusion that any digit can appear on any position. – 2017-01-07
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    IF BUSTED,,,,NO ANSWER THEN @TonyK – 2017-01-07
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    @CatalinZara that might be too strong. My most recent comment shows that I can get $\{6,4,5,2\}$ in the hundreds place. Not sure about the rest (but of course you may well be right...I just stopped looking after I got one partition that worked). – 2017-01-07
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    @jerry Can you stop using all caps? Is your caps lock key broken or something? – 2017-01-07
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    Another viable partition has $\{7,3,6,1\}$ in the even slots... – 2017-01-07
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    @lulu Any digit can appear on a sum of four or five digits equal to 17 or 28. – 2017-01-07
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    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. – 2017-01-07
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    @CatalinZara Indeed you are correct. Just took me longer to persuade myself. – 2017-01-07
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    Also, there is no explicit restriction on the number and place(s) of $0$ digits. – 2017-01-07

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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.

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    How can you say that it have more than $2880$ solution. Can you suggest at least $1$ different from the above mentioned. – 2017-01-08
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    @JustinBieber Why not just interchange $6+5+3+2+1=17$ and $9+8+7+4=28$ those are also $2880$ again – 2017-01-08
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    OK, Thats good.Nice answer +1 – 2017-01-08