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**Hint:**In this type of question we will try to observe its pattern which it follows. Now if we talk about \[{\left( {10} \right)^n}\]then we know that \[{10^1} = 10,{10^2} = 100,{10^3} = 1000\]. We observe that the number which is in power that no. of zeroes are written after digit 1. So, in that way we can write \[{\left( {10} \right)^n} = 1000...................n{\text{ }}terms\]. Now, in the same way we can write \[{\left( {10} \right)^{1000}}\] that is after digit 1, \[1000\] zero must be written. Then we have to subtract 1 from it. So

\[\begin{array}{c}

{\left( {10} \right)^{1000}} = {10^1} \times {10^2} \times {10^3}...............................{\text{ }}upto{\left( {10} \right)^{1000}}\\

= 10000.......................................000 = 1001{\text{ }}digits

\end{array}\]

Now, subtract it by 1 we will get \[999999.......upto{\text{ }}1000{\text{ }}digits\]

Then we can check the divisibility of \[9{\text{ }}and{\text{ }}11\] and can reach upto our answer.

**Step by step Solution:**

So, applying above concept we will get

\[\begin{array}{c}

{\left( {10} \right)^{1000}} = {10^1} \times {10^2} \times {10^3}...............................{\text{ }}upto{\left( {10} \right)^{1000}}\\

= 10000.......................................000 = 1001{\text{ }}digits

\end{array}\]

Now, subtract it by 1 we will get \[1000\]digits that is

(i)For divisibility of 9

We can see that each digit is fully divisible by 9 without leaving remainder. So, if we write it as

\[\dfrac{{999.........99}}{9} = 111......111 = 1000{\text{ }}digit\], here each digit will get divided.

So we can say that \[{\left( {10} \right)^{1000}} - 1\] is divisible by \[9\]

(ii)Divisibility by 11

As we have seen above \[{\left( {10} \right)^{1000}} - 1 = 9999..............99 = 1000{\text{ }}digits\]. We can see that 11 divides the first two digits that is 99 fully without leaving any remainder.

So, we can also write \[99999.........99 = as{\text{ }}500{\text{ }}pairs{\text{ }}of{\text{ }}99{\text{ }}to{\text{ }}get\]

Now, dividing it by 11, we get

\[\dfrac{{999........999}}{{11}} = 909090........09\]

0 comes in between because, after dividing the first two digits we need to add zero to get the next two digits for division at the same time.

**So, from above calculations we can write \[{\left( {10} \right)^{1000}} - 1{\text{ }}is{\text{ }}divisible{\text{ }}by{\text{ }}both{\text{ }}9{\text{ }}and{\text{ }}11\].**

**Note:**

One more method to check divisibility by 9 is that the total sum of the digits of the number must be a multiple of 9 or we can say that it must be fully divisible by 9 without leaving remainder.while expanding any power we must write all digits very carefully.

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