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# Is ${\left( {10} \right)^{1000}} - 1$ divisible by both $9$ and $11$ ?

Last updated date: 02nd Aug 2024
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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.