Answer

Verified

415.2k+ views

**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.

Recently Updated Pages

what is the correct chronological order of the following class 10 social science CBSE

Which of the following was not the actual cause for class 10 social science CBSE

Which of the following statements is not correct A class 10 social science CBSE

Which of the following leaders was not present in the class 10 social science CBSE

Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE

Which one of the following places is not covered by class 10 social science CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

The states of India which do not have an International class 10 social science CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

How do you graph the function fx 4x class 9 maths CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Why is there a time difference of about 5 hours between class 10 social science CBSE

Name the three parallel ranges of the Himalayas Describe class 9 social science CBSE