Question

The 300-digit number with all digits equal to 1 is
A.Divisible by neither 37 nor 101
B.Divisible by 37 but not by 101
C.Divisible by 101 but not by 37
D.Divisible by both 37 and 101

Answer 1

Hint: In this question, a 300 digit number is given whose all digits are equal to 1; since it is a 300 digit number, it will be difficult to check whether it is divisible by 37 and 101; hence we will try to break the number into the simplest form and then we will check its divisibility.

Complete step-by-step answer:
Given that the number is a 300 digit number where all digits are equal to 1
Let the number be n; hence we can write
\[n = 111.........11\](300 digits)
We can write this number as
\[n = \dfrac{{999.....99}}{9}\](300 digits)
Now this number can be written as
\[
  n = \dfrac{{999.....99}}{9} \\
   = \dfrac{{{{10}^{300}} - 1}}{9} \\
   = \dfrac{{{{10}^{3 \times 100}} - {1^{100}}}}{9} \\
 \]
Now let \[{1^{100}} = X\], hence by substituting this, we can write the number as
\[
  n = \dfrac{{{{10}^{3 \times 100}} - {1^{100}}}}{9} \\
   = \dfrac{{\left( {{{10}^3} - 1} \right)X}}{9} \\
 \]
Since \[{10^3} - 1 = 999\] is divisible by 37 (\[37 \times 27 = 999\]), hence we can say the number\[n = 111.........11\]will be divisible by 37.
Now let’s check for divisibility by 101
We can also write the number as
\[
  n = \dfrac{{999.....99}}{9} \\
   = \dfrac{{{{10}^{300}} - 1}}{9} \\
   = \dfrac{{{{10}^{4 \times 75}} - {1^{100}}}}{9} \\
 \]
Now let\[{1^{75}} = Y\], hence by substituting this, we can write the number as
\[
  n = \dfrac{{{{10}^{4 \times 75}} - {1^{75}}}}{9} \\
   = \dfrac{{\left( {{{10}^4} - 1} \right)Y}}{9} \\
 \]
Since \[{10^4} - 1 = 9999\] is divisible by 101 (\[101 \times 99 = 9999\]), hence we can say the number\[n = 111.........11\] will be divisible by 101.
Therefore, a 300-digit number with all digits equal to 1 is divisible by both 37 and 101.
Option D is correct.

Note: Students can also write this number as
\[n = 111.........11 = 111 + 111 \cdot 1000 + 111 \cdot {1000^2} + ... + 111 \cdot {1000^{99}}\]
Now we can divide take out 111 as common and check whether the number is divisible by 37 or not; similarly, the number is written in the form
\[n = 111.........11 = 11 + 11 \cdot 100 + 11 \cdot {100^2} + ... + 11 \cdot {100^{149}}\]