Question

If consecutive whole numbers from $1$ onwards are written one after another to the right of$1$, find the digit at the $31^{\text{st}}$place.
A. $1$
B. $2$
C. $3$
D. $0$

Answer 1

Hint:
We write down our numbers as $1{\text{ 2 3 4 }}...................{\text{upto 31st place }}$then count which digit falls at that place then we will get our answer.

Complete step by step solution:
Here we are given that the consecutive whole numbers from $1$ onwards are written one after another to the right of$1$ and we need to find which digit is at the $31^{\text{st}}$ place. There is no other way rather than counting the terms first of all we should know what the whole numbers are. Whole numbers are the numbers starting from zero till infinity like $0,1,2,3,4,.........{\text{ and so on}}$that means the positive integers containing zero are called the whole numbers.
So we must know that every whole number is the integer but every integer is not the whole number as there are also the negative terms in the integers while not in the whole numbers.
Here in the question we are given that if the consecutive whole numbers from $1$ onwards are written one after another to the right of $1$ and we need to find the digit at the $31^{\text{st}}$ place.
So let us write the given statement in the form of the numerical digits so the numbers would be
$1{\text{ 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20}}$
So here we get that 1st term is 1 and the second term is 2 and so on at the 31st place we get that there is the term 20 ending so we get that $0$ is the term at the $31^{\text{st}}$ place
So we can say that $0$ is the term at the $31^{\text{st}}$ place

So D is the correct option.

Note:
Every natural number is the whole number but every whole number is not the natural number as the natural numbers do not contain zero which is contained in the whole numbers.