Question

Using the digits from $0$ to $9$, how many $3$-digit numbers can be constructed such that the number must be odd and greater than $500$ and digits may be repeated?

Answer 1

This question involves the arithmetic operation of addition/ subtraction/ multiplication/ division. We need to know the possibilities for the first term in $3$ digit number, the possibilities for the second term in $3$ digit number, and the possibilities for the third term in $3$ digit number according to the condition that is given in the problem. Also, we need to know the difference between odd numbers and even numbers.

Complete step by step solution:
In this question, we have to find how many $3$ digit number can be constructed within the
following conditions,

1. The term of the answer would be from $0$ to $9$.
2. All $3$ digit numbers would be odd numbers.
3. All the$3$ digit numbers would be greater than $500$.
   Let’s assume the $3$ digit number is $ABC$
   According to the condition $\left( 1 \right)$ and $\left( 3 \right)$ the value of A would be $5,6,7,8,9 \to$ Five possibilities
   According to the condition $\left( 2 \right)$ the value of $C$ would be,
   $1,3,5,7,9 \to$ Five possibilities
   According to the condition $\left( 1 \right)$ the value of$B$would be,
   $0,1,2,3,4,5,6,7,8,9 \to$ Ten possibilities
   So, the total possibilities of $ABC$ is = 5 \times 5 \times 10$$$$ = 250 possibilities
   So, the final answer is,
   There are $250$ possibilities of $3$ digit numbers can be constructed within the given conditions that are mentioned in the question.

Note: This question involves the arithmetic operations like addition/ subtraction/ multiplication/ division. Also, note that the odd numbers always end with $1,3,5,7,9$ and the prime numbers always end with $0,2,4,6,8$. Take care while using the conditions in the problem to solve these types of questions.