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?

Hint: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.