How many three digit numbers are there in which all three digits are odd?

Introduction / Context:
This is a counting and basic combinatorics question. It asks for the number of three digit numbers whose every digit is odd. The problem tests whether the learner can apply the fundamental principle of counting by identifying independent choices for each digit position and then multiplying the number of choices.

Given Data / Assumptions:

• We are looking at three digit numbers, so the hundred's digit cannot be zero.
• Each digit must be odd.
• The odd digits available in the decimal system are 1, 3, 5, 7, and 9.
• Repetition of digits is allowed unless restricted, and the question does not restrict it.

Concept / Approach:
A three digit number has three positions: hundred's, ten's, and unit's digits. Each position must be filled with an odd digit. There are five odd digits available. Since the question does not forbid repetition, each position can independently take any of the five odd digits. The total number of such numbers is the product of the number of choices for each position.

Step-by-Step Solution:
Available odd digits are 1, 3, 5, 7, and 9.Hundred's place: cannot be zero, but all five odd digits are allowed, so there are 5 choices.Ten's place: also must be odd, again 5 choices (1, 3, 5, 7, 9).Unit's place: must be odd, so 5 choices.The choices for different positions are independent.Therefore, total number of three digit numbers with all digits odd = 5 * 5 * 5 = 5^3.Compute 5^3 = 125.So there are 125 such three digit numbers.

Verification / Alternative check:
We can think of the count as the size of the set of all triples (d1, d2, d3) where each digit is chosen from {1, 3, 5, 7, 9}.Since there are 5 choices for each digit and three positions, the total count is 5^3 = 125, which matches the earlier calculation.No restriction on repetition is mentioned, so combinations like 111 or 999 are included.

Why Other Options Are Wrong:
100, 250, and 500 correspond to different mistaken combinations of choices, such as assuming fewer or more options for each place or misapplying the multiplication principle.For instance, 100 could come from incorrectly thinking some odd digits are not allowed in a particular position.Only 125 correctly realizes that there are 5 independent odd digit choices for each of the three positions.

Common Pitfalls:
Some learners mistakenly exclude 1 or 9 for certain positions without any logical reason.Another mistake is to treat the problem as if digits cannot repeat, which would give a permutation count instead of a simple product of choices.Confusing two digit and three digit numbers can also lead to the wrong exponent in 5^n.

Final Answer:
The number of three digit numbers in which all digits are odd is 125.