Q18·CSAT · Prelims 2022

Permutation of odd digits for a 3-digit number divisible by 5

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Question

How many 3-digit natural numbers (without repetition of digits) are there such that each digit is odd and the number is divisible by 5?

Options

Answer

Explanation

We need to form a 3-digit number using exclusively odd digits without repetition. The available pool of odd digits is: \{1, 3, 5, 7, 9\}.

The number must be divisible by 5. A number is divisible by 5 if it ends in 0 or 5. Since all digits must be odd, the units digit is rigidly locked to 5. This consumes the digit 5, leaving 4 available odd digits \{1, 3, 7, 9\} for the remaining two slots (hundreds and tens).

Step 1: Choose the hundreds digit from the 4 remaining options. (4 ways)

Step 2: Choose the tens digit from the 3 remaining options. (3 ways)

Step 3: The units digit is fixed as 5. (1 way)

Total valid numbers = 4 × 3 × 1 = 12.

When building numbers under restrictive divisibility conditions, lock the units digit first to satisfy the divisibility rule, then arrange the remaining available digits in the higher-order slots.

Answer: (b).

Question details

Year

2022

Paper

CSAT

Question

Q18

Section

Numerical Ability

Sub-topic

Permutations & Combinations