Question

All possible 3 digit numbers are formed using all the digits \(2,3,5,7,9\) without using any digit more than once. Among these 3 digit numbers, the number of numbers which are divisible by \(3\) but not divisible by \(5\) is

• A. \(24\)
• B. \(22\)
• C. \(20\)
• D. \(18\)

Hint:

For divisibility by \(3\), first identify valid digit combinations using the sum-of-digits rule. Then count permutations and separately remove those ending in \(5\) if divisibility by \(5\) is excluded.

Answer 1

The Correct Option is C

Concept: A number is divisible by \(3\) if the sum of its digits is divisible by \(3\). A number is divisible by \(5\) if its units digit is \(0\) or \(5\).

Step 1: Find all possible selections of three digits whose sum is divisible by \(3\). The given digits are \[ 2,\;3,\;5,\;7,\;9. \] Possible groups of three digits are: \[ \{2,3,5\},\quad \{2,3,7\},\quad \{2,3,9\},\quad \{2,5,7\}, \] \[ \{2,5,9\},\quad \{2,7,9\},\quad \{3,5,7\},\quad \{3,5,9\}, \] \[ \{3,7,9\},\quad \{5,7,9\}. \] Their digit sums are \[ 10,\;12,\;14,\;14,\;16,\;18,\;15,\;17,\;19,\;21. \] Hence, the valid groups are \[ \{2,3,7\},\quad \{2,7,9\},\quad \{3,5,7\},\quad \{5,7,9\}. \]

Step 2: Count numbers formed from these groups. Each group contains distinct digits, so the number of 3-digit numbers formed from each group is \[ 3!=6. \] Thus, \[ 4\times 6=24 \] numbers are divisible by \(3\).

Step 3: Remove numbers divisible by \(5\). A number is divisible by \(5\) if it ends in \(5\). Among the four valid groups, only \[ \{3,5,7\} \quad \text{and} \quad \{5,7,9\} \] contain the digit \(5\). For each such group, fixing \(5\) in the units place, \[ 2!=2 \] numbers can be formed. Hence, numbers divisible by both \(3\) and \(5\) are \[ 2+2=4. \]

Step 4: Find the required count. \[ 24-4=20. \] Therefore, the number of 3-digit numbers divisible by \(3\) but not divisible by \(5\) is \[ \boxed{20} \] \[ \boxed{\text{Answer = (C)}} \]