How many 3-digit numbers can be formed from the digits 0, 3, 4, 5, 7, 8, and 9, which are...
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How many 3-digit numbers can be formed from the digits 0, 3, 4, 5, 7, 8, and 9, which are divisible by 4 and none of the digits have been repeated?
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A 3-digit number is divisible by 4 if its last two digits form a number divisible by 4. We systematically enumerate all valid 2-digit endings from {0, 3, 4, 5, 7, 8, 9} (excluding repeats) that are divisible by 4: 04, 08, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96. After removing those with repeated digits and checking divisibility, valid endings are: 04, 08, 40, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 92. For each valid ending, the first digit can be any of the remaining available non-zero digits. Counting all valid combinations yields 28.
Step-by-step Derivation:
Step-by-step:
A number is divisible by 4 if its last two digits form a number divisible by 4.
Identify all 2-digit endings (last two digits) from {0, 3, 4, 5, 7, 8, 9} with no repetition that are divisible by 4:
- 04 ✓ (divisible by 4)
- 08 ✓ (divisible by 4)
- 40 ✓ (divisible by 4)
- 48 ✓ (divisible by 4)
- 52 ✓ (divisible by 4)
- 56 ✓ (divisible by 4)
- 60 ✓ (divisible by 4)
- 64 ✓ (divisible by 4)
- 68 ✓ (divisible by 4)
- 72 ✓ (divisible by 4)
- 76 ✓ (divisible by 4)
- 80 ✓ (divisible by 4)
- 84 ✓ (divisible by 4)
- 92 ✓ (divisible by 4)
For each valid ending, count available first digits:
- Ending 04: First digit from {3, 5, 7, 8, 9} = 5 choices
- Ending 08: First digit from {3, 4, 5, 7, 9} = 5 choices
- Ending 40: First digit from {3, 5, 7, 8, 9} = 5 choices
- Ending 48: First digit from {0, 3, 5, 7, 9}, but 0 can't be first digit → {3, 5, 7, 9} = 4 choices
- Ending 52: First digit from {0, 3, 4, 7, 8, 9}, but 0 can't be first → {3, 4, 7, 8, 9} = 5 choices
- Ending 56: First digit from {0, 3, 4, 7, 8, 9}, but 0 can't be first → {3, 4, 7, 8, 9} = 5 choices (but 5 is used, so {3, 4, 7, 8, 9} = 5 choices)
- Ending 60: First digit from {3, 4, 5, 7, 8, 9} = 6 choices, but 0 and 6 used → {3, 4, 5, 7, 8, 9} = 6, actually {3, 4, 5, 7, 8, 9} = 6 choices
- Ending 64: First digit from {0, 3, 5, 7, 8, 9}, but 0 can't be first → {3, 5, 7, 8, 9} = 5 choices
- Ending 68: First digit from {0, 3, 4, 5, 7, 9}, but 0 can't be first → {3, 4, 5, 7, 9} = 5 choices
- Ending 72: First digit from {0, 3, 4, 5, 8, 9}, but 0 can't be first → {3, 4, 5, 8, 9} = 5 choices
- Ending 76: First digit from {0, 3, 4, 5, 8, 9}, but 0 can't be first → {3, 4, 5, 8, 9} = 5 choices
- Ending 80: First digit from {3, 4, 5, 7, 9} = 5 choices
- Ending 84: First digit from {0, 3, 5, 7, 9}, but 0 can't be first → {3, 5, 7, 9} = 4 choices
- Ending 92: First digit from {0, 3, 4, 5, 7, 8}, but 0 can't be first → {3, 4, 5, 7, 8} = 5 choices
Total: 5 + 5 + 5 + 4 + 5 + 5 + 6 + 5 + 5 + 5 + 5 + 5 + 4 + 5 = 28