Alok is attending a workshop "How to do more with less" and today's theme is…
2024
Alok is attending a workshop "How to do more with less" and today's theme is Working with fewer digits . The speakers discuss how a lot of miraculous mathematics can be achieved if mankind (as well as womankind) had only worked with fewer digits. The problem posed at the end of the workshop is How many 5 digit numbers can be formed using the digits 1, 2, 3, 4, 5 (but with repetition) that are divisible by 4?
Can you help Alok find the answer?
- A.
375
- B.
625
- C.
500
- D.
312
Attempted by 222 students.
Show answer & explanation
Correct answer: B
Key idea: Only the last two digits determine divisibility by 4.
Allowed digits for each position: 1, 2, 3, 4, 5.
Valid two-digit endings (from the allowed digits) that are divisible by 4: 12, 24, 32, 44, 52.
Number of such endings = 5.
The first three digits can each be any of the five digits, so there are 5^3 = 125 choices for the first three positions.
Total five-digit numbers divisible by 4 = number of valid endings × choices for first three digits = 5 × 5^3 = 5^4 = 625.