Question:
A five-digit number is formed using the digits 1, 2, 3, 4, and 5 without repetition. What is the probability that the number is divisible by 4?
A five-digit number is formed using the digits 1, 2, 3, 4, and 5 without repetition. What is the probability that the number is divisible by 4?
Show Hint
Only the last two digits decide if a number is divisible by 4, so check which digit pairs work.
Updated On: Jul 14, 2026
- \(\frac{1}{5}\)
- \(\frac{5}{6}\)
- \(\frac{4}{5}\)
- None of these
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The Correct Option is A
Solution and Explanation
Step 1: Recall the divisibility rule for 4.
A number is divisible by 4 exactly when the number formed by its last two digits is divisible by 4. So only the last two digits of the five-digit number decide divisibility here.
Step 2: List the possible last-two-digit pairs.
Using two different digits from \(1,2,3,4,5\), the two-digit endings divisible by 4 are 12, 24, 32, and 52. Check: \(12/4=3\), \(24/4=6\), \(32/4=8\), \(52/4=13\), all whole numbers, so these 4 endings work, and no other pair from these digits does.
Step 3: Count the favourable numbers.
For each of these 4 endings, the remaining 3 digits can fill the first three places in \(3! = 6\) ways. So the count of 5-digit numbers divisible by 4 is \(4 \times 3! = 4 \times 6 = 24\).
Step 4: Count the total numbers and check the wrong options.
The total number of 5-digit arrangements of these 5 distinct digits is \(5! = 120\). Option B, \(\frac{5}{6}\), and option C, \(\frac{4}{5}\), are both far too high for how restrictive the divisibility condition is, so they are wrong.
Final Answer:
Probability \[ \boxed{\frac{24}{120} = \frac{1}{5}} \]
A number is divisible by 4 exactly when the number formed by its last two digits is divisible by 4. So only the last two digits of the five-digit number decide divisibility here.
Step 2: List the possible last-two-digit pairs.
Using two different digits from \(1,2,3,4,5\), the two-digit endings divisible by 4 are 12, 24, 32, and 52. Check: \(12/4=3\), \(24/4=6\), \(32/4=8\), \(52/4=13\), all whole numbers, so these 4 endings work, and no other pair from these digits does.
Step 3: Count the favourable numbers.
For each of these 4 endings, the remaining 3 digits can fill the first three places in \(3! = 6\) ways. So the count of 5-digit numbers divisible by 4 is \(4 \times 3! = 4 \times 6 = 24\).
Step 4: Count the total numbers and check the wrong options.
The total number of 5-digit arrangements of these 5 distinct digits is \(5! = 120\). Option B, \(\frac{5}{6}\), and option C, \(\frac{4}{5}\), are both far too high for how restrictive the divisibility condition is, so they are wrong.
Final Answer:
Probability \[ \boxed{\frac{24}{120} = \frac{1}{5}} \]
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