How many five-digit prime numbers can be obtained by using all the digits 1, 2, 3, 4 and 5 without repetition of digits?
- (a) Zero ✓ UPSC's answer
- (b) One
- (c) Nine
- (d) Ten
Why the answer is (a)
• The sum of the digits 1, 2, 3, 4, and 5 is 1 + 2 + 3 + 4 + 5 = 15.
• A fundamental divisibility rule states that a number is divisible by 3 if and only if the sum of its digits is divisible by 3.
• Since 15 is divisible by 3, every five-digit number formed by permuting the digits 1, 2, 3, 4, 5 (in any order, without repetition) will have a digit-sum of 15 and therefore be divisible by 3.
• A prime number is a number greater than 1 whose only positive divisors are 1 and itself; any number greater than 3 that is divisible by 3 has at least three divisors (1, 3, and itself) and is therefore composite.
• Because every such five-digit number is divisible by 3 and is far greater than 3, none of them can be prime, so the count of five-digit primes obtainable this way is zero.
• Hence the answer is option (a).
Why the other options are wrong
- (b) One
- One prime would require at least one permutation of the digits to yield a number not divisible by 3, but the fixed digit-sum of 15 makes every permutation divisible by 3, so no single prime exists.
- (c) Nine
- Nine primes would require nine of the 120 permutations to be prime, yet the digit-sum argument shows all 120 are divisible by 3, leaving none prime.
- (d) Ten
- Ten primes would require ten of the 120 permutations to be prime, but the same digit-sum divisibility by 3 eliminates every single one of them.
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2020, held on 4 October 2020. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.