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UPSC Prelims 2025 CSAT, question 47

Topic: Number system

Q47

Let P=QQQP=QQQ be a 3-digit number. What is the HCF of PP and 481 ?

  1. (a) 1
  2. (b) 13
  3. (c) 37Correct answer
  4. (d) 481

Solution.

Answer: (c) 37

  1. Any 3-digit repdigit P=QQQP = QQQ (where Q∈{1,2,…,9}Q \in \{1, 2, \dots, 9\}) can be factored as P=111×Q=3×37×QP = 111 \times Q = 3 \times 37 \times Q.
  2. The prime factorization of 481481 is 481=13×37481 = 13 \times 37.
  3. Since QQ is a single non-zero digit (1≤Q≤91 \le Q \le 9), QQ cannot be divisible by 1313, so HCF⁡(P,481)=HCF⁡(3×37×Q,13×37)=37\operatorname{HCF}(P, 481) = \operatorname{HCF}(3 \times 37 \times Q, 13 \times 37) = 37.