CSAT Solved Papers/ 2025/Q64

2025 CSAT — Q64

Quant Number theory 2.5 marks Medium

Consider the following statements:

I. There exists a natural number which when increased by 50%50\% can have its number of factors unchanged.

II. There exists a natural number which when increased by 150%150\% can have its number of factors unchanged.

Which of the statements given above is/are correct?

  1. A I only
  2. B II only
  3. C Both I and II Answer
  4. D Neither I nor II

Worked rationale

Both are existence claims — one witness each settles them. Let d(⋅)d(\cdot) denote the number of factors.

Statement I (increase by 50%50\%: N→3N2N \to \tfrac{3N}{2}). Try N=2N = 2:

2→2×1.5=3,d(2)=2 {1,2},d(3)=2 {1,3}.2 \to 2 \times 1.5 = 3,\qquad d(2) = 2\ \{1,2\},\quad d(3) = 2\ \{1,3\}.

Same factor count. So such a number exists — I is TRUE.

Statement II (increase by 150%150\%: N→5N2N \to \tfrac{5N}{2}). Try N=2N = 2 again:

2→2×2.5=5,d(2)=2,d(5)=2.2 \to 2 \times 2.5 = 5,\qquad d(2) = 2,\quad d(5) = 2.

Same factor count. So such a number exists — II is TRUE.

(Any prime pp with 3p2\tfrac{3p}{2} or 5p2\tfrac{5p}{2} landing on another prime works; p=2p=2 does both.)

Answer: (c) Both I and II.

Why the other options miss

  • A
    missed a case: finds a witness for 50%50\% but assumes none exists for 150%150\%, not testing N=2→5N=2 \to 5.
  • B
    missed a case: the mirror error — tests 150%150\% but not 50%50\%.
  • D
    solved the wrong question: reads “unchanged” as “for all NN” and, finding it fails generally, wrongly rejects both, missing that existence needs just one example.

Specialist insight

The make-or-break reading is “there exists” — these are not “for all” claims, so a single example proves each, and there is no need to characterise all such numbers. The fast witness is the smallest prime N=2N=2: ×1.5→3\times1.5 \to 3 and ×2.5→5\times2.5 \to 5 both land on primes (factor count stays 22). Candidates lose this by hunting for a general rule or by testing large composite NN where the factor count does change, then wrongly concluding non-existence. Existence claims reward the smallest clean witness, not a proof of universality.

The trap, in one line

Both are existence claims: N=2N=2 works for both (2→32\to3 and 2→52\to5, factor count stays 22), so both hold.

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