CAT 2025

Slot 1

QA

Question & Solution

1. Concept Used

• Topic: Number Systems — Digit Constraints & Number of Factors
• Formula: $$ \text{Number of Factors} = (a_1 + 1)(a_2 + 1)(a_3 + 1)\cdots \text{ where } N = p_1^{a_1} \cdot p_2^{a_2} \cdot p_3^{a_3}\cdots $$

2. Calculation

We need to find the digits that satisfy ALL of the following conditions simultaneously: the digits must be non-zero, distinct, none is a perfect square, and exactly one is a prime.

First, let us identify which single-digit numbers are perfect squares: $$1, 4, 9$$ — these are eliminated.

The remaining non-zero digits are: $$2, 3, 5, 6, 7, 8$$.

Among these, the prime digits are: $$2, 3, 5, 7$$ and the non-prime digits are: $$6, 8$$.

Since exactly one digit must be prime and the other two must be non-prime, and the only available non-prime digits (excluding perfect squares) are $$6$$ and $$8$$, the two non-prime digits are fixed as $${6, 8}$$.

The prime digit can be any one of $${2, 3, 5, 7}$$. To minimize the 3-digit number N, we should:

• Choose the smallest prime digit: $$2$$
• Arrange the digits $${2, 6, 8}$$ in ascending order to form the smallest number: $$N = 268$$

Now, we factorize $$268$$: $$268 = 2 \times 134 = 2 \times 2 \times 67 = 2^2 \times 67$$

Since $$67$$ is a prime number, the prime factorization is complete.

Using the factor-count formula: $$\text{Number of factors} = (2 + 1)(1 + 1) = 3 \times 2 = 6$$

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3. Solution

Answer = 6 ✅

The minimum possible value of N is 268, and it has 6 factors: $$1, 2, 4, 67, 134, 268$$.