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CAT 2025Slot 1QAQues & Sol

Number SystemsMedium

Question

In a 3-digit number N, the digits are non-zero and distinct such that none of the digits is a perfect square, and only one of the digits is a prime number. Then, the number of factors of the minimum possible value of N is

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: $\text{Number of factors} = (2 + 1)(1 + 1) = 3 \times 2 = 6$

3. Solution

Answer = 6 ✅

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