1512 = 2³ × 3³ × 7

Given:
The HCF of the 3-digit integer and 1512 = 21

Now 21 = 3 × 7

Since the HCF is 21, we know that the integer must only have one prime factor 3 and one prime factor 7 , that is in common with 1512.

The powers of these prime factors can only be 1. (otherwise the HCF will be higher)

We also realise that the 3-digit integer cannot have prime factor of 2 (otherwise the HCF will be higher since 1512 contains prime factor 2)

This also means it cannot be a multiple of 2 and powers of 2.

To get the answer, we need to multiply 21 (3¹×7¹) by a prime factor that is not 2,3 or 7.


The smallest possible is 5.

The smallest 3-digit integer possible would be:

3 × 7 × 5 (or 3¹ × 7¹ × 5¹ if you prefer to see the powers listed)
= 21 × 5
= 105

Excellent explanation, J. Salute.

Dear @J,
Great explanation always. ✨✫✰