The required number when divides 626, 3127 and 15628, leaves remainder 1, 2 and 3. This means
626 – 1 = 625, 3127 – 2 = 3125 and
15628 – 3 = 15625 are completely divisible by the number
∴ The required number = HCF of 625, 3125 and 15625
First consider 625 and 3125
By applying Euclid’s division lemma
3125 = 625 × 5 + 0
HCF of 625 and 3125 = 625
Now consider 625 and 15625
By applying Euclid’s division lemma
15625 = 625 × 25 + 0
∴ HCF of 625, 3125 and 15625 = 625
Hence required number is 625