Using Euclid’s division algorithm, find the largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3, respectively.

Solution:
Since, 1, 2 and 3 are the remainders of 1251, 9377 and 15628 respectively.
So, 1251 – 1 = 1250 is exactly divisible by the required number,
9377 – 2 = 9375 is exactly divisible by the required number,
15628 – 3 = 15625 is exactly divisible by the required number.
So, required number = HCF of 1250, 9375 and 15625.
By Euclid’s division algorithm,
15625 = 9375 x 1 + 6250
9375 = 6250 x 1 + 3125
6250 = 3125 x 2 + 0
=> HCF (15625, 9375) = 3125
3125 = 1250 x 2 + 625
1250 = 625 x 2 + 0

HCF(3125, 1250) = 625
So, HCF (1250, 9375, 15625) = 625
Hence, the largest number is 625.

Since, 1, 2 and 3 are the remainders of 1251, 9377 and 15628 respectively. Thus after subtracting these remainders from the numbers , we get

1251 – 1 = 1250, 9377 − 2 = 9375 and 15628 − 3 = 15625 which is divisible by the required number.

Now, required number = HCF (1250, 9375, 15625)

By Euclid’s division algorithm, b = a × q + r, 0 ≤ r < a

Here, b is any positive integer .

Firstly put b = 15625 and a = 9375

⇒ 15625 = 9375 × 1 + 6250

⇒ 9375 = 6250 × 1 + 3125

⇒ 6250 = 3125 × 2 + 0

So, HCF (9375, 15625) = 3125

Now, put a = 1250 and b = 3125

⇒ 3125 = 1250 × 2 + 625

⇒ 1250 = 625 × 2 + 0

So, HCF (1250, 3125) = 625

Hence, 625 is the largest number which divides 1251, 9377 and 15628 leaving remainder 1, 2 and 3, respectively.