# Find the largest number which on dividing 1251, 9377 and 15628 leaves remainders 1, 2 and 3 respectively.

By BYJU'S Exam Prep

Updated on: September 25th, 2023

625 is the largest number which on dividing 1251, 9377 and 15628 leaves remainders 1, 2 and 3 respectively. To know how we get this answer, refer to the detailed solution below. The method used to define the answer is Euclid’s Dilemma which is used to find out the highest common factor between two or more numbers.

## Largest Number Divisible By 1251, 9377 and 15628

to calculate the largest number dividing 1251, 9377 and 15628 and leaving remainders 1, 2 and 3 respectively, we will use Euclid’s dilemma.

For it, we will first subtract the numbers with their respective remainders –

1251 – 1 = 1250

9377 – 2 = 9375

15628 – 3 = 15625

Now, from these newly derived numbers, we will find the HCF using Euclid’s Dilemma –

9375 = 1250 × 7 + 625

Here, we have the remainder which is not zero,

Thus, now we will divide 1250 and 625 –

1250 = 625 × 2 + 0

We now have the zero as a reminder when we divide 1250 by 625. Thus, HCF (1250, 9375) is 625.

The next step is to calculate the HCF of 15625 and 625 which we will divide in the following way –

15625 = 625 × 25 + 0

The remainder is again zero. Hence, HCF (15625, 625) is 625.

From these calculations, we have found that the HCF (1250, 9375, 15625) is 625.

## Euclid’s Method

Euclid’s dilemma is a division method that helps find the HCF of two or more numbers. As per this method, there are two positive integers, let’s say, a and b. With them, there also exist unique integers q and r where q is the quotient and r is the remainder, with both being distinct. According to this method, a = bq + r, 0≤ r

Summary:

## Find the largest number which on dividing 1251, 9377 and 15628 leaves remainders 1, 2 and 3 respectively.

The largest number that divides 1251, 9377 and 15628 leaving 1, 2 and 3 as remainders is 625. We calculated this answer by finding out the HCF of all these numbers by using Euclid’s Dilemma.

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