Question

using euclid division algorithm find the largest no. that divides 1251,9377&15628 leaving remainder 1 ,2,3 .

Answer 1

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Finding HCF of :-

=>1251-1=1250

=>9377-2=9375

=>15628-3=15625

# H•C•F of 1250 and 9375 :-

=>9375=1250×7+625

=>1250=625×2+0

: Now H•C•F will be 625

: Find H•C•F of all three numbers :-

=>15625=625×25+0.

# Since the H•C•F will be 625........ Answer

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Answer 2

Hey friend, Harish here.

Here is your answer:

Given ,

i)Three numbers, 1251 , 9377 & 15628 leaves remainder 1,2,3 respectively when divided by a number.

To find, 

The largest number that divides them leaving those remainder.

Solution,

Let the largest number that divides them b 'x'. 

Now, When we subtract the remainders from the number x will divide them perfectly.

Then,

→ 1251 - 1 = 1250

→ 9377 - 2 = 9375

→ 15628 - 3 = 15625 .

As 1250, 9375 & 15625 are exactly divisible by x.Then x must be the HCF of them.

So, First (HCF 15625 & 9375 ) 

   15625 = 9375 × 1 + 6250        (using,  a = b(q) + r )

⇒9375 = 6250 × 1 + 3125 

⇒6250 = 3125 × 2 + 0 . 

So, HCF of ( 15625 & 9375 ) is 3125.

Now, We must find HCF of (3123 & 1250 ) to get HCF of all three numbers.

Then,

3125 = 1250 × 2 + 625.

1250 = 625 × 2 + 0.

So, HCF of all three numbers is 625.

Therefore 625 is the largest number which divides 1251, 9377, 15628 leaving remainders 1, 2 , 3 respectively.
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Hope my answer is helpful to you.