Math, asked by mailtokamal9888, 1 year ago

Find the greatest number which is such that when 697, 909 and 1227 are divided by it, the remainders are all the same?

Answers

Answered by Soniapati1967
0

Answer:

answer is 106.

Step-by-step explanation:

I hope it will help u

Answered by steffiaspinno
0

The greatest number is 106

Explanation:

Given Numbers:

1. 697, 909 and 1227

2. The numbers which are divided by it, the remainders are all the same.

To find :

The HCF of the given numbers

Solution:

==> We have to find the difference in the numbers

==> let x=697

==> y=909

==> z=1227

==>  we know that, (x-y,y-z,z-x)

==> Change the number to avoid negative values

==> x-y = 909-697 = 212

==> y-z = 1227-909= 318

==> z-x = 1227-697 = 530

==> HCF of 212, 318, 530

==> Prime factorization of 212

==> 212÷2 = 106

==> 106÷2 = 53

==> 53÷53 =1

==> Prime factorization of 318

==> 318÷2 = 159

==> 159÷3 = 53

==> 53÷53 =1

==> Prime factorization of 530

==> 530÷2 = 265

==> 265÷5 = 53

==> 53÷53 =1

Prime factorization of 212  is 2×2×53

Prime factorization of 318  is 2×3×53

Prime factorization of 530  is 2×5×53

The HCF of 212, 318, 530 is 2×53

The HCF of 212, 318, 530 is 106

The greatest number is 106

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