Find the greatest number which is such that when 697, 909 and 1227 are divided by it, the remainders are all the same?
Answers
Answer:
answer is 106.
Step-by-step explanation:
I hope it will help u
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