Question

find the hcf long divisible method of 506,1155​

Answer 1

this is your answer

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Answer 2

Approach 1: Integer numbers prime factorization

506 = 2 × 11 × 23;

1,155 = 3 × 5 × 7 × 11;

Take all the common prime factors, by the lowest exponents.

Greatest (highest) common factor (divisor):   gcf, gcd (506; 1,155) = 11

Approach 2: Euclid's algorithm

Step 1. Divide the larger number by the smaller one:  1,155 ÷ 506 = 2 + 143;

Step 2. Divide the smaller number by the above operation's remainder:  

506 ÷ 143 = 3 + 77;

Step 3. Divide the remainder from the step 1 by the remainder from the step 2:  

143 ÷ 77 = 1 + 66;

Step 4. Divide the remainder from the step 2 by the remainder from the step 3:  

77 ÷ 66 = 1 + 11;

Step 5. Divide the remainder from the step 3 by the remainder from the step 4:  

66 ÷ 11 = 6 + 0;

At this step, the remainder is zero, so we stop:  

11 is the number we were looking for, the last remainder that is not zero.  

This is the greatest common factor (divisor).

Greatest (highest) common factor (divisor):  

gcf, gcd (506; 1,155) = 11