say the co prime numbets of 51 and 76
Answers
Step-by-step explanation:
Prime Factorization of a number: finding the prime numbers that multiply together to make that number.
51 = 3 × 17;
51 is not a prime, is a composite number;
76 = 22 × 19;
76 is not a prime, is a composite number;
* Positive integers that are only dividing by themselves and 1 are called prime numbers. A prime number has only two factors: 1 and itself.
* A composite number is a positive integer that has at least one factor (divisor) other than 1 and itself.
Calculate greatest (highest) common factor (divisor):
Multiply all the common prime factors, by the lowest exponents (if any).
But the two numbers have no common prime factors.
gcf, hcf, gcd (51; 76) = 1;
coprime numbers (relatively prime)
Coprime numbers (relatively prime) (51; 76)? Yes.
Numbers have no common prime factors.
gcf, hcf, gcd (51; 76) = 1.
>> Integer numbers prime factorization
Approach 2. Euclid's algorithm:
This algorithm involves the operation of dividing and calculating remainders.
'a' and 'b' are the two positive integers, 'a' >= 'b'.
Divide 'a' by 'b' and get the remainder, 'r'.
If 'r' = 0, STOP. 'b' = the GCF (HCF, GCD) of 'a' and 'b'.
Else: Replace ('a' by 'b') & ('b' by 'r'). Return to the division step above.
Step 1. Divide the larger number by the smaller one:
76 ÷ 51 = 1 + 25;
Step 2. Divide the smaller number by the above operation's remainder:
51 ÷ 25 = 2 + 1;
Step 3. Divide the remainder from the step 1 by the remainder from the step 2:
25 ÷ 1 = 25 + 0;
At this step, the remainder is zero, so we stop:
1 is the number we were looking for, the last remainder that is not zero.
This is the greatest common factor (divisor).
gcf, hcf, gcd (51; 76) = 1;
Why is the answer a factor (a divisor) of the initial 'a' and 'b'?
Note: 'a' ÷ 'b' = 'q' + 'r' is equivalent to the equation: 'a' = 'q' × 'b' + 'r', where 'q' is the quotient of the operation.
When the final value of 'r' = 0, the final value of 'b' is a factor (a divisor) of the final value of 'a', since 'a' = 'q' × 'b' + 0.
Go backwards the previous division steps, through each equation, 'a' = 'q' × 'b' + 'r', and notice that at each step the final value of 'b' is a factor (a divisor) of each value of 'r' and of each value of 'b' and therefore is a factor of each value of 'a'. So the final value of 'b', which is the last remainder in our list that is not zero, is a factor of the initial values of ('a' and 'b'), or in other words, is a divisor of the intial values of ('a' and 'b').
Why is the answer equal to the CGF (HCF, GCD)?
Look at all the equations: 'a' = 'q' × 'b' + 'r'. As we saw above, the final value of 'b' is a factor of all the values of 'a', 'b', and 'r'.
Therefore the final value of 'b' must also be a factor of the last value of 'r', the one that is not zero. And the final value of 'b' couldn't be larger than that value. But the final value of 'b' is actually equal to that value of 'r', therefore the final value of 'b' is the largest factor (divisor) of the initial values of 'a' and 'b'. And by definition it's called the greatest (highest) common factor
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