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Can you use a number twice when working out HCF using prime factors?
Wiki User
∙ 11y ago
Yes.
Wiki User
∙ 11y ago
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What number is divisible by 1 2 3 4 5 6 10?
List the prime factors of each of these numbers, then multiply all the prime factors. The prime factor 2 has to be included twice, since it appears twice in the number 4. Other prime factors, in this case, only have to be included once.
How can you prove that a perfect number can never be a prime number?
In a perfect number, the sum of all the factors (including the number itself) is twice the number. E.g., the sum of the factors of 6 is 1 + 2 + 3 + 6 = 12 (equal to 2 x 6). Every prime number has two factors: 1, and itself. So, the sum of the factors is only one more than the prime number itself; for any number greater than 1, this can't be twice the number. For example, the prime number 7 has the factors 1 and 7, which add up to 8.
What is the largest perfect square factor of 210?
Except in this case, the largest square factor is 1.
What are factors of square numbers?
The prime factors are the factors of its square root, each one appearing twice as often.
can fifteen over twenty six simplified?
your face lolNo because they have no factors in common, because 26 is twice 13, a prime number.
What is a number whose prime factors are 3 squared x 11 x 7 x 5 x 2?
6930. The answer is obtained by multiplying the factors together (3 being used twice).
What two prime numbers equal 44?
To find the primes that equal a number, start with any factor pair of the number and keep factoring the composite factors until all factors are prime. 44 2 x 22 2 x 2 x 11 = 44 The two prime numbers that equal 44 are 2 (twice) and 11.
What is a composite number between 60b and 70c that has prime factors that have a sum of 12a?
The answer will depend on a, b and c. For example, if b = 7 and c = 6 then 60b and 70c are the same number, 420. That has prime factors 2 (twice), 3, 5, 7 which sum to 17. So a would have to be 17/12.
What is largest twice digit prime number?
That would be 11.
Give a proof that the square root of 7 is an irrational number?
Proof by contradiction: suppose that root 7 (I'll write sqrt(7)) is a rational number, then we can write sqrt(7)=a/b where a and b are integers in their lowest form (ie they are fully cancelled). Then square both sides, you get 7=(a^2)/(b^2) rearranging gives (a^2)=7(b^2). Now consider the prime factors of a and b. Their squares have an even number of prime factors (eg. every prime factor of a is there twice in a squared). So a^2 and b^2 have an even number of prime factors. But 7(b^2) then has an odd number of prime factors. But a^2 can't have an odd and an even number of prime factors by unique factorisation. Contradiction X So root 7 is irrational.
A number for which sum of all its factors is equal to twice the number is called?
composite number
What is the number called when all its factors add up to twice the number?
multiply
