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Q: What are the prime number b?

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180 is not a prime number.

14th Prime Minister

It is: B 31 is a prime number because it has only two factors which are itself and one

6 is not a prime number b\c, it is divisible to 2 and 3

yes b/c 1 is neither prime or composite

21 is not a prime.

Any number of the form a*b^4 where a and b are different prime numbers, or c^9 where c is a prime, will have exactly 10 factors.

He was the 11th Prime Minister, since 1930 - 1935.

293 is a prime number, therefore the prime factorization of it would be 293, you cant do 293x1 because 1 is NOT a prime number therefore you cant do it because it wouldn't b a prime factorization that is your answer

If √7 is rational, then it can be expressed by some number a/b (in lowest terms). This would mean: (a/b)² = 7. Squaring, a² / b² = 7. Multiplying by b², a² = 7b². If a and b are in lowest terms (as supposed), their squares would each have an even number of prime factors. 7b² has one more prime factor than b², meaning it would have an odd number of prime factors. Every composite has a unique prime factorization and can't have both an even and odd number of prime factors. This contradiction forces the supposition wrong, so √7 cannot be rational. It is therefore irrational.

The only number which is not a prime number in that list is 49 - it is equal to seven squared.

B, as a variable, can stand for any number. The factor possibilities are infinite.

62 is not a prime number and you cannot get a relitavely prime number it either is a prime number or it isnt a prime number! The definition of a prime number is a number that can only be divided b itself and one. No other number 62 can be divided by 2 as well as itself and 1 so NO IT IS NOT A PRIME NUMBER even numbers tend not to be prime numbers because they can be divided by two the only even number that is prime is 2 because it can only be divided by itself and 1 I hope this helped!

c.23

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 factor of a integer is an integer that divides the second integer into a third integer exactly; i.e. A is a factor of B if B/A is exactly C, where all of A, B and C are integers. A prime factor is a factor as above, but is also a prime number. This means that the only factors of that factor are one and the number itself; i.e. A is a prime factor of B if B/A is exactly C andthe only factors of A are 1 and A.

Any number of the form n = a*b*c*d*e*f where a, b, c, d, e and f are different prime numbers. n has 26 = 64 factors in total, of which 1 is the number 1 (neither prime nor composite), 6 are prime, and the remaining 57 are composite.

B may be any prime number. The fact that C is at least 3A has no bearing on the limits of B.

91 = 13*751 = 17*321 = 3*731 is prime. â–

a factor is a factor , a prime number is a prime number, but a prime factor is a factor which is a prime number. it has to depend on what number number. exp 13 is a prime number

You is called a prime number!You is called a prime number!You is called a prime number!You is called a prime number!

2310

17 = 23 + 32

Prime numbers only have two factors, one and themselves. If Player A picks the prime number, Player B gets the 1.

No.