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How many factors does 2p have if p is an odd prime number?
It has four factors: 1, 2, p, and 2p.
Why is the product of two odd numbers is odd?
Here's the following proof that "product of two odd numbers is odd". Proof: Any odd number can be written in the form 2p+1. Let your first odd number be written in this form. Let your second odd number be written as 2q+1 - essentially the same form as above, but since p may not equal q, separate variables are used. Thus: First odd number x second odd number = (2p+1)(2q+1) = 2pq + 2p +2q +1. The plus one on the end indicates that the product is odd, as required.
Is odd times odd always odd?
Yes, it's always odd, and here's the proof: All odd numbers can be expressed as 2p + 1, where p is any integer. Multiply two of those together: (2n + 1)(2p + 1) = 4np + 2n + 2p + 1 = 2(np + n + p) + 1. Since both np, n, and p are integers, that means np + n + p is an integer; and since that integer is being multiplied by 2, it must be even. Thus, by adding 1 to that even number, the result will be odd.
What is a mersume prime number?
Let p = any prime number. (2p -1) is called a Mersenne number. Any such number that is prime is called a Mersenne Prime. Father Mersenne wrote a list of numbers of this type which he thought were prime, but a few were not. In fact, most of the large Mersenne numbers are not prime, but all the really large numbers that have been proved to be prime are Mersenne Primes.
What is a prime number that is one more than a square number?
A prime number that is one more than a square number is known as a Sophie Germain prime. Sophie Germain primes are of the form 2p+1, where p is a prime number. For example, 17 is a Sophie Germain prime because it is one more than the square of 4, and both 4 and 17 are prime numbers. Sophie Germain primes have applications in number theory and cryptography.
How many factors does 2p have if p is an odd prime number?
It has four factors: 1, 2, p, and 2p.
How many perfect numbers are known?
So far 47. Euler proved that every even perfect number will be of the form 2p−1(2p−1), where p is prime and 2p−1 is also prime. If 2p−1 is prime it is known as a Mersenne prime. Since 47 Mersenne primes are known, 47 even perfect numbers are known. As for odd perfect numbers, none are known, nor has it been proven yet that there aren't any.
Why is the product of two odd numbers is odd?
Here's the following proof that "product of two odd numbers is odd". Proof: Any odd number can be written in the form 2p+1. Let your first odd number be written in this form. Let your second odd number be written as 2q+1 - essentially the same form as above, but since p may not equal q, separate variables are used. Thus: First odd number x second odd number = (2p+1)(2q+1) = 2pq + 2p +2q +1. The plus one on the end indicates that the product is odd, as required.
Which three prime numbers make 121 if 1 is 10 more than the other?
Because only two prime numbers are multiplied together to make 121, I assume you mean which three prime numbers can be added together to make 121. I will also assume that all three prime numbers must be different.Since one number is 10 more than another, we have the equation p + (p + 10) + q = 2p + 10 + q = 121, which can be changed to 2p + q = 111. Since 2p will be an even number, q must be an odd number in order to sum to 121 which is an odd number. The smallest odd prime is 3. If we substitute 3 for q, we have 2p + 3 = 111, so 2p = 108, and p would be 54. Thus, the correct value for p must be less than 54. So, let's look at primes less than 54 to find ones that will produce primes when 10 is added to them.The prime numbers less than 54 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, and 53, but it cannot be the even prime 2. Our possible pairs are {3, 13}, {7, 17}, {13, 23}, {19, 29}, {31, 41}, {37, 47}, and {43, 53}. Any number ending with 3, will have the value of 2p ending in a 6, which when subtracted from 111 will have a final digit of 5, which will not be prime, so we can eliminate {3, 13} and {13, 23}. Try the first of each remaining pair as p in the equation q = 111 - 2p to see if the resulting value for q is prime. For the value of p = 7, q is the prime number 97. For the value of p = 19, q is the prime number 73. For the value of p = 37, q is the prime number 37, which we will exclude as a possibility because then there is duplication of the prime number. So, the two possibilities are p = 7 and p = 19.Thus the possible answers are121 = 7 + 17 + 97121 = 19 + 29 + 73
Is odd times odd always odd?
Yes, it's always odd, and here's the proof: All odd numbers can be expressed as 2p + 1, where p is any integer. Multiply two of those together: (2n + 1)(2p + 1) = 4np + 2n + 2p + 1 = 2(np + n + p) + 1. Since both np, n, and p are integers, that means np + n + p is an integer; and since that integer is being multiplied by 2, it must be even. Thus, by adding 1 to that even number, the result will be odd.
Why do two even numbers always equal an even number?
On a number line, adding an even number to another number (or zero) results in an even displacement, which must end in the same type of number as the original. If the beginning number is odd, adding an even number produces an odd sum. If the beginning number is even, adding an even number produces an even number. A corollary is that: Adding two like numbers produces an even number. Adding two unlike numbers produces an odd number. ---------------------------------- Algebraically: let x be an even number, Then y = 2x for some value of x Two even numbers would be 2m and 2n Adding them gives: 2m + 2n = 2(m + n) = 2p where p = m + n; 2p is of the form y = 2x, so 2p is an even number. Thus adding two even numbers results in an even number. Similarly for odd numbers: If y = 2x is an even number then z = y + 1 = 2x +1 is an odd number. Two odd numbers would be 2m+1 and 2n+1 Adding them gives: 2m+1 + 2n + 1 = 2m + 2n + 2 = 2(m + n + 1) = 2p where p = m + n + 1 Thus adding two odd numbers results in an even number. Similarly for one even and one odd number. An even number would be 2m and an odd number would be 2n+1 Adding them gives: 2m + 2n + 1 = 2(m + n) + 1 = 2p + 1 where p = m + n Thus adding an even number and an odd number results in an odd number.
What is an odd number multiplied by an odd number?
A definition of an odd number is 2n-1 where n is any integer. The product of two odd numbers is thus (2n-1)(2p-1) for any numbers n and p. Expanding the expression yields 2(2pn -(p+n)) + 1. The term on the left is even by definition. Adding 1 to any even number is, by definition, an odd number. Therefore, the conclusion is elementary (Watson); the product of two two odd numbers is an odd number
What is a mersume prime number?
Let p = any prime number. (2p -1) is called a Mersenne number. Any such number that is prime is called a Mersenne Prime. Father Mersenne wrote a list of numbers of this type which he thought were prime, but a few were not. In fact, most of the large Mersenne numbers are not prime, but all the really large numbers that have been proved to be prime are Mersenne Primes.
Is 13 a perfect number?
No, it is not.Incidentally, is a number p is prime then its 2 factors are 1 and p. Their sum is p+1.If p is a perfect number then this sum must be 2p.That is, p+1 = 2p or p = 1.But 1 is not a prime. So no prime can be a perfect number and since 13 is prime, it cannot be a perfect number.
How many factors does 2p have?
The number 2p has 4 factors: 1, 2, p, and 2p.
When you multiply an odd number and an even number is your answer always odd?
No, it's always even, and here's the proof: All even numbers can be expressed as 2n, where n is any integer. All odd numbers can be expressed as 2p + 1, where p is any integer. Multiply those two together: 2n(2p + 1) = 2(2np + n). Since both 2np and n are integers, that means 2np + n is an integer; and since that integer is being multiplied by 2, it must be even.
Do perfect numbers have to be prime numbers?
Perfect numbers cannot be prime numbers. Here's why:A number N is perfect if σ(N) = 2N (σ is the sum of divisors function). If there is a prime p that is a perfect number, then σ(p) = 2p. However, the only factors of p are 1 and p, so σ(p) is also equal to p+1. If 2p = p+1, then p=1, which is not prime, and 1 is defined to have only one factor, 1.
