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The first Prime number is 2
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Can you prove If m n and p are three consectuive integers then mnp is even?
If m, n, and p are three consecutive integers, then one of them must be even. Let's say the even number is m. Since m is even, it is divisible by two, and so can be written as 2*k, where k is some integer. This means that m*n*p = 2*k*n*p. Since we are multiplying the quantity k*n*p by 2, it must be divisible by two, and therefore must be even.
The variance of first n natural numbers is?
1 Sum of first n natural numbers = n(n+1)2[Formula.]2 Arthmetic mean of first n natural numbers = Sum of the numbers n[Formula.]3 = n(n+1)2n = n+124 So, the Arthmetic mean of first n natural numbers = n+12
How do you calculate prime factors?
Let n be the number whose prime factors we so desire to know. Required knowledge: All prime numbers less than sqrt(n).Test n for divisibility by each such prime numbers, starting with 2:If a prime number, p, is found to divide n, divide n by p, record p and continue (test for divisibility by p again) using n/p in the place of n.The recorded prime factors are the prime factors of n.
How can I how that for any prime number p square root of p is an irrational number?
Suppose not. Let p be prime and n = Sqrt[p]. Since p is an integer, if n is rational, then n is also an integer. So we have n.n = p. But since p is prime, only 1 and p divide p. -><- therefore n must not be rational
If you have a calculate the original number from a percent increase?
Let's set this equation up. Call the original number No and the number you have N and the percentage increase P. The equation to get the number you have ( N ) is No + No x P = N and we want to solve for No so No ( 1 + P ) = N No = N / ( 1 + P )
