It is easy to find a rule based on a polynomial of order 6 such that the first six numbers are as listed in the question followed by the chosen next number. There are also non-polynomial solutions. Short of reading the mind of the person who posed the question, there is no way of determining which of the infinitely many solutions is the "correct" one.The simplest rule, based on a polynomial of order 5 isU(n) = (24*n^5 - 595*n^4 + 5310*n^3 - 21185-n^2 + 36726*n - 19920)/120 for n = 1, 2, 3, ...
How about: 3*7 = 21 or 1*21 = 21
20 and 1
Because 21 is divisible by numbers other than 1 and 21. 21 is divisible by 1, 3, 7, 21
The numbers 1 and 3.
1. The two numbers are coprime.
1. The two numbers are coprime.
No.Prime numbers are numbers that can only be divided evenly by itself and 1.For example, the number 7 can only be divided by 7 and 1.Odd numbers are numbers that end in (or are) 1, 3, 5, 7, or 9.For example, the number 21 is odd since it ends in a 1, but it is not prime because itcan be divided by 1, 21, 7, and 3. (7 times 3 = 21.)2 is a prime number and is even not odd - it is the only even prime number.
No, the numbers 19, 21, and 23 are not composite numbers. A composite number is a positive integer greater than 1 that has more than two distinct positive divisors. In the case of these numbers, 19 and 23 are prime numbers because they are only divisible by 1 and themselves, while 21 is a composite number because it has divisors other than 1 and itself (3 and 7).
Each new number in the series is the sum of the previous two numbers. That sequence is part of an infinite series called the Fibonacci series.
The factors of 21 are 1, 3, 7, and 21. The only factor of 21 that is a composite number is 21.
They are: 25-4 = 21
A prime number has exactly two factors, 1 and the number itself. The factors of 21 are 1, 3, 7, and 21, and the factors of 27 are 1, 3, 9, and 27. Since 21 and 27 have more than 2 factors, they are composite numbers.