14 is deficient. It is less than the sum of it's divisors. In mathematics, a deficient number or defective number is a number n for which σ(n) < 2n. Here σ(n) is the sum-of-divisors function: the sum of all positive divisors of n, including n itself Proof.. divisors of 14 are 1,2, and 7 and 14. Now, 2n=28 and and the sum the all the divisors including 14 is 24<28
the divisors of 6 are: 1,2,3, and 6.
The divisors of 21 are: 1, 3, 7, 21.
No it is not. The closest perfect number to it is 28. The divisors of 20 are 1,2,4,5,10 those add up to more than 20 so 20 is an abundant number.
The divisors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24.
1,2,4,7,14,28
An example - the divisors of 28 are 1 -2 - 4 - 7 - 14 and 28. In other words 1+2+4+7+14 = 28
Yes
28 has six divisors: 1, 2, 4, 7, 14, 28.
Actually, 56 has eight divisors: 1, 2, 4, 7, 8, 14, 28, 56.
1, 2, 7, and 14 are common divisors (and common factors) of 28 and 42.
14 is deficient. It is less than the sum of it's divisors. In mathematics, a deficient number or defective number is a number n for which σ(n) < 2n. Here σ(n) is the sum-of-divisors function: the sum of all positive divisors of n, including n itself Proof.. divisors of 14 are 1,2, and 7 and 14. Now, 2n=28 and and the sum the all the divisors including 14 is 24<28
It isn't a perfect number.A perfect number is a number which is the sum of its proper divisors (including 1, but excluding itself).The divisors of 20 are 1, 2, 4, 5, 10.1 + 2 + 4 + 5 + 10 = 22 => 20 is not a perfect number.An example of a perfect number is 28:The divisors of 28 are 1, 2, 4, 7, 141 + 2 + 4 + 7 + 14 = 28 => 28 is a perfect number.
The divisors of 280 are: 1, 2, 4, 5, 7, 8, 10, 14, 20, 28, 35, 40, 56, 70, 140, 280.
They are: 1, 2, 4, 7, 8, 14, 28, 56.
A perfect number is a number, which, when adding all of its proper divisors (all divisors except himself) give the number itself Divisors of 6: 1,2,3 ------>1+2+3=6 The next proper number is 28 Divisors of 28:1,2,4,7,14 -------> 1+2+4+7+14=28 All perfect numbers found so far are found using this formula: 2^(p-1)*((2^p)-1), where p, and ((2^p)-1) are prime numbers
1, 2, 4, 7, 8, 14, 16, 28, 56, 112.