Show all factor pairs, prime factorization (factor tree), sum of factors (divisors), aliquot sum, and prime power decomposition of 465.

We do this by listing out all pairs of numbers greater than 0 and less than or equal to 465 who have a product equal to 465:

465 = 1 x 465

465 = 3 x 155

465 = 5 x 93

465 = 15 x 31

There are 4 factor pairs of 465.

1, 3, 5, 15, 31, 93, 155, 465

1, 3, 5, 15, 31, 93, 155, 465

Proper factors are all factors except for the number itself, in this case 465

1, 3, 5, 15, 31, 93, 155

Now, show the prime factorization (factor tree) for 465 by expressing it as the product of ALL prime numbers.

465 = 3 x 155 <--- 3 is a prime number

Next step is to reduce 155 to the product of prime numbers:

155 = 5 x 31 <--- 5 is a prime number

Next step is to reduce 31 to the product of prime numbers:

No prime power decomposition exists since there are no duplicate prime numbers in the prime factorization:

1 + 465 + 3 + 155 + 5 + 93 + 15 + 31 =

The aliquot sum is the sum of all the factors of a number except the number itself

1 + 3 + 155 + 5 + 93 + 15 + 31 =