
Finding sum of factors of a number using prime factorization
It's factors are $1, 5, 7, 25, 35, 49, 175, 245, 1225 $ and the sum of factors are $1767$. A simple algorithm that is described to find the sum of the factors is using prime factorization.
How to determine the smallest numbers with $n$ factors
Oct 23, 2017 · In this case, we want to create the lowest numbers with 12 factors, so we can use the first prime numbers $2$, $3$, and $5$, and arrange them in such a way that we get the smallest value, …
"Divisors" vs "factors" - Mathematics Stack Exchange
Nov 1, 2020 · Is there a difference between divisors and factors? Are they the same or are there differences? Example: Find the factors and divisors of 18. What I have found is answers that …
How to get all the factors of a number using its prime factorization ...
May 16, 2018 · The factors of $9=3^2$ are $3^0, 3^1, 3^2$ aka $1,3,9$. Now the factors of $72$ can be made up by combining two numbers chosen, one from each of these two sets:
factoring - Determine all factors of zero (divisors of $0 ...
Mar 14, 2023 · How many integer factors of $0$ are there, and what are they? I'm just curious, but what counts as a factor of $0$? My guess is that there are an infinite number of factors of $0$, but is there …
Sum of all the factors of $33333333$ - Mathematics Stack Exchange
Jun 9, 2022 · 5 What is the sum of factors of factors of $33333333$ (that's $3$ eight times)? Here's how I tried to attempt this question yet failed: Factors of …
Finding Invariant Factors of a Matrix - Mathematics Stack Exchange
Nov 18, 2020 · Finding Invariant Factors of a Matrix Ask Question Asked 5 years ago Modified 11 months ago
Factor $z^4 +1$ into linear factors - Mathematics Stack Exchange
May 5, 2015 · $z$ is a complex number, how do I factor $z^4 +1$ into linear factors? Do I write z in terms of $x+yi$ so that $z^4+1=(x+yi)^4+1?$
Is there a formula to calculate the sum of all proper divisors of a ...
Feb 19, 2011 · A positive integer factor is the product of 0, 1, 2, or 3 factors of 2, 0 or 1 factor of 3, and 0 or 1 factor of 5. Expanding $ (1+2+2^2+2^3) (1+3) (1+5)$ gives the sum of all possible such products.
probability - Weighted average of multiple weighted factors ...
Mar 26, 2016 · Weighted average of multiple weighted factors Ask Question Asked 9 years, 8 months ago Modified 9 years, 8 months ago