What is the method to find maximum possible factors of any no. like 999?

Question

Please answer it as soon as possible

Answer 1

To find all factors of a number N: First you need to find the prime factors. Divide by prime numbers up to square root of N (you don't need to go further since if there's a bigger divisor then there will also be a smaller one). Note some factors may be repeated!

For example, 999 = (3)(333) = (3)(3)(111) = (3)(3)(3)(37) and 37 is prime so you can stop. So 999 = (3^3)(37) as prime factorization.

Now find all combinations of the factors. In the example you have 1, 3, 3^2, 3^3, 37, (3)(37), (3^2)(37), (3^3)(37) so the factors are 1, 3, 9, 27, 37, 111, 333, 999. (You did ask for ALL the factors, not just prime factors??)

Make sure you combine all the powers of all the primes. For example, 225 = (3^2)(5^2) and so its factors are 1, 3, 3^2, 5, (3)(5), (3^2)(5), 5^2, (3)(5^2), (3^2)(5^2) or 1, 3, 9, 5, 15, 45, 25, 75, 225.

Check problem statement to see if you are asked for "proper factors" of N, meaning that you leave out 1 and N from the list.

Answer 2

Find divisors of 999,
Divisors are 1, 3, 9, 27, 37, 111, 333, 999
999 = 1 × 999
= 3 × 333
= 9 × 111
= 27 × 37
Factor of 999 = 27 × 37
Prime factors = 3 × 3 × 3 × 37

Answer 3

Keep dividing until you have nothing left but prime numbers!

999 = 3x3x3x37 = 3^3x37

some quick techniques:

Even numbers always have 2 as a factor.
If the sum of the digits is divisible by 3 then 3 is a factor.
If the number ends on 5 then 5 is a factor,
If the number ends in 0 both 2 and 5 are factors.

Answer 4

I have to say, this is probably the most unique optimization problem I've seen in a long time. The first step is to count the number of fights that can occur based on x. We have three sets of children x of John's children who fight with (24-x) other children (x+1) of Mary's children who fight with (23-x) other children (23-2x) of John and Mary's children who fight with (2x+1) other children Now, if we add all of these together, we'll actually be counting every fight twice. (I'll just leave the 2 out of the derivative though, as it applies to everything) Thus, double the number of fights, F(x), is F(x)=x(24-x)+(x+1)(23-x)+(23-2x)(2x+1) =46+90x-6x^2 F'(x)=90-12x So, x=15/2 is the critical point. Checking the sign line assures you this is a maximum. However, we can't have a non-integer number of children, so our answers must be 7 or 8. Check F(7)=F(8)=382. So 382/2=196 is our answer (as we counted twice the number of fights).

Answer 5

by trying all primes smaller than floor ( sqrt(999) )

Answer 6

LCM