Loading [MathJax]/jax/output/SVG/jax.js
 
+0  
 
0
950
2
avatar+166 

435462

How many distinct, natural-number factors does it have

 Apr 24, 2018
 #1
avatar
0

4^3. 5^4. 6^2 =1,440,000.

1,440,000 = 2^8×3^2×5^4 (14 prime factors, 3 distinct)

 Apr 24, 2018
 #2
avatar+2234 
+3

Solution:

A divisor of a number is an integer that divides the number with a remainder of zero. All integers greater than one (1) have at least two (2 )divisors. If a number has only two divisors then it is a prime number.

 

To find the total number of divisors of a number, factor out the primes and add one (1) to the exponent of each prime. (43)(54)(62)=(28)(32)(54)(8+1)(2+1)(4+1)=135 divisors for (43)(54)(62) Additional notes: (28)(32)(54)All factors will be in this form (2x)(3y)(5z)where x,y,z are0 and{8,2,4} the highest prime factor exponents of the number itself.As such, the calculation of all individual divisors may be found by iterating (stepping through) each prime’s exponent from zero to the maximum and multiplying by the other primes. 8x=02y=04z=0((2x)(3y)(5z))=5,188,183 (sum of divisors) 

 

 

GA

 Apr 24, 2018

4 Online Users

avatar