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# HELP DIVISOR STUFF

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$$4^3 \cdot 5^4 \cdot 6^2$$

How many distinct, natural-number factors does it have

Creeperhissboom  Apr 24, 2018
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#1
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4^3. 5^4. 6^2 =1,440,000.

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

Guest Apr 24, 2018
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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.

$$\small \text{To find the total number of divisors of a number, factor out the primes }\\ \small \text{and add one (1) to the exponent of each prime. }\\ (4^3) \cdot (5^4) \cdot (6^2) = (2^8) \cdot(3^2)\cdot(5^4)\\ (8+1)(2+1)(4+1) = \mathbf {135} \text { divisors for } (4^3) \cdot (5^4) \cdot (6^2)\\ \text{ }\\ \text{Additional notes: }\\ (2^8) \cdot(3^2)\cdot(5^4)\\ \small \text{All factors will be in this form } (2^x) \cdot(3^y)\cdot(5^z) \\ \small \text {where x,y,z are} \geq 0 \text{ and} \leq \{8,2,4\} \small \text{ the highest prime factor exponents of the number itself.} \\ \small \text {As such, the calculation of all individual divisors may be found by iterating (stepping through) each }\\ \small \text{prime’s exponent from zero to the maximum and multiplying by the other primes. }\\ \sum_\limits {x=0}^{8} \sum_\limits {y=0}^{2}\sum_\limits {z=0}^{4} ((2^x)(3^y)(5^z)) =5,188,183 \text{ (sum of divisors) }\\$$

GA

GingerAle  Apr 24, 2018