Explanation:
<em>Your premise and conclusion are both incorrect</em>.
120 has the prime factorization ...
120 = 2³ × 3 × 5
The exponents of the prime factors are {3, 1, 1}. The number of divisors of 120 is the product of the increments of these numbers: (3+1)(1+1)(1+1) = 4·2·2 = 16.
120 has 16 natural-number divisors
__
1200 has the same prime factor and two more: 2·5, so the factorization is ...
1200 = 2⁴ × 3 × 5²
These exponents are {4, 1, 2} and the product of their increments is 5·2·3 = 30.
1200 has 30 natural-number divisors.
__
When a number is multiplied by 2, its number of natural-number divisors will be (n+1)/n times the previous number of divisors, where n is the increment of the exponent of 2 that is a factor of the original number.
For 120, 2^3 is the power of 2 that is a factor. So, multiplying the number by 2 multiplies the number of divisors by (4+1)/4 = 5/4.
240 will have 20 divisors versus the 16 that 120 has.
Likewise, multiplying this number (240) by 5 will multiply its number of divisors by (2+1)/(1+1) = 3/2, where 1 is the power of 5 that is a factor of 240.
1200 will have 20(3/2) = 30 divisors, versus the 20 that 240 has, or the 16 that 120 has.
__
<em>Summary</em>
The number of divisors of an integer is the product of the increments of the exponents of its prime factors.
_____
<em>For reference</em>
Here are the divisors of 120:
1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120
Here are the divisors of 240:
1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 40, 48, 60, 80, 120, 240
Here are the divisors of 1200:
1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 25, 30, 40, 48, 50, 60, 75, 80, 100, 120, 150, 200, 240, 300, 400, 600, 1200
:
⇒
:
![b=\sqrt[3]{\frac{95}{4}}](https://tex.z-dn.net/?f=b%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B95%7D%7B4%7D%7D)
![y=4(\sqrt[3]{\frac{95}{4}})^{x}](https://tex.z-dn.net/?f=y%3D4%28%5Csqrt%5B3%5D%7B%5Cfrac%7B95%7D%7B4%7D%7D%29%5E%7Bx%7D)
