Question 1207622: Let $m$ be a positive integer. If $m$ has exactly $18$ positive divisors, then how many positive divisors does $m^2$ have?
Found 2 solutions by greenestamps, math_tutor2020:
Answer by greenestamps(13200)   (Show Source):
Answer by math_tutor2020(3816)  (Show Source):
You can put this solution on YOUR website!
The prime p has exactly two divisors: 1 and itself.
Example: the prime 7 has factors 1 and 7 only.
p^k has k+1 divisors and they are:
p^0, p^1, p^2, ..., p^(k-1), p^k
Note that p^0 = 1 and p^1 = p.
Example: p = 3 and k = 4 yields p^k = 3^4 = 81 having k+1 = 4+1 = 5 divisors.
Those five divisors are 1, 3, 9, 27, 81.
Let p and q represent two different primes.
Let k and m be positive integers.
We can construct the following composite number
n = p^k*q^m
How many divisors does n have?
Imagine we can form a table that has k+1 rows and m+1 columns.
Along the left hand side we will have the labels p^0, p^1, p^2, ..., p^(k-1), p^k.
Along the top header we will have the labels q^0, q^1, q^2, ..., q^(m-1), q^m.
This table generates (k+1)*(m+1) different divisors.
Each entry into the table is the result of multiplying the headers. Eg: multiply p^2 with q^3 to get p^2q^3 as one divisor of n.
This logic can be extended to as many unique prime factors as you want.
Example: n = p^k*q^m*r^s will have (k+1)*(m+1)*(s+1) divisors.
We add 1 to each exponent, then multiply those results.
More info found here
https://mathworld.wolfram.com/DivisorFunction.html
Refer specifically to equation (3) on that page.
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So this is why tutor greenestamps had written various ways to factor 18.
His goal is to write an equation like
18 = (k+1)(m+1)
or
18 = (k+1)(m+1)(s+1)
Let me know if you have any questions.
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