Suppose the number n = 90k is the smallest integer that has exactly ninety divisors (all positive integers) which includes 1 and n. Find k Suppose the number n = 90k is the smallest integer that has exactly ninety divisors (all
positive integers) which includes 1 and n. Find k.
I know this question has been answered, but I am wondering if there is an elementary way to answer this using modular arithmetic and/or Euler's totient function?
 A: We have, $N=90k=2\cdot3^2\cdot5\cdot(k)$
Number of divisors of a number is product of the $(\text{powers}+1)$ of each of the it's prime factors.
Since we want exactly $90$ divisors, so the product would look like:
$$\begin {align} 90 &=2\cdot3^2\cdot5 \\ &= 2\cdot3\cdot3\cdot5 \tag A \\ &= 6\cdot3\cdot5 \tag B \\ &= 2\cdot9\cdot5 \tag C \\ &= 2\cdot3\cdot15 \tag D\end{align}$$
Note: $(A),(B),(C),(D)$ limit our prime factors to 3 or 4 only.
Case (A):
$N=ab^2c^2d^4$
Here, we need to assign lowest prime number to the factor with highest power.
Thus, $a=7,b=5,c=3,d=2$
Thus, $N=25200$
Case (B):
$N=a^5b^2c^4$
Here, we need to assign lowest prime number to the factor with highest power.
Thus, $a=2,b=5,c=3$
Thus, $N=64800$
Case (C):
$N=ab^8c^4$
Here, we need to assign lowest prime number to the factor with highest power.
Thus, $a=5,b=2,c=3$
Thus, $N=103680$
Case (D):
$N=ab^2c^{14}$
Here, we need to assign lowest prime number to the factor with highest power.
Thus, $a=5,b=3,c=2$
Thus, $N=737280$
So, the smallest number satisfying the given conditions is $25200$.
$\therefore 90k=25200$
$$\Rightarrow \boxed {k=280}$$
