For a positive integer $n$ we denote by $d(n)$ the number of positive divisors of $n$ and by $s(n)$ the sum of these divisors. For example, $d(2018)$ is equal to $4$ since $2018$ has four divisors $(1, 2, 1009, 2018)$ and $s(2018) = 1 + 2 + 1009 + 2018 = 3030$. Determine all positive integers $x$ such that $s(x) \cdot d(x) = 96$. (Richard Henner)
Problem
Source: Austria Beginners' Competition 2018 p4
Tags: number theory, sum of divisors, number of divisors
zonee4
25.02.2020 01:05
The issue is that I have no idea where to start like, we could just plug spam, but that's no fun
Alculator11
25.02.2020 01:18
For a start, I think that we must have $d(x)<10,$ because $s(x)\ge d(x),$ and so $s(x)d(x)$ would be greater than $96$ otherwise.
v--
25.02.2020 01:24
Ans : 14,15,47 Note $x \leq 48$ and just plug in
FailedJerk101
05.01.2021 08:33
Sorry to bump an old post but I’d like to have a crack at this
If the prime decomposition of an positive integer $N$ is
$$N = p_1^{a_1}p_2^{a_2}.....p_n^{a_n}$$Where $p_i$ is prime.
Thus
$$D(N) = (a_1 + 1)(a_2 + 1)....(a_n + 1)$$
$$S(N) = \frac{p_1^{a_1 + 1}- 1}{p_1 - 1} \frac{p_2^{a_2+1}-1}{p_2-1}.......\frac{p_n^{a_n+1}-1}{p_n-1}$$
Now we see that $$96 = 1 X 96 = 2 X 48 = 3 X 32 = 4 X 24 = 6 X 16 = 8 X 12$$Now take note that $2 X 48, 4 X 24$ are possible only Take $d(x) = 2, s(x) = 48$ and $d(x) = 4, s(x) = 24$ to realise that $14 = 2 X 7, 15 = 3 X 5, 47 = 1 X 47$ Are possible Hence, Ans = $14,15,47$ Do let me know if i missed any other cases
TheMathCruncher_007
01.02.2021 18:04
FailedJerk101 wrote: Now take note that $2 X 48, 4 X 24$ are possible only I used exactly the same method as you did , but I didn't jump to this conclusion straightaway . How did you do so ? Also you didn't miss any cases
megarnie
15.08.2021 06:37
The answer is $14,15,47$.
Obviously $d(x)$ must be a divisor of $96$. If $d(x)\ge6$, then $s(x)\le16$, so $x<16$, and $12$ is the only positive integer less than $16$ with at least $6$ divisors. $s(12)\ne16$, so $d(x)<6$.
Therefore, $d(x)=2,3,4$.
Case 1: $d(x)=2$
$x$ is prime. So $s(x)=x+1$, $(x+1)2=96, x=47$.
Case 2: $d(x)=3$
$x$ is the square of a prime number and $s(x)=32$. $x=4,9,25$ don't work so this case has no solutions.
Case 3: $d(x)=4$
$s(x)=24$. We can test out values of $x$ less than $24$ with $4$ divisors and get that only $14$ and $15$ are solutions.