Proportional factors?

Relevant wiki: Factors / Divisors - Problem Solving

Let $n$ be the given integer, and $(d_1, d_2, ..., d_n)$ be the set of positive divisors of $n$ . Then the positive divisors of $3n$ would include all the divisors of $n$ $(d_1, d_2, ..., d_n)$ and any new unique number formed by multiplying the original divisors by $3$ $(3d_1, 3d_2, ..., 3d_n)$ . Even if all the new divisors are unique, we can at most double the size of the list, so tripling the given positive integer can never result in triple the number of positive divisors.

If doubling $n$ doubles the number of divisors, $n$ must be a prime number which is not $2$ : $n$ has 2 divisors ( $1$ and $n$ ) while $2n$ would have 4 ( $1$ , $2$ , $n$ and $2n$ ).

So if $n$ is prime, then $3n$ would also always have 4 divisors ( $1$ , $3$ , $n$ and $3n$ ), with an exception if $n=3$ . But either way, $d$ has no way of tripling in this case.

Introducing any new base into a number's prime factorization doubles the amount of factors.

Therefore, if $n$ exist, it must already by divisible by $3^e$ .

Consider the number $m = n / 3^e$ and suppose it had $k$ factors.

• $3*m$ would have $2*k$ factors (100% increase in factors)
• $3*3*m$ would have $3*k$ factors (50% increase in factors)
• $3*3*3*m$ would have $4*k$ factors (33% increase in factors)
• ...

So multiplying by 3 will never triple the number of factors.

Let $n$ be the given number, $n=p_1^{s_1}p_2^{s_2}3^{s_3}...p_N^{s_N}$ . Then the number of positive divisors $d = (s_1 + 1)(s_2 + 1)(s_3 + 1)....({s_N} + 1)$ . So $3 \cdot n=3 \cdot p_1^{s_1}p_2^{s_2}3^{s_3}...{p_N}^{s_N}=p_1^{s_1}p_2^{s_2}3^{s_3 + 1}...{p_N}^{s_N}$ , and number of positive divisors for $3 \cdot n$ is $(s_1 + 1)(s_2 + 1)(s_3 + 2)....({s_N} + 1)$ , where is $s_3$ is original number for $3^{s_3}$ . Let imagine that hypothesis is true, i.e. $3 \cdot d = (s_1 + 1)(s_2 + 1)(s_3 + 2)....({s_N} + 1)$

it should be $3(s_1 + 1)(s_2 + 1)(s_3 + 1)....({s_N} + 1)=(s_1 + 1)(s_2 + 1)(s_3 + 2)....({s_N} + 1)$ all ${s_N}\geqslant0$ and we can divide both sides by non-zero expressions, as a result we have $3(s_3 + 1)=(s_3 + 2)$ or $2s_3=-1$ we cannot solve it for ${s_3}\geqslant0$ . Conflict occurred, it seems our initial hypothesis was wrong.

Whenever you multiply a number by a prime, the number of divisors of the new number is at most twice the number of divisors of the previous number. That is easy to verify: if $n$ is not divisible by 3, then the divisors of $3n$ will consist of every divisor $n$ already had plus those same divisors multiplied by 3. If, however, $n$ is divisible by 3, suppose $3^k$ is the highest power of 3 that divides $n$ : then the number of divisors that will be added by multiplying $n$ by 3 is the number of divisors of $n$ that have $3^k$ as a factor (the new divisors will consist of those divisors times 3).

Simple Consider 1 for example 1 has1 as divisor 2×1 has 2 and 1 as. Divisir 3×1 has 3 and 1 as divisor Same for other

Given number $n$ has $d$ number of divisors. We have to make number of divisors $p_k d$ by multiplying the number $n$ by $p_k$ .
We can write $n$ as $p^{e_1}_1 \times p^{e_2}_2 \times p^{e_k}_2 \times p^{e_k}_k \times \ldots$ .
So the value of $d$ will be $(e_k + 1) \times (e_1 + 1) \times (e_2 + 1) \times (e_3 + 1) \times \ldots$ .
After multiplying the number $n$ with $p_k$ the $p_k d$ = $(e_k + 2) \times (e_1 + 1) \times (e_2 + 1) \times (e_3 + 1) \times \ldots$

$(e_k + 2) \times (e_1 + 1) \times (e_2 + 1) \times (e_3 + 1) \times \ldots$ = $p_k (e_k + 1) \times (e_1 + 1) \times (e_2 + 1) \times (e_3 + 1) \times \ldots$ .
$(e_k + 2) = p_k (e_k + 1)$
We can plot the values for $p_k$ and $e_k$ and see that for ( $p_k , e_k \in N$ ), only satisfying values are. $e_k= 0$ and $p_k =2$ .

We denote the number of divisors of $n$ by $\tau(n)$ Now, since $\tau$ is a multiplicative function, we have $\tau(2n)=\tau(2)\tau(n)=2\tau(n)$ Similarly, $\tau(3n)=\tau(3)\tau(n)=2\tau(n)$ Hence, after tripling, it still will have twice as many divisors as there were of $n$ . (This, in general is true when $n$ is multiplied by $p$ , a prime number. In our case, $p=3$ )

The only multipliers for which this can happen are 1 (always) and 2 (if n is odd)

If we multiplied $n$ by $3^2$ , then it is possible.