Immensely Compound Figure

The number of positive divisors of a number $p_1^{e_1} \cdot p_2^{e_2} \cdot \dots \cdot p_n^{e_n}$ for prime factors of $p$ is $(e_1 + 1) \cdot (e_2 + 1) \cdot \dots \cdot (e_n + 1)$ . For example, since $1\times2\times3\times4\times5\times6\times7 = 2^4 \cdot 3^2 \cdot 5 \cdot 7$ , it has $(4 + 1) \cdot (2 + 1) \cdot (1 + 1) \cdot (1 + 1) = 5 \cdot 3 \cdot 2 \cdot 2 = 60$ positive divisors.

We can also see that for a number to have exactly $x$ positive divisors, it must be in the form of $p_1^{f_1 - 1} \cdot p_2^{f_2 - 1} \cdot \dots \cdot p_n^{f_n - 1}$ , where $x = f_1 \cdot f_2 \cdot \dots \cdot f_n$ . In addition, the minimum number with exactly $x$ positive divisors would have the smallest prime number ( $2$ ) raised to the highest exponent, the next smallest prime number ( $3$ ) raised to the second highest exponent, and so on.

Therefore, $2^4 \cdot 3^2 \cdot 5 \cdot 7$ is the minimum number to have exactly $60$ positive divisors for the factorization of $60 = 5 \cdot 3 \cdot 2 \cdot 2$ , but we still need to consider all of the factorizations of $60$ , which are summarized in the table below. (We can also reduce the number of calculations needed by eliminating any factorization that gives a number that is obviously greater than $2^{13}$ , since $2^4 \cdot 3^2 \cdot 5 \cdot 7 = 5040 < 2^{13}$ .)

Therefore, $2^4 \cdot 3^2 \cdot 5 \cdot 7 = 5040 = 1\times2\times3\times4\times5\times6\times7$ is the smallest number with exactly $60$ divisors, so there is not a positive integer smaller than it that has exactly $60$ divisors.

Notice the following $2\times 3\times 5 \times \cdots\times 13 =30030 > 7!$ This shows that number less than $7!$ can have at most the factors of $\left\{2,3,5\cdots, 11\right\}$ in its prime factorization. As $2\times 3\times 5\times \cdots \times 11 = 2301k = 5040\Rightarrow k =\left\lfloor \dfrac{5040}{2301}\right\rfloor=2$ This means the number less 7! that holds highest number divisors is $2301\times 2= 4610$ having
$48$ divisors.

Since $7!$ has the distict prime factors less 11 so let us assume that there exist such number $N <7!$ with 60 divisors with distinct prime divisors less than 11. Following same working from above $2\times 3\times 5\times 7 = 210k =5040\implies k = \left\lfloor \dfrac{5040}{210}\right\rfloor=24 =2^3\times 3$ Since $N<7!$ so from $210\times 2^3\times 3$ either we can eliminate $2$ or $3$ . TO maximize the product we clearly eliminate 2 giving us $210\times 2\times 3^2 =2510$ . Having $d(2510) = d(2^2\cdot 3^2\cdot 5\cdot 7) = 48$ Thus the answer is No .

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With the same concept I had set a problem almost 3 months back here .