Thirty nines!

If the expression $n!(n+1)!(2n+1)!-1$ is to end in a string of thirty 9's, we need $N=n!(n+1)!(2n+1)!$ to end in thirty zero's, i.e. to be a multiple of $10^{30}$ . This will happen if the three factors of $N$ collectively contain at least thirty 2's and thirty 5's in their prime factorizations.

For $p$ a prime, the prime factorization of $n!$ will contain at least one $p$ for every multiple of $p$ in its expansion, so for $n \ge 2$ , $n! \color{#333333} = 1 \cdot \color{#20A900}2 \color{#333333} \cdot 3 \cdot \color{#20A900}{4} \color{#333333} \cdot \color{#D61F06}5 \color{#333333} \cdot \color{#20A900}{6} \color{#333333} \cdot 7 \cdot \ldots \cdot n$ will always contain more 2's than 5's; thus for our purposes it suffices to find the minimum $n$ so that the prime factorization of $N$ contains thirty 5's.

For $p$ a prime, every multiple of $p$ in the expansion of $n!$ will contribute one $p$ to the prime factorization of $n!$ ; however every multiple of $p^2$ will contribute two $p$ 's rather than one, and in general every multiple of $p^k$ will contribute $k$ $p$ 's. Thus the number of $p$ 's in the prime factorization of $n!$ is given by

$\displaystyle\sum_{k=1} \left\lfloor \dfrac{n}{p^k} \right\rfloor$

Roughly, the prime factorizations of $n!$ and $(n+1)!$ will contain the same number of 5's, while the prime factorization of $(2n+1)!$ will contain twice as many. Thus a good starting point for our minimum $n$ is a value for which $n!$ yields roughly a quarter of the required 5's, i.e. approximately seven. A quick calculation shows that $n=30$ is the lowest value that yields seven 5's, as

$\left\lfloor \dfrac{30}{5} \right\rfloor + \left\lfloor \dfrac{30}{5^2} \right\rfloor = 6+1=7$

For $n=30$ we have $N=(30!)(31!)(61!)$ , so the total number of five's in the prime factorization of $N$ is

$\left( \left\lfloor \dfrac{30}{5} \right\rfloor + \left\lfloor \dfrac{30}{5^2} \right\rfloor \right) + \left( \left\lfloor \dfrac{31}{5} \right\rfloor + \left\lfloor \dfrac{31}{5^2} \right\rfloor \right) + \left( \left\lfloor \dfrac{61}{5} \right\rfloor + \left\lfloor \dfrac{61}{5^2} \right\rfloor \right) = 7+7+14 = 28$

If we slowly increase n, we will obtain an additional 5 every time one of $n, n+1$ or $2n+1$ becomes another multiple of 5. The first time this happens is for $n=32$ when $2n+1=65$ ; the second is for $n=34$ , when $n+1=35$ . Thus when $n=34$ , the prime factorization of $N$ will contain thirty 5's, so $N$ will end in a string of thirty zeroes, meaning that $n!(n+1)!(2n+1)!-1$ will end in a string of thirty 9's.

Using $Mathematica$

Min@Select[Range@100,Mod[#!(#+1)!(2#+1)!-1,10^30]==FromDigits@Table[9,30]&]

returns $34$