Master the factorial

The number of trailing zeroes in any number is the same as the exponent of the largest power of $10$ that divides the number; this in turn means that the number of trailing zeroes in any number is the same as the exponent of the either the largest power of $2$ or the largest power of $5$ (whichever is lower) that divides the number.

For $n \ge 2$ , $\ n!$ has a larger of number of multiples of $2$ than multiples of $5$ , which means that the number of trailing zeroes in $n!$ is the same as the exponent of the largest power of $5$ which divides $n!$ . We can calculate this power using the formula

$\left\lfloor \dfrac{n}{5} \right\rfloor + \left\lfloor \dfrac{n}{5^2} \right\rfloor + \left\lfloor \dfrac{n}{5^3} \right\rfloor + \ ....$

In our case we find that the largest power of $5$ which divides $258!$ is

$\left\lfloor \dfrac{258}{5} \right\rfloor + \left\lfloor \dfrac{258}{5^2} \right\rfloor + \left\lfloor \dfrac{258}{5^3} \right\rfloor = 51 + 10 + 2 = 63$

so $258!$ has $63$ trailing zeroes. So the digit we're seeking is the first non-zero digit to the left of the trailing zeroes.

$$

Consider the product of four consecutive natural numbers between multiples of $5$ . These could either be of the form $(10k + 1)(10k + 2)(10k + 3)(10k + 4)$ or $(10k + 6)(10k + 7)(10k + 8)(10k + 9)$ . Then

$\begin{array}{rl} (10k + 1)(10k + 2)(10k + 3)(10k + 4) &\equiv 1 \cdot 2 \cdot 3 \cdot 4 \pmod{10} \\ \\ &\equiv 24 \pmod{10} \\ \\ &\equiv 4 \pmod{10} \end{array}$

and similarly

$\begin{array}{rl} (10k + 6)(10k + 7)(10k + 8)(10k + 9) &\equiv 6 \cdot 7 \cdot 8 \cdot 9 \pmod{10} \\ \\ &\equiv \text{-}4 \cdot \text{-}3 \cdot \text{-}2 \cdot \text{-}1 \pmod{10} \\ \\ &\equiv 24 \pmod{10} \\ \\ &\equiv 4 \pmod{10} \end{array}$

Any number congruent to $4 \bmod {10}$ can be written as $2m$ , where $m$ is congruent to either $2 \bmod {10}$ or $7 \bmod {10}$ ; however, our products above contain exactly two even factors so if we were to factor out a single $2$ the remaining factor would have to be congruent to $2 \bmod {10}$ .

Thus any product of five consecutive natural numbers ending in a multiple of $5$ can be written as

$(10k + 1)(10k + 2)(10k + 3)(10k + 4)(10 k + 5) = (10k + 5) \cdot 2 \cdot a$

or

$(10k + 6)(10k + 7)(10k + 8)(10k + 9)(10 k + 10) = (10k + 10) \cdot 2 \cdot b$

where $a$ and $b$ are both congruent to $2 \bmod{10}$ .

$$

Now $258!$ can be split into $51$ blocks of five consecutive natural numbers, with three remaining factors:

$\begin{array}{rl} 258! &= (1 \cdot 2 \cdot 3 \cdot 4 \cdot 5) \cdot (6 \cdot 7 \cdot 8 \cdot 9 \cdot 10) \cdot (11 \cdot 12 \cdot 13 \cdot 14 \cdot 15) \cdot (16 \cdot 17 \cdot 18 \cdot 19 \cdot 20) \cdot \ .... \ \cdot 256 \cdot 257 \cdot 258 \\ \\ &= (5 \cdot 2 \cdot a) \cdot (10 \cdot 2 \cdot b) \cdot (15 \cdot 2 \cdot c) \cdot (20 \cdot 2 \cdot d) \cdot \ .... \ \cdot 256 \cdot 257 \cdot 258 \\ \\ &= (5 \cdot 10 \cdot 15 \cdot 20 \ \cdot \ ....) \cdot (2 \cdot 2 \cdot 2 \cdot 2 \ \cdot \ ....) \cdot (a \cdot b \cdot c \cdot d \ \cdot \ ....) \cdot 256 \cdot 257 \cdot 258 \\ \\ &= 5^{51} \cdot (1 \cdot 2 \cdot 3 \cdot 4 \cdot \ ....) \cdot 2^{51} \cdot (a \cdot b \cdot c \cdot d \ \cdot \ ....) \cdot 256 \cdot 257 \cdot 258 \\ \\ &= 10^{51} \cdot 51! \cdot (a \cdot b \cdot c \cdot d \ \cdot \ ....) \cdot 256 \cdot 257 \cdot 258 \end{array}$

We know that the factor $(a \cdot b \cdot c \cdot d \cdot ....)$ above will be congruent to $2^{51} \bmod{10}$ and we can calculate that $256 \cdot 257 \cdot 258 \equiv 6 \cdot 7 \cdot 8 \equiv 336 \equiv 6 \bmod{10}$ .

Thus we can write

$\dfrac{258!}{10^{51}} \equiv \left( 51! \cdot 2^{51} \cdot 6 \right) \pmod{10}$

If we repeat the above process for $51!$ , we get that

$\dfrac{51!}{10^{10}} \equiv \left( 10! \cdot 2^{10} \cdot 1 \right) \pmod{10}$

and

$\dfrac{10!}{10^2} \equiv \left( 2! \cdot 2^{2} \right) \pmod{10}$

Combining these we find that

$\begin{array}{rl} \dfrac{258!}{10^{63}} &\equiv \left( 2! \cdot 2^2 \cdot 2^{10} \cdot 2^{51} \cdot 6 \right) \pmod{10} \\ &\equiv \left( 2^{64} \cdot 6 \right) \pmod{10} \end{array}$

Finally, using the facts that $2^4 \equiv 6 \bmod{10}$ and $6^n \equiv 6 \bmod{10}$ we get

$\begin{array}{rl} \dfrac{258!}{10^{63}} &\equiv \left( (2^4)^{16} \cdot 6 \right) \pmod{10} \\ \\ &\equiv 6^{17} \pmod{10} \\ \\ &\equiv 6 \pmod{10} \end{array}$

so $258!$ will have $63$ trailing zeroes, and the $64$ th digit will be $\boxed{6}$