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A week ago i made a similar question but nobody help me, i´ve been trying but i still don't get it.

I want to know how to find the last non-zero digit of $50!$.

my try:

First i have to know how much Zeros $50!$ has so i did this:

$$E_{5}50 = \sum _{1\leq k <n} \Bigg[\frac{50}{5^{k}}\Bigg] =\Bigg[\frac{50}{5}\Bigg] + \Bigg[\frac{50 }{25}\Bigg] = 12$$

So $50!$ has $12$ zeros which means that the last digit of $\quad\frac{50!}{10^{12}}\quad$ is the number that i´m looking for.

so if $x = \frac{50!}{10^{12}}$ i need to find $x (mod 10)$ to get it but this is such a big number and i still don't know how to reduce it.

Thanks in advance!!

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It is enough to find $x \mod 2$ and $x \mod 5$ in order to know $x \mod 10$. It is easy to show that $x \mod 2 = 0$, so all you really need to do is find $x \mod 5$. Since $2^{12}$ is invertible modulo $5$, it is enough to do this for $2^{12} x \mod 5$ and then divide by $2^{12}$. The number $2^{12} x = \frac{50!}{5^{12}}$ is not too hard to write as a product of several numbers, and after reducing modulo $5$, you'll see that there is a lot of repetition in these numbers, which simplifies the calculation.

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Here's a different approach: find the last non-zero digit of $10!$, then find the last non-zero digit of $20!/10!$, and so on for $30!/20!$, $40!/30!$, and $50!/40!$, then multiply them. It's easier to cancel in $5$ groups of $10$ numbers rather than one group of $50$.

Here's $10!$:

$$1\times2\times3\times4\times5\times6\times7\times8\times9\times10$$

Divide out the two factors of $5$, along with two factors of $2$ (the $4$):

$$1\times2\times3\times6\times7\times8\times9\times2$$

(We took out the $4$ and the $5$, and replaced the $10$ with a $2$). It's easy to compute the last digit of this, which is an $8$.

For $20!/10!$, we have

$$11\times12\times13\times14\times15\times16\times17\times18\times19\times20$$

Divide out the two factors of $5$ along with two factors of $2$ to get

$$11\times3\times13\times14\times3\times16\times17\times18\times19\times4$$

which, taking the units digit only, is

$$1\times3\times3\times4\times3\times6\times7\times8\times9\times4$$

and has units digit $8$. Can you continue?

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  • $\begingroup$ Yes, i guess i can continue...Thanks!!!1 $\endgroup$ – Jearson Narvaez Rojas Sep 29 '14 at 3:40
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My Solution:: calculation of last non zero digit in $50!$

Now first we will calculate number of zeros at the end of $50!$, which is obtained

when we multiplied $2$ and $5$. So we will calculate no. of $2$ and $5$ in $50!$

$\displaystyle 50! = \lfloor \frac{50}{2}\rfloor +\lfloor \frac{50}{2^2} \rfloor+\lfloor \frac{50}{2^3}\rfloor+\lfloor \frac{50}{2^4}\rfloor+\lfloor \frac{50}{2^5}\rfloor+........... = 25+12+6+3+1 = 47$

$\displaystyle 50! = \lfloor \frac{50}{5}\rfloor+\lfloor \frac{50}{5^2}\rfloor+..... = 10+2 = 12$

So Total no. of zeros at the end of $50! = 12$

Now $\displaystyle 50! = \lfloor \frac{50}{3}\rfloor+\lfloor\frac{50}{3^2}\rfloor+\frac{50}{3^3}\rfloor+..... = 16+5+1 = 22$

Now we can write $\displaystyle 50! = 1\times 3^{22}\times 7^8\times 11^4 \times 13^3 \times 17 \times 19\times 23\times 29\times 31 \times 17\times 37 \times 41 \times 43 \times 23 \times 47 \times 10^{12}\times 2^{35} = \bf{last digit\; (6)} = 6$

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    $\begingroup$ It's not $6$, it's $2$. $\endgroup$ – Edward Jiang Sep 29 '14 at 3:43
  • $\begingroup$ Your logic is unclear! :-( $\endgroup$ – xxx--- Oct 8 '14 at 13:33
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After reducing the zeroes, what you have is $2^{35}3^{22}7^{8}\cdots \equiv 8 \cdot 9 \cdots \equiv 2 \pmod {10}$.

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