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Is there any formula for finding the last digit of the factorials? How to approach these type of questions? Thanks in advance.

Community
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PARTH
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    Hint, look the last digit of 1!+2!+3!+4! and 5! separately. – Someone Jun 06 '15 at 07:03
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    What do $n!$ for $n\ge5$ contribute to the last digit? – David Mitra Jun 06 '15 at 07:03
  • notice that all factorials are even except for $1!$ and so you are adding odd + even + even + even + ............ + even which is odd.

    And so the last digit will be one of $1,3,5,7,9$. this is kinda easier

     – alkabary Jun 06 '15 at 07:05

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First of all, when you see this sort of seemingly intractable problem, don't despair. There's usually a very simple "trick" that makes the problem trivial.

In this case, you have to realise two things:

1) only the sum of last digits contributes to the last digit of the final sum.

2) factorials of larger numbers have a lot of zeroes at the end.

So your problem reduces to deciding the final term you have to consider. Luckily this is a very easy problem. Because:

$5! = 120$

$6! = 720$

and so forth, every factorial after that ending with a zero.

So you only have to consider the sum $1! + 2! + 3! + 4!$.

Even that's simplified by recognising that $3!$ ends with a $6$ and $4!$ with a $4$, so they will sum up to give $0$ as the last digit.

Turns out all you have to consider is $1! + 2!$, which is just $3$.

I wanted to put an exclamation point at the end of the last line to emphasise how easy the whole thing was, but decided not to because it might look like a factorial! :)

Deepak
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  • Nice bro! So can we say that the last digit of $\sum\limits_{x=1}^{n} x!$ where $n\geq 4$ is always $3$ ? – NeilRoy Jun 06 '15 at 07:16
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    As for proof that $n!$ has a zero for a final digit for all $n\geq 5$, note that for $n\geq 5$ you have $2|n!$ and $5|n!$, (seen either by induction using the definition of factorial that $n!=n\cdot (n-1)!$, or directly using definition that $n!=\prod\limits_{i=1}^n i$). – JMoravitz Jun 06 '15 at 07:17
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    @Neil not quite. Change the statement to be $n\geq 4$ and it will be true. When $n=2$ you have the final digit is $3$, but when $n=3$ you have the final digit is $9$ (since $1!+2!+3!=1+2+6=9$) – JMoravitz Jun 06 '15 at 07:18
  • @JMoravitz Yes, good catch! – Deepak Jun 06 '15 at 07:18
  • @JMoravitz Oh yeah...sorry my bad! – NeilRoy Jun 06 '15 at 07:19
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well as the comments stated, you should observe that for $n \geq 5$ we have $n!$ is a multiple of $5$. For example $$5! = 5 \times 4 \times 3 \times 2 \times 1 = 24 \times 5$$

and $$6 ! = 6 \times 5! = 6 \times 24 \times 5$$

and so on. Now you should know that any multiple of $5$ is either $0$ or $5$ for instance $5,10,15,20,25,.........$ as you can see.

now $$1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33$$

and $5! + 6! + ..........+49!$ is a mutliple of $5$ as we said.

However, $$5! + 6! + ..........+49!$$ is an even number because they are also a multiple of $2$

And so we have $1! + 2! + 3! + 4!$ is odd

and $5! + 6! + ..........+49!$ is even

and you should know that odd + even = odd. $$(2k + 1) +2m = 2k +2m + 1 = 2(k +m) + 1$$ which is odd.

And so the sum is odd.

and so $5! + 6! + ..........+49!$ ends with $0$ because it is a multiple of $5$ and an even number.

and $33$ + a number that ends with $0$ is will result in a number with last digit = $3$

alkabary
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