Question

Q.What is the last digit of the sum 1!+2!......49! ?

Answer 1

Hey...!! HERE IS YOUR ANSWER...!!

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!=1205!=120

6!=7206!=720

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

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

Even that's simplified by recognising that 3!3! ends with a 66 and 4!4! with a 44, so they will sum up to give 00 as the last digit.

Turns out all you have to consider is 1!+2!1!+2!, which is just 33.

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!

Answer 2

HERE IS YOURS SOLUTION;

◆ 1 ! + 2! + 3! + ...........+ 49!

Now,

1! = 1

2! = 2

3! = 6

4! = 24

5! = 120

6! = 720

◆ Now, the last digit of 5! = 0 and also next term has 0 last digit.

Therefore, on adding first four terms, that is, 1+2+6+24=33 and the last digit of 33 is 3.

★ So, the last digit is 3 ★

HOPE IT HELPS