Latest Elitmus Aptitude Question SOLUTION: N= 1!+2!+3!.....+10!. what is the last digit of N^N?

as you see 5! till 10 ! each unit digit is zero.
so 1!+2!+3!+4! =33
so unit digit 3+ 0 =3
N=3
N^N= 3^3=27
so last digit is 7.

N= 1!+2!+3!.....+10!
Summing up the factorials gives,
N=1+2+6+24+120+720+5040+40320+362880+3628800=4037913
Now we have to find last digit of N^N
N^N=(4037913)^4037913
for finding out last digit, we can write only the last digit of 4037913 which is 3, hence
(3)^4037913
[(3)^(4*1009478)]*3 ............using cyclicity of 3
1*3=3 ans

1!--1
2!--2
3!--6
4!--24
5!--120
6!--__0
7!--__0
8!--__0
9!--__0
10!--__0

1+2+6+24+0+0+0+0+0+0
=33

=33^33
=(33^4)^8 * 33
=(1)^8 * 33
=1*33
=33

Last digit of N^N==3

last digits of
1!=1
2!=2
3!=6
4!=-4
5!=--0
6!=--0
7!=---0
8!=----0
9!=-----0
10!=----0
hence last digits of N=1+2+6+4+0+0+0+0+0+0=13
ANS=3

1! + 2! + 3! + 4! + 5! + 6! +7! +8! +9! +10!
1 + 2 + 6 + 24 + 120+........(from here all the last digits are zeros)................
The last digit of above sum is 3
So if we cube 3 ==> (3^3)=27
So the answer is 7

the last digit of 5!,6!,7!,8!,9!10! is zero bcoz it has 5*2. and 1!+ 2! +3!+4!=33 so last digit is 3.

5! till 10 ! each unit digit is zero.
so 1!+2!+3!+4! =33
so unit digit 3+ 0 =3
N= 3(unit digit)
N^N= 3^33
3 has the cyclicity of 4 i.e.
3^1=3
3^2=9
3^3=7
3^4=1
3^5=3
3^6=9
3^7=7
3^8=1.....so on
then divide 33by4 , rem= 1
1st term of cycle will be the unit digit....i.e. 3

7 will be the answer

3 will be the answer

As we know that the power of 3 follows 4 remainder only that is
3^1=3
3^2=9
3^3=7
3^4=1

Divisibility of 4 is always consern with last two digit I.e.
1!+2!+3!......10!= _____13


=> ______13^________13 = 3^1 = 3

And 3 will be the exact correct answer....

Given: N= 1!+2!+3!.....+10!
To find: The last digit of N^N

Solution:
we know that the last digit of x^y only depends on the last digit of x...
so, the last digit of x^y = (last digit of x)^y

therefore the last digit of N is as follows:
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120// since 0 in the last digit it won't add anything to my sum for last digits. further, all factorials will be having a zero at the end, therefore we don't need to add them.
..
So coming back to our problem:

the last digit of N is just the addition of last digits of each term of N.

>> 1+2+6+4(not 24)+0+0.... = _3 //(actual addition is 33 for (1!+2!+3!+4!) but we are focussed on last digit only)

and the last digit of N^N is as follows:
=> 3^N
where N is a large number (here it is 4037913 )

So, Using the concept of cyclicity of 3, to find-- 3^(anything)

we get the last digit of 3^N = 3^(N/4) // as 3 has got a power cycle of 4

now, We know the divisibility of 4 only depends on last two digits of any number, So here We have to find out the last two digits of N

hence, 3^N = 3^( [last two digits of N] / 4 )

last two digits of N can be found by adding all the last two digits, and yes if last two digit of any term again comes to be 00 then the term and there after term are not of our concern.

1!=1, 2!=2, 3!=6, 4!=24 & 5!=120.
After 4, factorials of all numbers end in 0.
So, unit digit = Unit digit of (1+2+6+4) = Unit digit of 13 = 3.
6!=720, 7!= --40, 8!=--20,
9!=--80, 10! = ---800.
After 9, factorials of all numbers end in 00.
Tens digit = Unit digit of (1[carry from 13]+2+2+2+4+2+8) = Unit digit of 21 = 1.

Last 2 digits = 13.

so here is the final answer:

the last digit of N^N = 3^(13/4) // 3 is the last digit of N and 13 is the last two digit of N

the remainder of 13/4 = 1.
3^(13/4) = 3^1 // same the concept of cyclicity says to get the remainder of 13/4 and that is to be raised to 3

= 3 (ANS)

@naresh_chandra it's not that simple we have to use the concept of cyclicity when dealing with large powers insted of using the unit digit of the power to reach the answer

1 will be answer

Last digit is 7
From 1 to 4! = 33
unit digit = 3
And from 5 till 10! each unit digit = 0
So 3^3 = 27

remainder is 22

here last digit of n will be 3 is right but in this question they ask for last digit of N^N. that means 3^3=27.
hence last digit will be 7 is right ans.

1!+2!+3!+4!=33+0(bcz last digit from 5!=0)=33
i.e last digi is 3.
so 3^3=27 .
last digit is 7

N=1!+2!+3!+........ +10!
Break the factorials first
N=1+2+6+24+120+........+(Last digit of every number after 3! will be zero,so no need of considering the remaining numbers,as the question asks about the last digit only)
Now add the last digit of every number i.e. 1+2+6+4=13.
As we are concerned about only the last number,we will just consider N=3 and apply N^N which gives,
N^N= 3^3=27
So our answer comes out to be 7.