Question

[Solved by @dg2 ] find the last digit of this sum

\(\LARGE \sum \limits_{n=1}^{100} n^n \)

Answer 1

1

Answer 2

nice try haha! but no :
http://www.wolframalpha.com/input/?i=%5Csum+%5Climits_%7Bn%3D1%7D%5E100+n%5En

Answer 3

;)

Answer 4

\(1^1+2^2+3^3+4^4+....+ 100^{100} \mod 10 \)

Answer 5

yep !

Answer 6

this might be hard as i was not bale to find any good solution anywhere online...

Answer 7

brb

Answer 8

20

Answer 9

so 0

Answer 10

how did u get 20 ?

Answer 11

1^1-last digit1
2^2-4
3^3=7
4^4=6
5^5=5
6^6=6
7^7=3
8^8=6
9^9=9
10^10=0

Answer 12

interesting, could you explain a bit more... :)

Answer 13

thinking!.give me some more minutes

Answer 14

i think you're on right track, take your time.. .

Answer 15

power cycle of 1^1 is 1 ok

Answer 16

yes

Answer 17

power cycle of 3 is 3,9,7,1
power cycleof 9 is 7,9,3,1
.......................7 is same as above

Answer 18

you know power cycle of 2 is ,6,8,10,12

Answer 19

4 is 6 always

Answer 20

it should be :
2,4,8,6,2...

right?

Answer 21

cool ^^

Answer 22

yes

Answer 23

5^5 is always 5

Answer 24

oh yes!

Answer 25

6^6,and 8 s4*2 cycles 10 is 0

Answer 26

so add all we get the answer i think

Answer 27

brilliant !
i think we can simply add first 10 powers and multiply them by 10

Answer 28

1 is 10 times=10
2 is 10 times 2,4,6,8etc =80

Answer 29

1^1-last digit1
2^2-4
3^3=7
4^4=6
5^5=5
6^6=6
7^7=3
8^8=6
9^9=9
10^10=0
----------------
7

Answer 30

7*10 = 70
so units place is 0

fantastic work @dg2 !!

Answer 31

well
iwould like to do
sum mod 2 shold be = sum mod 5

sum mod 2
1+0+1+0+1+0+1+0 .....( 50 times) mod 2
so
50 mod 2 = 0 mos 2 = 0 mod 10 mmm

for sum mod 5
1+2+3+4+5 +1+2+3+4+5 + ....... ( without ordering )** mod 5 = 5 k mod 5 =0

** trick with fermat mmm

Answer 32

10+80+60+60+50+60+60+50+60+0=490 like this also i think

Answer 33

wait i need to work more on this part

for sum mod 5 1+2+3+4+5 +1+2+3+4+5 + ....... ( without ordering )** mod 5 = 5 k mod 5 =0 ** trick with fermat mmm

Answer 34

oh i dont understand that mod what @ikram002p is using.and idk about mod .give some examples

Answer 35

:3

Answer 36

@ikram002p are you using chinese remaidner thm ?

Answer 37

not really ,
will i used this identity first

if a,b primes

x=m mod a
x=m mod b

then
x= m mod ab

10=2*5
its more simple to work with primes instead of composite numbers

Answer 38

I see... you want to work the remainders by first dividing the sum by 2 and 5 ?

Answer 39

however ,
\(\LARGE \sum \limits_{n=1}^{100} n^n \mod 2 \)
since terms are odd/even
then
it would be like this
1+0+1+0+1+0+.... +0 mod 2=50 mod 2 = 0 mod 2
exactly ganesh

Answer 40

that way mod2 is easy, but working mod5 is very painful ^^

Answer 41

now lets see fermats apply that

\(a^4=1 \mod 5\)

Answer 42

^^

Answer 43

Wow! so that kills 25 terms in the sum, what aboutthe remaining 75 terms ?

Answer 44

Fermat reduces all exponents of form 4k to 1, right ?

Answer 45

from 1 to 3 we will check by hand

when power n is :-

n=4k ,then mod =0
25 elements
the others will be from 1 to 0 xD
the mod sum will be 1+2+3+4+0+1+2 ( without ordering ) mod 5 :)

Answer 46

@dg2 method is also awesome :3

Answer 47

but i dont understand

Answer 48

from 1 to 4 **

Answer 49

wait made a typo there i meant mod when its 4k is 1
when its 5k is zero

Answer 50

help me.some what undertood when divided by 10 we have two factors 2*5 but when divided by 2 we get 50 number as remainder zero,5 means 20 number as zero we got 60 numbers as remainder zero

Answer 51

a^4=1 mod 5?

Answer 52

this is a theorm , Fermat little theorms
ok ill write down in neat way :P
which im not good in :P

Answer 53

okay

Answer 54

i think u should also hav patience

Answer 55

hehe i also dont have that as well

Answer 56

create it

Answer 57

lol okay they spend one full semister studying number theory @dg2

Answer 58

:3 patience can be neither created nor destroyed, but can change form

Answer 59

haha! the key thing to notice is this :
when you divide by 2, you get only two numbers as remainders : {0, 1}

Answer 60

yes

Answer 61

so performing division by 2 is a piece of cake :

1^1 + 2^2 + 3^3 + 4^4 + 5^5 + .... + 100^100 mod 2

is same as :

1 + 0 + 1 + 0 + .... + 0 mod 2

Answer 62

who they ?

sry Os laging to me

Answer 63

which is same as 50 mod 2

Answer 64

which is same as 0 mod 2

Answer 65

then 60 ones and 40ones

Answer 66

:o

Answer 67

Is that clear why the reaminder of the entre sum is 0 when divided by 2 ?

Answer 68

srry 60 zeros,40 ones

Answer 69

50 zeroes ,50 ones

Answer 70

make it 50 zeroes, 50 ones :P

Answer 71

when divided by 5 we get 20 zeros

Answer 72

every odd number leaves a remainder 1
every even number leaves a remainder 0

50 odd numbers, 50 even numbers between 1 & 100
so 50 1's and 50 0's

Answer 73

thats right !

Answer 74

we will be left with 80 terms in the sum, which @ikram002p has worked by using Fermat and other advanced number theory tricks which i am also not able to understand fully

Answer 75

:-):-):-):)

Answer 76

ohh ,ok ill go through it step by step :)
tbh when u get through it its the same as dg did

so we wanna find :-
\(\LARGE \sum \limits_{n=1}^{100} n^n \mod 5\)

Answer 77

he told he will explain neatly

Answer 78

Fermat set that for any number a >2 , then

\(a^4 =1 \mod 5\)

Answer 79

so
\(a^{4k}=1 \mod 5\)

thus
\(a^{4k+1}=a \mod 5\)

\(a^{4k+2}=a^2 \mod 5\)

\(a^{4k+3}=a^3 \mod 5\)

Answer 80

ok so far ?

Answer 81

okay

Answer 82

\[a^{p-1} \equiv 1 \pmod{p}.\]

Answer 83

PS , we do not count a =5n sence mod would be zero
@ganeshie8 reaso these step for me
i understand it but cant explain hehe

like we would have

a it self might be
4k
4k+1
4k+2
4k+3

Answer 84

formula right?

Answer 85

yeah Fermat :)

Answer 86

its quite confusing .i will see it later.after working some problems using fermat

Answer 87

ok so the mod sum well be like this :-( this might clear doubts )

\(\ \sum \limits_{n=1}^{100} n^n=1+2^2+3^3+1 +0 + 6^2+7^3+1+9+0 +11^3 +1+13+14^2 +15^3+1+17+18^2+19^3+0+....+ ect xD\)

Answer 88

\(\ \sum \limits_{n=1}^{100} n^n=1+2^2+3^3+1 +0 + 6^2+7^3\)
\(+1+9+0 +11^3 +1+13+14^2 +15^3+1+17+18^2+19^3+0+....+ \) ect xD

Answer 89

mmm thats it -.-

we would have
1+2+3+4+0+.....(without ordering ) =10 k mod 5 = 0

Answer 90

nice xD

Answer 91

<3

Answer 92

good