Find the remainder on division of the following expressions....
1) 502008 by 7
2) 2008 by 7
3) 20082008 by 13
4) (1!+2!+3!+..........2008!) by 2008
5) Last 3 digits of (1!+2!+3!+..........2008!)
6) (1!+2!+3!+..........2008!) by 2009
Some of these even I dont know right now ;)
1plz explain this part :--
"3rd last digit of (10! + 11! + 12! + 13! + 14!) [since 3rd last digit of (1! + 2! + 3! +...+8! + 9!) & (15! + 16! +....) is 0] = 2 "
how did we find out that the 3rd last digit is a zero [7]
if all three last would be zero its fine but how only the 3rd last digit
especially in "(1! + 2! + 3! +...+8! + 9!)" [7]
plz... [2]
13 > 200812 = 1 ( mod 13) ( using Eulers theorem)
so 20082004 = 1 ( mod 13)
again , 20082008 = 20084 ( mod 13 )
but 2008 = 6 (mod 13 )
so 20084 = 64 ( mod 13 )
again 62 = -3 (mod 13 )
so 64 = 9 ( mod 13 )
so we get 20082008 = 9 (mod 13)
1a ≡ b(mod)c
means a-b is divisible by c
or a leaves remainder b upon division by c
1one silly doubt may be : wat is this (mod) dat u use?
i did it earlier... but now cudn recall wat it is... [3]
help me.. :)
62no.. by just keeping the last 3 digits and "throwing" everything else away!
1!+2!+3!+.....6!= 1+2+6+24+120+720=873
7!+8!+9! = 720.7(1+8+8.9) = 5040.81
Now it seems "less" dirty?
62You both are right..
You have to do that by "dirtying your hand"
11Sorry, philip [2]
rkrish or anybody else.........how can we get the last 2 digits of 1! + 2! + 3! +...+8! + 9! ???
1who asked what would be there after 15!
i mentioned clearly
"(1! + 2! + 3! +...+8! + 9!)"
sorry if i am blabbering [2]
11philip, after 1! all the other factorials are even. So on multiplying by a multiple of 5, you will get a 0. So at 5!, 10! AND 15!, we will get an extra 0 at the end. So after 15!, there will be 3 0's for all other factorials after it.... HOPE U UNDERSTAND [1]
62this 2008^2008 by 13 is becoming massive..
Rem=(2002+6)2008
Rem=(6)2008
Rem=(6)2008
612 ≡ 1 (mod 13)
612*167 ≡ 1167 (mod 13)
62004 ≡ 1 (mod 13)
Thus 62008 ≡ 64 (mod 13)
Thus the answer is probably 9 :)
1this 2008^2008 by 13 is becoming massive..
it wud be fine if u gv a power which is a multiple of three [3]
115) Last digit of (1! + 2! +...+ 2008!) = Last digit of (1! + 2! +3! + 4!)by [since last digit of (5! + 6! +....) is 0] = 3
2nd last digit of (1! + 2! +...+ 2008!) = 2nd last digit of (1! + 2! +...+ 8! + 9!) [since 2nd last digit of (10! + 11! +....) is 0] = 1
3rd last digit of (1! + 2! +...+ 2008!) = 3rd last digit of (10! + 11! + 12! + 13! + 14!) [since 3rd last digit of (1! + 2! + 3! +...+8! + 9!) & (15! + 16! +....) is 0] = 2
[ NOTE : You dont have to compute the exact values of 10!,11!,...14! etc.You only need to know their respective 3rd last digits,since their last two digits are 0's. ]
Hence,the last 3 digits of (1! + 2! +...+ 2008! ) are 2 , 1 & 3.
1sorry for the disturbance.[2] U seem to be reminding me how much time I've wasted in 2008 . :P . only the last one speaks of this year . I wonder why????[12]
