11(7!+.........2007!) = 7a
1+2+6+24+120+720= 873 = 868+5
thus, give series (1+2!+3!+.......+2007!)500= (7n+5)500
What to do after this?
Find the remainder when (1!+2!+3!+...+2007!)^500 is divided by 7...
#This question is directly copied from a standard buk... juzz wanna know different ways of solving it...
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54 Answers
13i'm not satisfied wid nishant bhaiya's method... i even posted my query...
n sry gagar... answer is not 1... [2]
111sorry the ans will be 4 now the value of r as suggested by nishant is 5
therefore the question will reduce to remainder in 5500
we can say 25*5498
→ 25*(56)83 and 56=15625which is of the form 7k+1 so the remainder in (56)83 =1
so the remainder in 25*(56)83=25 which is greater than 7 again calculating the ans will be 4
1is the ans 4
sorry i din see that as i was typing the same time
1nothing just modular arithmetic
(a+b)mod(n) = (a(mod)n+b(mod)n)mod(n)
and the same with multiplication and exponentiation
[1]
62all numbers above 7! will be divisible by 7
write the summation as
7k+r where r is less than 7
and then try..
1i must have to increase the level of my mathematics to understand the soln
1Aragorn arre mod the % almost in all programming lang
oops sorry forgot you were a bio student
x%y== remainder when x is divided by y
11What does this come under? Btw I heard talk of some modulo theory. Something to do wth this?
1see aragorn its a bit difficult to explain under the light that you are not used to this modulus
see we need an operator to gives us the remainder we cannot always write
rem=b-[b/a]*a where [] is the GINT
1hey Gagar bhaiyya you re kidding aren't you
is the method wrong?? please tell
1no philip this is the weakest part of mine and even i got two consecutive f's in the same topic so just trying my hands
1im not a genius at it either [6]
was the 1st prob i solved by the method you see.
learnt it today only
though the statement i wrote above about addition must be clear. [7]
1i learnt it today in 20 mins everyone of you can read it somewhere
annd it can be extremely useful
i think now everyone is content with the answer
but hey gagar bhiyya give me something back! [4]
i cant get your method
plz explain [2]
13well... the answer is not even 4... sorry...
ur interpretation is correct till 5^500...
but i didnt get how d remainder is 4 when 5^500 is divided by 7...?
gagar... can u explain it clearly plzzz...?
11Gagar bhaiyyah, MAQ's wondering how u thought of 25.[56]83?
1i just wanted the term in the form of 7k+1 thats all and 56on div with 7 gives a remainder of 1 so i arrange the term accordingly
and still i am getting the ans 4 plz tell me if not correct