first of all 1+2^2+2^3..+2^10=2045,its not required thou.that's bcoz 7!=5040 is completly divisible by 24. so by the known technique that for getting the remainder of a*b*c*d...to n terms ,when divided by say R ,we multiply the remainder of (a/R),(b/R)...(n/R) .so here since 7! is itself divisible by 76 hence the remainder will b simply
to see the remainder 7! is divided by 76
first divide 76 in 19*4
then see the remainder when we divide 7! by 19.>>>we get remainder 1
and 7! is perfectly divisible by 4
so we can write 7!= 76k+4
but now i m stuck up...
but my guess is remainder may be 4
i am getting 5 as ans....wat say ???
no wait.. i'll try again !!!
Q)find the remainder when (7!)^k is divided by 76,
where k = 1+2^2+2^3+2^4+.........+2^10
ANS...= 24
7! = 5040, now 5040 divided by 96 gives u a remainder of 24, so nw qn becomes:-
(24^k)/76 = [24 x 24^(k-1)]/76 = [24 x 576^(k-1)/2]/76 = 24/76 (Since 576 is completely div by 76)
Hence 24/76 will give a remainder of 24..& that is the ans
any doubts ne1????
iam sry 2 bothr u but my humble doubt is ....
hows 576 completely div by 76!!!!!!!!!!!!!!!!!!!!!!!
oh man..F...K !!! wat m dng..i tuk it as 96..DAMN...
first of all 1+2^2+2^3..+2^10=2045,its not required thou.that's bcoz 7!=5040 is completly divisible by 24. so by the known technique that for getting the remainder of a*b*c*d...to n terms ,when divided by say R ,we multiply the remainder of (a/R),(b/R)...(n/R) .so here since 7! is itself divisible by 76 hence the remainder will b simply
0*0*....2045 times=0..
so the ANS is 0.
7! is itself divisible by 76 ?????????? hw is dat possible..??
its giving a rem of 24...chk it again !!!! Mr. INCOGNITO !!!!
n y dont u jzt get logged in first !!!!
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