find the sum 1/2 + 2^2/2^2 + 3^2/2^3 + 4^2/2^4+..........
@raopradeep said:yaar i had done little mistake in typing i edited my question plz see they worked from 1st jan 1991 to 31st dec 2000 sorry for my silliness plz now put your solution
arey yar approach i have explained.. carry it that way..u will get the answer..if u dont then pm us...
@vbhvgupta said:are bhai 32 * 20 is twice the work which needs to be completed.works still need to be completed is 8 days and x no of mensso32 * 20 = 2( 8 * X)X =40so we need to employ 20 extra mensclear hua???Quantum cat
aur question discuss kar...mere bi kuch questions ho jayege tere saath..... 
Q1
@raopradeep said:k =1^2 +2^2 +3^2....................m^2k= m*(m+1)*(2m+1)/6and k is divisible by 4k =m*(m+1)*(2m+1)/24putting m=1,2,3,4,5,6,7,8we see 7 ,8 value satisfy the condtn and this trend goes on in every 8 numbers only 2 are satisfying the condtn so mm=6 so 6*2=1212 values satisfy
Can u give a hint about 55 / 8 ?
@vbhvgupta said:Q1
16 days...
convert the work into manhr..
i.e. 10men for 9 days = 90 manhr
6women for 9 days = 6*7.5 = 45womenhr = 45*2/3(to convert their equivalent effort into a man's effort) = 30 manhr
total manhr for 1 day = 120manhr
for 18 days = 120*18
2nd case :
10men *9+ 9*7.5*2/3 = 135manhr
120*18 = 135*X
X= 16 days...
whats the OA??
arey simple 2/3 women ki efficiency.. no of women got increased for same job from 6 to 9 so the days will get reduced by 2...
@saurav205 said:16 days...convert the work into manhr..i.e. 10men for 9 days = 90 manhr6women for 9 days = 6*7.5 = 45womenhr = 45*2/3(to convert their equivalent effort into a man's effort) = 30 manhrtotal manhr for 1 day = 120manhrfor 18 days = 120*182nd case :10men *9+ 9*7.5*2/3 = 135manhr120*18 = 135*XX= 16 days...whats the OA??
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@ananyboss said:Can u give a hint about 55 / 8 8*6 =48 which is just less then 55 so in every 8 numbers we get 2 numbers therefore for 8*6 we get 6*2 =12 numbers
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hello puys Can anyone plz explain or provide a method
"To find the last two non zero numbers of a factorial"???????????????/
@harshit12
36!-24!=24!(25*26*.............36 - 1)last 2 non zero digits of 24! are 36.last 2 digits of the term within brackets are 99(WHY????)So multiply to get last 2 digits which will be 64
i stumbled upon this question of yours can you plz elaborate the solution if at all you have understood it??
@ani6 said:@harshit1236!-24!=24!(25*26*.............36 - 1)last 2 non zero digits of 24! are 36.last 2 digits of the term within brackets are 99(WHY????)So multiply to get last 2 digits which will be 64i stumbled upon this question of yours can you plz elaborate the solution if at all you have understood it??
but the overall answer of this question would be 36 not 64.I calculated it from scientific calculator.but i don't know the method to solve this question.If anyone knows the method.kindly explain.
@ani6 said:hello puys Can anyone plz explain or provide a method "To find the last two non zero numbers of a factorial"???????????????/
@saurabhlumarrai said:but the overall answer of this question would be 36 not 64.I calculated it from scientific calculator.but i don't know the method to solve this question.If anyone knows the method.kindly explain.
i do not know how this is derived. learnt this somewhere in PG
Last two non-zero digit of n!
Z(ab) = 66 * Z(ab/5)! * Z(b!)
=> Z(24) = 66* Z(4!)* Z(4!) = 66*24*4 = xxxx36.
Let me tell you what have I done.
ab is our number 24.
Z(ab/5)! is the last two non-zero digits of highest multiple of 5 in ab.
Z(b!) is the last non-zero digit of b!.
Z(ab) = 66 * Z(ab/5)! * Z(b!)
=> Z(24) = 66* Z(4!)* Z(4!) = 66*24*4 = xxxx36.
Let me tell you what have I done.
ab is our number 24.
Z(ab/5)! is the last two non-zero digits of highest multiple of 5 in ab.
Z(b!) is the last non-zero digit of b!.
@chandrakant.k said:i do not know how this is derived. learnt this somewhere in PG Last two non-zero digit of n!Z(ab) = 66 * Z(ab/5)! * Z(b!)=> Z(24) = 66* Z(4!)* Z(4!) = 66*24*4 = xxxx36.Let me tell you what have I done.ab is our number 24.Z(ab/5)! is the last two non-zero digits of highest multiple of 5 in ab.Z(b!) is the last non-zero digit of b!.
i dont think the method is correct because-
fact(10)=3628800,means last two non zero digits are 88.
according to formula-
fact(10)=66*fact(10/5)*fact(0)
=66*fact(2)*1
=66*2*1=xxx132
so,last two digit comes out to be 32.....
correct me if i am wrong....
fact(10)=3628800,means last two non zero digits are 88.
according to formula-
fact(10)=66*fact(10/5)*fact(0)
=66*fact(2)*1
=66*2*1=xxx132
so,last two digit comes out to be 32.....
correct me if i am wrong....