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Numerical Ability
Permutation and Combination
How many zeros will be there at the end of expression (2!)^2! + (4!)^4! + (8!)^8! + (9!)^9! + (10)^10! + (11!)^11!?
Read Solution (Total 11)
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- to find the numbr of zeroes in this series we need to have look on first two term.As the third term onward is going to be very high.so numbr of zeroes will be given by.4+24^24..as 24^24 gives 6 as its unit digit so number of zeroes in this expression will be 1.
- 11 years agoHelpfull: Yes(35) No(13)
- None because (2!)^2! does not have any zero at the end.
- 11 years agoHelpfull: Yes(22) No(12)
- none. because (2!)^2+(4!)^4 donot have any zero at the end
- 11 years agoHelpfull: Yes(6) No(9)
- 1 zero as (2!)^2 + (4!)^4! will have a zero at the last digit.
as last digit of 1st exp is 4 and last digit of 2nd exp is 6. - 11 years agoHelpfull: Yes(5) No(2)
- last digit of 2!^2! = 4
last digit of 4!^4! = 6 so total wud be 10 and 0 is in units place and 1 is carried
now 8!^8! contains 8! zeroes so
somenumber ....................000000000000000000000....00000000+ 10 = .................0000000.....000000010
so at the end it wud be 10
so at the end of the given expresion there wud be only one zero - 11 years agoHelpfull: Yes(5) No(1)
- None at the last digit of the expression will be determined by 2!^2! which results into a 4. Hence at the end of expression there will no zero.
- 11 years agoHelpfull: Yes(4) No(4)
- 0 as 2^2 will yield 4 and hence the unit's digit will b 4
- 11 years agoHelpfull: Yes(2) No(2)
- no zero bcz 2^2 is 4..nd end digits is not zero
- 11 years agoHelpfull: Yes(2) No(1)
- ca u explain it @amisha ????????
- 11 years agoHelpfull: Yes(1) No(1)
- no zeros is answer
- 11 years agoHelpfull: Yes(0) No(0)
- one zero
2^2 =4
4!^4! = 24^24 ---> 6 is last digit
so 1 zero at last - 10 years agoHelpfull: Yes(0) No(0)
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