special numbers question??????

[ankursanghi]

going by the same logic as given by eccentric,

instead of taking 7 to divide 999999, i take 777 to divide 999999, i get 1287 as dividend, meaning thereby that it will be 18 such numbers... looks like this answer is also not correct, i have a gut feeling that the answer would be 9.... don't know how.

[khanna_sumit]

my question got wasted... ......no one except 2 or 3 persons seems to have tried it ....

[DesiGuru]

Originally Posted by khanna_sumit

my question got wasted... ......no one except 2 or 3 persons seems to have tried it ....

The answer is 8.
Its the question that drives us Neo.

K

[khanna_sumit]

Originally Posted by DesiGuru

The answer is 8.
Its the question that drives us Neo.

K

your answer is correct.....approach????????

[made_for_iims]

Originally Posted by khanna_sumit

your answer is correct.....approach????????

check this approach........for forming those special numbers...........
n also check the sum should reach

special numbers are
..................................
0.000000000000000000
..................................

now consider first digit after decimal
in all special numbers only one special number wil have 7 as next digit.....rest will have 0

....................................
0.700000000000000000
...................................

now you are left with 0.3.........
two digits after decimal....
in all special number four (28 <30 && 35>30) special number will have 7 as 2nd digit after decimal.....rest wil have 0.....

....................................
0.77000000000000000
0.07000000000000000
0.07000000000000000
0.07000000000000000
...................................

now u r left with 0.02
similarly.........at 3rd digit you will have 2 7's
leaving you with 0.006
on 4th digit you wil have 8 7's.............as 56<60 && 63>60...

proceed till u get something like 0.00000001.....................
there u stop & find max number of 7's you found..............

in this case it will be..
1st digit ------------ 1
2nd digit-------------4
3rd digit--------------2
4th digit-------------8
5th digit---------------5
6th digit----------------7.....

among them 8 is highest..........so u can make 1 in 8 such possible special numbers....
you can make many combination of special numbers...................................

i hope it clears.........one combination i am writing

0. ( 777777 ) bar
0. ( 077777 ) bar
0. ( 070777 ) bar
0. ( 070777 ) bar
0. ( 000777 ) bar
0. ( 000707 ) bar
0. ( 000707 ) bar
0. ( 000700 ) bar

[khanna_sumit]

Originally Posted by made_for_iims

check this approach........for forming those special numbers...........
n also check the sum should reach

special numbers are
..................................
0.000000000000000000
..................................

now consider first digit after decimal
in all special numbers only one special number wil have 7 as next digit.....rest will have 0

....................................
0.700000000000000000
...................................

now you are left with 0.3.........
two digits after decimal....
in all special number four (28 <30 && 35>30) special number will have 7 as 2nd digit after decimal.....rest wil have 0.....

....................................
0.77000000000000000
0.07000000000000000
0.07000000000000000
0.07000000000000000
...................................

now u r left with 0.02
similarly.........at 3rd digit you will have 2 7's
leaving you with 0.006
on 4th digit you wil have 8 7's.............as 56<60 && 63>60...

proceed till u get something like 0.00000001.....................
there u stop & find max number of 7's you found..............

in this case it will be..
1st digit ------------ 1
2nd digit-------------4
3rd digit--------------2
4th digit-------------8
5th digit---------------5
6th digit----------------7.....

among them 8 is highest..........so u can make 1 in 8 such possible special numbers....
you can make many combination of special numbers...................................

i hope it clears.........one combination i am writing

0. ( 777777 ) bar
0. ( 077777 ) bar
0. ( 070777 ) bar
0. ( 070777 ) bar
0. ( 000777 ) bar
0. ( 000707 ) bar
0. ( 000707 ) bar
0. ( 000700 ) bar

good work

but there may be a systematic solution as well
any special number of this type can be written as 7*(another type of special number which contains only 1s and 0s) so if 7*a1,7*a2......7*aN are such special no.s which satisfy the condition
7*a1+7*a2+7*a3+.........7*aN=1 where a1,a2,a3......are special numbers containing 0s and 1s only......or we can write the same as
a1+a2+a3+.........=1/7=0.142857
lets say f(n) is number of numbers in which 1 comes at nth place after the decimal...
using the condition we can say that one number can have 1 at first place, 4 numbers can have 1 at the second place, 2 numbers can have 1 at third place and so on.......
here we can see that 8 numbers will have 1 at the fourth place which is the largest amongst the others....so at least there should be 8 numbers of this type