[nagi_nov3]

@ hanuman

u have elaborated too much i guess

@siddhesh_j

buddy try to sit 2.5 hours on one go, it is not as useful as it seems



Problem :#
the natural no. 555..5 (upto 65 fives) is equla to

1. 5x[(10^65)-1/9]
2. 5x(1+10^13+10^26+...+10^65)(1+10+10^2+...+10^12)
3.5x(1+10^5+10^10+...+10^60)(1+10+...+10^4)
4.none of these

i too dont have answer

P:S: please dont post ur answer without explanation , thats for no use

[warrior]

Originally Posted by nagi_nov3

Problem :#
the natural no. 555..5 (upto 65 fives) is equla to

1. 5x[(10^65)-1/9]
2. 5x(1+10^13+10^26+...+10^65)(1+10+10^2+...+10^12)
3.5x(1+10^5+10^10+...+10^60)(1+10+...+10^4)
4.none of these

i too dont have answer

P:S: please dont post ur answer without explanation , thats for no use

x=5555555555.............

9/5x=999999999999.........=10^65-1

x=5*[{(10^65)-1}/9]

[hanuman]

Originally Posted by nagi_nov3

@ hanuman

u have elaborated too much i guess

@siddhesh_j

buddy try to sit 2.5 hours on one go, it is not as useful as it seems



Problem :#
the natural no. 555..5 (upto 65 fives) is equla to

1. 5x[(10^65)-1/9]
2. 5x(1+10^13+10^26+...+10^65)(1+10+10^2+...+10^12)
3.5x(1+10^5+10^10+...+10^60)(1+10+...+10^4)
4.none of these

i too dont have answer

P:S: please dont post ur answer without explanation , thats for no use

I feel option 1 and 3 are holding for this

1) 5*10^64+.....5*10^0
= 5*(10^64+...10^0) (sum of gp)=5*(10^65-1)/(9)

2) Option 3 is quite obivious with direct multiplication (of (1+10^5+10^10+...+10^60) and(1+10+...+10^4))..

Please correct me if i am wrong (Hoping my *improvement* stops at some stage)

[mave]

Originally Posted by nagi_nov3

@ hanuman

u have elaborated too much i guess

@siddhesh_j

buddy try to sit 2.5 hours on one go, it is not as useful as it seems



Problem :#
the natural no. 555..5 (upto 65 fives) is equla to

1. 5x[(10^65)-1/9]
2. 5x(1+10^13+10^26+...+10^65)(1+10+10^2+...+10^12)
3.5x(1+10^5+10^10+...+10^60)(1+10+...+10^4)
4.none of these

i too dont have answer

P:S: please dont post ur answer without explanation , thats for no use

clearly, the answer is 3.
@warrier - your approach is right but check for the brackets. it's (10^65) - 1/9

5555...65 times
= 5(111....65 times)
= 5(1 + 10 + 10^2 + .... 10^64)
= 5(1+10^5+10^10+...+10^60)(1+10+...+10^4)
.

[anupam_khanna]

Originally Posted by hanuman

Originally Posted by hanuman
Let us consider a 2 digit number d1d0
we want to check whether this two digit number is divisible by 7
(i.e) d1d0 on division by 7 should leave a remainder 0..........

perfect.... thanks a lot dude.... this cleared my doubts...

[catfever]

Originally Posted by nagi_nov3

here u go

we need 0's that is nothing but 5*2 multiples

we know there are so many even numbers which are more than 5 in number

so our task is reduced to calculate no. of 5's

we will have 20 5's till 102 and 4 25's till 102

means 20 + 4 (every 25 muliple will have additional 5)

so total 24 zeros

going by that logic 432! should have 86+17=103 0s right??? but the ans i guess is 106 0s pls guide me??:

[vinit456]

Originally Posted by catfever

going by that logic 432! should have 86+17=103 0s right??? but the ans i guess is 106 0s pls guide me??:

U missed out on the last 3

432/5=86
86/5=17
17/5=3
86+17+3=26

[nagi_nov3]

Originally Posted by catfever

going by that logic 432! should have 86+17=103 0s right??? but the ans i guess is 106 0s pls guide me??:

125 contains additional 5

so 125,250,375 so 3 more

432/5 = 86 5's

86/5 = 17 5's ( which are from 25 multiples)

17/5 = 3 5's ( which are 125 multiples)

so 86+17+3 = 106


now try for 100!/12^x what is the maximum value x can take?

[vinit456]

Originally Posted by nagi_nov3

now try for 100!/12^x what is the maximum value x can take?

32?

100!/4=>32

100!/3=>47

x=32

[nagi_nov3]

Originally Posted by vinit456

32?

100!/4=>32

100!/3=>47

x=32

Amswer is 47, try again..

use this for example 12!/12