Solve the given puzzle 2

[entrana]

1000000008

 

[QwertyManiac]

Lol entrana, where is the count for number 8 then?

 

[navjotjsingh]

Correct answer has already been posted and there are very few solutions for this problem in the world and yours is not one of them.

 

[The_Devil_Himself]

better luck next time entrana.

 

[nithinks]

[EDITED] wrong solution

 

[contactpraven2001]

devil is right i got the same .......

 

[The_Devil_Himself]

^^nice try dude.You guys are damn smart.

 

[adi007]

new puzzle added.Thread updated!!

 

[The_Devil_Himself]

puzzle 3 answer is:2

logic is to keep on adding the digits till you get a single digit


5+6+2+3+4+6+4+6+5+4+1+5+1=52=>5+2===7
4+6+4+8+6+6+4+9+8+8+2+1+2=68=>6+8=14=>1+4===5
5+4+6+5+4+6+4+6+4+5+6+4+6=65=>6+5=11=>1+1===2

now Don't tell me I solved 3 in a row.

 

[adi007]

puzzle 3 answer is:2

logic is to keep on adding the digits till you get a single digit


5+6+2+3+4+6+4+6+5+4+1+5+1=52=>5+2===7
4+6+4+8+6+6+4+9+8+8+2+1+2=68=>6+8=14=>1+4===5
5+4+6+5+4+6+4+6+4+5+6+4+6=65=>6+5=11=>1+1===2

now Don't tell me I solved 3 in a row.

you are right!!

 

[entrana]

aww man devil got it again, i knew this was the answer but u just had to post it before mee

 

[The_Devil_Himself]

No class man all the puzzles are damn easy and for a seasoned player like me they are no match.Maybe I should take a break now and let others catch up.

 

[piyush gupta]

^^ may be u can solve this

Divisible from 1 to 9

Find a number consisting of 9 digits in which each of the digits from 1 to 9 appears only once. This number should satisfy the following requirements:
a. The number should be divisible by 9.
b. If the most right digit is removed, the remaining number should be divisible by 8.
c. If then again the most right digit is removed, the remaining number should be divisible by 7.
d. etc. until the last remaining number of one digit which should be divisible by 1.

copied from a site (I am very new to this puzzling thing) but i likes this a lot

 

[adi007]

^^ may be u can solve this

Divisible from 1 to 9

Find a number consisting of 9 digits in which each of the digits from 1 to 9 appears only once. This number should satisfy the following requirements:
a. The number should be divisible by 9.
b. If the most right digit is removed, the remaining number should be divisible by 8.
c. If then again the most right digit is removed, the remaining number should be divisible by 7.
d. etc. until the last remaining number of one digit which should be divisible by 1.

copied from a site (I am very new to this puzzling thing) but i likes this a lot

987654321 ????

Puzzle 4and 5

Here is the 4th puzzle

here is an message written according to some rule.break it:

mncyhalwn lnhzzlt jkbruy gqd xrb hud yaotkm

Here is the 5th puzzle

2267256821 ----->10
8387281215 ----->80
6165728216 ----->?


Bonus: if you are the first to solve both the puzzle, then you will get 3 points instead of 2

Note:state your proof along with the answer.If there are multiple solutions ,then the first 3 unique solutions will be take into regard

Gallery:
Total number of puzzles solved:3
adi007 0 (If nobody solves the puzzle, then it will go to my credit.)
The_Devil_Himself 3
Vyasram 1
aditya.shevade 1
entrana 1

Leading :The_Devil_Himself

New puzzles added.Thread updated

 

[entrana]

puzzle 5 ill try
2-2+6-7+2-5+6-8+2-1 = -5, -5*-2 = 10
8-3+8-7+2-8+1-2+1-5 = -5, -5*-20 = 80
6-1+6-5+7-2+8-2+1-6=12, 12*2 = 24
24????

 

[piyush gupta]

987654321 ????

No

Also u tell me how u reached vaise this is wrong answer

 

[adi007]

puzzle 5 ill try
2-2+6-7+2-5+6-8+2-1 = -5, -5*-2 = 10
8-3+8-7+2-8+1-2+1-5 = -5, -5*-20 = 80
6-1+6-5+7-2+8-2+1-6=12, 12*2 = 24
24????

how it is -5*-2 ,-5*-20 and 12*2 ?

 

[entrana]

o ya got confused, nooooooooooooooooooooooooooooooooooo

 

[adi007]

no one able to solve 2 puzzles. where is devil ????

 

[piyush gupta]

Nopes

and what abut my puzzle