# Solve the given puzzle 2



## adi007 (Sep 17, 2007)

Puzzle 9

Here is the 9th puzzle

*
There are 9 iron balls.All look alike,but one ball weighs more than the other 8 balls.You have to find that ball.
You will be allowed to use weighing scale twice.
*

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:8
adi007                   3   (If nobody solves the puzzle, then it will go to my credit.)
The_Devil_Himself 3
QwertyManiac       2
Vyasram               1
aditya.shevade      1
entrana                 1
fun2sh                   1
Leading :The_Devil_Himself and adi007(me)


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## ahref (Sep 17, 2007)

1000000000


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## adi007 (Sep 17, 2007)

ahref said:
			
		

> 1000000000


wrong.
Actually you have not understood the problem itself


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## ahref (Sep 17, 2007)

1000000001


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## piyush gupta (Sep 17, 2007)

^^Ya its the right answer

I was about to post it

It will be better if u put  all ur puzzles in one thread

*www.thinkdigit.com/forum/showthread.php?t=68141

Mods please merge two threads


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## Kalyan (Sep 17, 2007)

No.. it cant be.. 


> 1st digit in the number must specify the number of 1's in the number,



in 





> 1000000001


 , there are 2 '1's...

The answer should be 2100000001.. 

two 1's
one 2
one 10


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## piyush gupta (Sep 17, 2007)

Yups point hai

urs is right answer


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## adi007 (Sep 17, 2007)

Kalyan said:
			
		

> No.. it cant be..
> 
> 
> in  , there are 2 '1's...
> ...


wrong , last digit[10th] should indicate the number of 0's


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## Kalyan (Sep 17, 2007)

Read your own puzzle dude..



> and the 10th digit should specify the number of 10's in the number.



you said it to be the no. of 10's


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## adi007 (Sep 17, 2007)

piyush gupta said:
			
		

> ^^Ya its the right answer
> 
> I was about to post it
> 
> ...


It will be more neat and tidy if i create separate thread for each puzzle.


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## piyush gupta (Sep 17, 2007)

But  u can create on thread saying "All puzzles here dare to solve"

that will be better


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## The_Devil_Himself (Sep 17, 2007)

2100010006?

did I just solve this one too?


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## adi007 (Sep 17, 2007)

Kalyan said:
			
		

> Read your own puzzle dude..
> 
> 
> 
> you said it to be the no. of 10's


oops! typing mistake.
10th digit should indicate the number of 0's


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## piyush gupta (Sep 17, 2007)

^^I thinks devil got it right?


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## adi007 (Sep 17, 2007)

The_Devil_Himself said:
			
		

> 2100010006?
> 
> did I just solve this one too?


Bravo! Puzzle solved!
Excellent!!!


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## The_Devil_Himself (Sep 17, 2007)

I am dangerous!!!
2/2.


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## Kalyan (Sep 17, 2007)

then this is the right answer.. Congrats.. Devil


> 2100010006?
> 
> did I just solve this one too?


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## adi007 (Sep 17, 2007)

The_Devil_Himself said:
			
		

> I am dangerous!!!
> 2/2.


yes,u are man
Because of you i have to find new new puzzles 

for the next puzzle,shall i create new thread or new post or edit existing thread.
Please suggest


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## navjotjsingh (Sep 17, 2007)

2100010006

is the solution. Am I correct? No mistake seems now. Oops..didn't saw solution by Devil.

Please continue puzzles in this thread only.


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## The_Devil_Himself (Sep 17, 2007)

adi007 said:
			
		

> yes,u are man
> Because of you i have to find new new puzzles
> 
> for the next puzzle,shall i create new thread or new post or edit existing thread.
> Please suggest



Thanks dude.Edit this thread itself.
I think I should give others a chance too.lols.


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## entrana (Sep 17, 2007)

1000000008


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## QwertyManiac (Sep 17, 2007)

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


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## navjotjsingh (Sep 17, 2007)

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.


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## The_Devil_Himself (Sep 17, 2007)

better luck next time entrana.


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## nithinks (Sep 17, 2007)

[EDITED] wrong solution


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## contactpraven2001 (Sep 18, 2007)

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


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## The_Devil_Himself (Sep 18, 2007)

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


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## adi007 (Sep 18, 2007)

new puzzle added.Thread updated!!


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## The_Devil_Himself (Sep 18, 2007)

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.


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## adi007 (Sep 18, 2007)

The_Devil_Himself said:
			
		

> puzzle 3 answer is:2
> 
> logic is to keep on adding the digits till you get a single digit
> 
> ...



you are right!!


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## entrana (Sep 18, 2007)

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


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## The_Devil_Himself (Sep 18, 2007)

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.


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## piyush gupta (Sep 18, 2007)

^^ 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


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## adi007 (Sep 19, 2007)

piyush gupta said:
			
		

> ^^ may be u can solve this
> 
> Divisible from 1 to 9
> 
> ...


987654321 ????



			
				adi007 said:
			
		

> Puzzle 4and 5
> 
> Here is the 4th puzzle
> 
> ...



New puzzles added.Thread updated


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## entrana (Sep 19, 2007)

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????


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## piyush gupta (Sep 19, 2007)

adi007 said:
			
		

> 987654321 ????


 
No

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


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## adi007 (Sep 20, 2007)

entrana said:
			
		

> 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 ?


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## entrana (Sep 20, 2007)

o ya got confused, nooooooooooooooooooooooooooooooooooo


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## adi007 (Sep 21, 2007)

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


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## piyush gupta (Sep 21, 2007)

Nopes

and what abut my puzzle


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## adi007 (Sep 24, 2007)

^^ just trying to create a VB program to solve your puzzle.Possibly i will get a answer by tommorow..


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## piyush gupta (Sep 24, 2007)

Best of luck


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## adi007 (Sep 26, 2007)

created a vb program but the problem is it is checking 10,000 possibilites in one sec.the 9th digit number is ending with 99999999.so it will take 99999 sec's 

no one has yet solved my puzzles. The solutions will be given tommorow.This is the last day to solve it.


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## Kalyan (Sep 26, 2007)

The answer to the 9 digit number is: 381654729. solved through elimination and used excel for results.


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## piyush gupta (Sep 26, 2007)

^^How did u got this result...
first let me know only then i can tell weather u r right or wrong


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## Kalyan (Sep 26, 2007)

I know that it is the right answer 'cause I checked the condition u gave.

  Process: I first eliminated the digits for each place value. then, according to the divisibility rules, I eliminated to some more extent. then, I got combinations like this:

divisible by 3
147xxxxxx
381xxxxxx      
387xxxxxx      
789xxxxxx      
783xxxxxx      
741xxxxxx      
981xxxxxx      
987xxxxxx    

divisible by 4

xx16xxxxx
xx12xxxxx
xx36xxxxx
xx32xxxxx
xx76xxxxx
xx72xxxxx
xx96xxxxx
xx92xxxxx

divisible by 6
xxx654xxx
xxx258xxx


divisible by 8

xxxxx816x     
xxxxx832x     
xxxxx872x
xxxxx896x
xxxxx416x
xxxxx432x
xxxxx472x
xxxxx496x 

Now, using the most appropriate combinations(eliminating the repeating digits) and using some functions in excel, I sorted out to these values.

1472583xx
3816547xx
7836549xx

using the divisibility by 8 combinations,

38165472x is the only appropriate combination. so, it gives me:381654729

381654729	/	9	=	42406081
38165472	/	8	=	4770684
3816547	/	7	=	545221
381654	           /	6	=	63609
38165	           /	5	=	7633
3816	           /	4	=	954
381	           /	3	=	127
38	           /	2	=	19
3	           /	1	=	3

ok???


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## adi007 (Sep 27, 2007)

^^ great!!!

Answers to puzzle 4 and 5:

Puzzle 4:

*A B C D E F G H I J K L M N O P Q R S T U V W X Y Z*

take individual words,count the number of words,subtract it by 1,
thus,
mncyhalwn  -->9 chars-->9-1-->8
repace each character by a character 8 previous to it.
thus m=d n=e c=t ...
thus mncyhalwn becomes detpyrcne
reverse the word --->encrpted
lnhzzlt --->egassem--->message
jkbruy  --->devlos    --->solved
gqd     --->dna --->and
xrb       --->uoy -->you
hud     ---->era --->are
yaotkm --->sucneg --->genius

the sentence is 
*"encrpted message solved and you are genius"*

Puzzle 5 is some what easy

2267256821

write the digits which are repeated more than once
2 repeated 4 times
6 --2 times
multiply the digit with it's number of occurence
thus 2*4=8
6*2=12
add this
8+12-->20
divide it by the maximum occurence of a number
thus in this number the maximum occurence is 4
20/4-->5
multiply it by the number which has occured most times
thus in this case it is 2
5*2-->10

similarly,
8387281215 
8*3=24 2*2=4 1*2=2
24+4+2=30
30/3=10
10*8=80

and 

6165728216
6*3=18
2*2=4
1*2=2
18+4+2=24
24/3=8
8*6=48

*48 is the right answer*

since no one was able to solve the puzzle i get 3 points 



			
				adi007 said:
			
		

> Puzzle 6
> 
> Here is the 6th puzzle
> 
> ...


Puzzle 6 added.Hope anyone solves this within one week.


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## piyush gupta (Sep 27, 2007)

Kalyan said:
			
		

> I know that it is the right answer 'cause I checked the condition u gave.
> 
> Process: I first eliminated the digits for each place value. then, according to the divisibility rules, I eliminated to some more extent. then, I got combinations like this:
> 
> ...


 
great

u r a beautiful mind


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## adi007 (Oct 8, 2007)

no one solved my puzzle


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## piyush gupta (Oct 8, 2007)

No solutions found


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## QwertyManiac (Oct 8, 2007)

I assume no two characters are equal in any way.

If we use numbers, R must be even cause at the end S + S = R ( Odd + Odd and Even + Even both equal Even, so R's got to be Even )

First conclusion: R is even ( Say, 2, 4, 6, 8, since solution set doesn't consist of any 0 ( i.e. 1-9 only, possibly ) )

Now CROSS
   +ROADS
   ------
   DANGER

As you'd notice, D is extra, like something carried over in addition. So D's got to be 1 since any sum of 1-9 will can can give a carry of 1 alone.

So, A will have a value of [2,3,4,5,6,7,8,9] cause D's already taken 1.

First value found, D = 1

C+R = Set of numbers ( 12,13,14,15,16,17,18 ( 19 not possible as 10 is not allowed ) )

Which equal only to sums of:

If A:		C+R    C+R
	Case 2:
		3+9 || 9+3
		4+8 || 8+4
		5+7 || 7+5

		( Other combo's not possible cause C!=R )

	Case 3:
		4+9 || 9+4
		5+8 || 8+5
		6+7 || 7+6

	Case 4:
		5+9 || 9+5
		6+8 || 8+6

	Case 5:
		6+9 || 9+6
		7+8 || 8+7

	Case 6:
		7+9 || 9+7

	Case 7:
		8+9 || 9+8

	Case 8:
		Not Possible as ( 9,9 ) is the only case and C!=R ( Assumed at the beginning of this puzzle solving )

Thus from the above equation, as R is even, we have R = Possiblity( 4, 6, 8 ) and C = Possiblity( 4, 5, 6, 7, 8, 9 )

So far,
A = ( 2, 3, 4, 5, 7 ) [6's totally Odd for values of R]
D = 1
C = ( 4, 5, 6, 7, 8, 9 )
R = ( 4, 6, 8 )

Now since R is Even and definitely in [4, 6, 8]

From R + O = N, we get R = N - O

Thus,

If R:		N-O

	Case 4:
		9-5
		7-3
		6-2

		( Other combo's not possible as in our assumption, we use R = 4 )

	Case 6:
		9-3
		8-2

	Case 8:
		Not possible as 1's already taken by D ( 9-1 being the only possible way )

Thus from the above,

N = ( 6, 7, 8, 9 )
O = ( 2, 3, 5 )
A = ( 2, 3, 4, 5 )
D = ( 1 )
C = ( 4, 5, 6, 7, 8, 9 )
R = ( 4, 6 )

Also, A doesn't have 7 since C + R = A but R has only ( 4, 6 ) while C is ( 4, 5, 6, 7, 8, 9 ) ( Adding C+R will never yeild 17 )

Now similarly, G = O + A

Thus, O = G - A

If O:		G-A

	Case 2:
		7-5
		6-4
		5-3

	Case 3:
		8-5
		7-4
		5-2

		( 6-3 not Possible as O!=A, thus I G can't be 6 and A can't be 3 )

	Case 5:
		9-4
		8-3
		7-2

This gives us two conclusions,

First, A is now ( 2, 4, 5 )
Second, G is ( 5, 7, 8, 9 )

Phew .. I need a break.

Continuing,

We now have:

A = ( 2, 4, 5 )
D = ( 1 )
C = ( 4, 5, 6, 7, 8, 9 )
R = ( 4, 6 )
O = ( 2, 3, 5 )
N = ( 6, 7, 8, 9 )
G = ( 5, 7, 8, 9 )

Good so far? I hope so, else my entire evening is lifeless. 

Now since R = S + S and R has ( 4, 6 )

If R:		S+S

	Case 4:
		2+2

	Case 6:
		3+3

Therefore, S = ( 2,3 )

Applying for E = S + D, we have:

E = 2+1 = 3 or
E = 3+1 = 4

Thus, E = ( 3,4 )

So this far,

D = ( 1 )
A = ( 2, 4, 5 )
C = ( 4, 5, 6, 7, 8, 9 )
R = ( 4, 6 )
N = ( 6, 7, 8, 9 )
O = ( 2, 3, 5 )
G = ( 5, 7, 8, 9 )
S = ( 2, 3 )
E = ( 3, 4 )

Not being able to proceed more as I did so long, I try to apply the smallest value sets to the solution of bigger sets.

If S is 2, R will be equal to 4 ( S + S ) and E will equal 3 ( S + D )

Thus, if R = 4 and E = 3 and S = 2 the only possible values of A will be ( 5 ) and O will be ( 5 ) too which makes this impossible since A cant equal O.

Thus S will NOT be 2 and is surely 3 instead.

Now if S is 3, we have:

E = S + D = 4
R = S + S = 6

Thus, we now have possible values as:

D = ( 1 )
A = ( 2, 5 )
C = ( 5, 7, 8, 9 )
R = ( 6 )
N = ( 6, 7, 8, 9 )
O = ( 2, 3, 5 )
G = ( 5, 7, 8, 9 )
S = ( 3 )
E = ( 4 )

Now A = C + R and has got to be either 12 or 15 since it gives D = 1 as carry.

Thus only 9 + 6 satisfies it for A = 5 and 12 isn't possible with the remaining values.

Hence, A = ( 5 ) and C = ( 9 ) 

[Proof: C = A - R = 15 - 6 = 9 not considering the Carry character. This proof is only additive to the above logical result]

So again, so far, removing duplicates as we've been doing,

D = ( 1 )
A = ( 5 )
C = ( 9 )
R = ( 6 )
S = ( 3 )
E = ( 4 )
N = ( 7, 8 )
O = ( 2 )
G = ( 7, 8 )

( As numbers 3, 5 and 9 are already used up. )

Now N = R + O = 6 + 2 = 8 and hence G = 7 ( Odd one out )

[Proof for G: G = O + A = 2 + 5 = 7, again an additive proof.]

Thus finally the solution is,

D = 1
A = 5
N = 8
G = 7
E = 4
R = 6
C = 9
O = 2
S = 3

Or in numeric order:

*D = 1
O = 2
S = 3
E = 4
A = 5
R = 6
G = 7
N = 8
C = 9*

Phew, took an hour. Hope am not wrong in my approach itself! If I am, I desperately need a life. 

P.s. Am attaching the TXT format in case the formatting I typed this in isn't showing well for reading here.

Attachments below:


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## harryneopotter (Oct 8, 2007)

great ................................


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## adi007 (Oct 9, 2007)

You are right QwertyManiac 



			
				adi007 said:
			
		

> Puzzle 7
> 
> Here is the 7th puzzle
> 
> ...



Puzzle 7 added!!Thread updated!!


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## Kalyan (Oct 9, 2007)

How many packs are there? Do you mean that one cigar is weighing 9gm or a complete pack with cigars weighing 9gm each? Please be clear in the qn.. I think some detail is missing..


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## fun2sh (Oct 9, 2007)

i will post the full explained sol to CIGARETTE problem as soon i get to my Lappy(Abhi me in aur using gprs in mobile.) 
but here is the simple solution (without explaination)
the problem is actually simple. 
the procedure is take 1 Cig from 1st 
2 cig from 2nd 
3 cig from 3rd 
and so on 
so 10 cig from 10th box and measure the weight for all these 55cig together.
for boxes without defect the total weigt would had been 55 grams.
but if there's defect then we will get difference in weigt as follows 
0.1 difference for 1st 
0.2 difffrence for 2nd
and so on 

i think i hav explained here only properly. if still any doubt then i will clear it wen i reach my Lappy 

one mistake i considered weigt of each CIGAR as 1grams instead of 10grams. 
but it doesnt matter and we will get 550grams for total undefected in my procedure 
and difference of 1g,2g, and so on

i will explain neatly from my Lappy. 

by the way ANY PLZ SORT ALL THE PUZZLES AT ONE PLACE. I SEE SOME OF SOLUTIONS HERE BUT CANT FIND THE QUESTIONS

ok here is the solution in explained way


GIVEN:- 10 BOXES OF CIG
        10 CIG IN EACH BOX...  SO TOTAL OF 100 CIG
	WEIGHT OF EACH CIG = 10 GRAM
	WEIGHT OF EACH BOX = 100 GRAM
        BUT ONE BOX HAS A DEFECTED CIG IN WHICH EACH CIG WEIGHT 9GRAMS

PROCEDURE :- TAKE OUT 1 CIG FROM 1st BOX
                      2 CIG FROM 2nd BOX
                      3 CIG FROM 3RD BOX
		      .
		      .
		      .
		      .
		      .
		and so on...
		     10 CIGS FROM 10th BOX

	   NOW MEASURE THE WEIGHT OF THESE 55 CIGS


IF THERE WAS *NO DEFECT* THEN WEIGHT WOULD HAD BEEN
  1x10 + 2x10 + 3X10 + .......... + 10x10 = 550 grams

BUT DUE TO DEFECT THERE WILL BE AN ERROR

case 1: IF 1st BOX HAS DEFECTED CIGS THE WEIGHT IS
  1x9 + 2x10 + 3X10 + .......... + 10x10 = 549 grams

DIFFERENCE FROM UNDEFECTED CIG WEIGHT = 550-549= 1 GRAM

case 2: IF 2nd BOX HAS DEFECTED CIGS THE WEIGHT IS
  1x10 + 2x9 + 3X10 + .......... + 10x10 = 548 grams

DIFFERENCE FROM UNDEFECTED CIG WEIGHT = 550-548= 2 GRAM

case 3: IF 3rd BOX HAS DEFECTED CIGS THE WEIGHT IS
  1x10 + 2x10 + 3X9 + .......... + 10x10 = 547 grams

DIFFERENCE FROM UNDEFECTED CIG WEIGHT = 550-547= 3 GRAM

.
.
.
.
and so on


case 10: IF 10th BOX HAS DEFECTED CIGS THE WEIGHT IS
  1x10 + 2x10 + 3X10 + .......... + 10x9 = 540 grams

DIFFERENCE FROM UNDEFECTED CIG WEIGHT = 550-540= 10 GRAM



hence IN THE DIFFERENCE IS TELLIN THE NAME OF THE DEFECTED BOX


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## fun2sh (Oct 11, 2007)

hello adi............. where r mine points


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## adi007 (Oct 15, 2007)

^^ excellent.Points added.

I have started a new thread devoted to C puzzles

*www.thinkdigit.com/forum/showthread.php?t=70697Here it is


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## adi007 (Oct 18, 2007)

adi007 said:
			
		

> Puzzle 8
> 
> Here is the 8th puzzle
> 
> ...



New puzzle added  thread updated...


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## QwertyManiac (Oct 18, 2007)

Hey this is easy, cause I've done it before somewhere! 

My answer's 'Notable'.

I.e.:

The *notable* doctor was *not able* to operate because there was *no table*.


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## pritish_kul2 (Oct 18, 2007)

notable.


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## adi007 (Oct 18, 2007)

Puzzle sucessfully solved by qwertymaniac.Any other answers???


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## pritish_kul2 (Oct 18, 2007)

Th unable doctor was un able to operate because there was una ble

The unstable doctor was un stable to operate bcoz there were uns table


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## adi007 (Oct 24, 2007)

^^^
what does una ble means
what does uns table means
un stable is a single word...



			
				adi007 said:
			
		

> Puzzle 9
> 
> Here is the 9th puzzle
> 
> ...


Puzzle9 added.Thread updated!!
From now onwards only one week time will be given


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## harryneopotter (Oct 24, 2007)

First Divide the Balls into 3 groups of three ball each ........ let their name be Group A (with A1, A2 and A3 ), Group B(B1....) and Group C (C1 ...)


Now take the weighing scale and put Gruop A on one side and Group B on 2nd side. Now if the defective ball is in Group A ..it will weigh more than group B ... or if its in Group B ..then group B will weigh more...if both are equal then its in gruop C ....

Now again take the group in which we have the defective ball (derrived from previous step) ...lets assume it was Group C ...

take C1 and C2 and put them on each side of weighing scale ...
if C1 weighs more then C1 is defective 
If C2 weighs more then C2 is defective ..
if both weigh equal ..then C3 is defective.

Hope my solution is right and easy to understand... srry for grammer.

??????????????


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