Solve the given puzzle 2
- Status
piyush gupta
Cyborg Agent
Best of luck
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.
no one has yet solved my puzzles. The solutions will be given tommorow.This is the last day to solve it.
Last edited:
piyush gupta
Cyborg Agent
^^How did u got this result...
first let me know only then i can tell weather u r right or wrong
first let me know only then i can tell weather u r right or wrong
Kalyan
Journeyman
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???
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???
^^ 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
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
Puzzle 6 added.Hope anyone solves this within one week.adi007 said:Puzzle 6
Here is the 6th puzzle
For some value of C R O S A D N G E the condition
CROSS+ROADS=DANGER holds good
CROSS
+ROADS
-------------
DANGER
find the value of C R O S A D N G E
THE RULES ARE:
VALUE OF C ,R ,O ,S ,A ,D ,N ,G ,E MUST BE DIFFERENT
IT SHOULD NOT HAVE '0' VALUE
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:5
adi007 3 (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 and adi007(me)
Last edited:
piyush gupta
Cyborg Agent
Kalyan said:
great
u r a beautiful mind
piyush gupta
Cyborg Agent
No solutions found
QwertyManiac
Commander in Chief
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:
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:
harryneopotter
Padawan
great ................................
You are right QwertyManiac
Puzzle 7 added!!Thread updated!!
adi007 said:Puzzle 7
Here is the 7th puzzle
There are 10 cigarates in a pack.Each cigar weighs 10 gram.But due to manufacture defect one pack is produced in which all cigars weigh 9 gram.You are allowed to use electronic weighing machine only once.You have to find the pack which is defective.(You can remove cigars from the pack)
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:6
adi007 3 (If nobody solves the puzzle, then it will go to my credit.)
The_Devil_Himself 3
Vyasram 1
aditya.shevade 1
entrana 1
QwertyManiac 1
Leading :The_Devil_Himself and adi007(me)
Puzzle 7 added!!Thread updated!!
Last edited:
fun2sh
Pawned!... Beyond GODLIKE
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
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
Last edited:
adi007 said:Puzzle 8
Here is the 8th puzzle
Fill in the blanks to make a meaningful sentense
THE __________________ DOCTOR WAS __________ _____________ TO OPERATE BECAUSE THERE WAS ________________________________ ___________________________.
RULES:
The first blank should be of a single word
You have to fill the remaining blanks such that if you add 2 consecutive blanks you will get the word filled in first blank
Ex:If the first blank is say pathologist
then it could be split as path ologist or patholo gist.Any way you like .these two split words should be filled in 2 blanks.It is not neccesary you have to split the same way for 4th and 5th blank.
EX:The pathologist doctor was path ologist to operate because there was pa thologist.
Of course this doesn't make a sense
Note:state your proof along with the answer.If there are multiple solutions ,then the first 3 unique solutions will be take into regard
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adi007 3 (If nobody solves the puzzle, then it will go to my credit.)
The_Devil_Himself 3
Vyasram 1
aditya.shevade 1
entrana 1
QwertyManiac 1
fun2sh 1
Leading :The_Devil_Himself and adi007(me)
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QwertyManiac
Commander in Chief
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.
My answer's 'Notable'.
I.e.:
The notable doctor was not able to operate because there was no table.
pritish_kul2
★★★★★★★★★
notable.
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