bhramarraj
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Please review.A1-Prashanth sharma
A2-Lalitha Mishra
A3-Sushila Tripathi ----Loud customer
A4-Sushma Verma
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Thank you. You are absolutely right. Could anybody please give logical explaination also.Works if you omit the clue "One of the teachers has his last name as Sharma."
Tripthni Lalita A3
Verma Sushila A1
Sharma Sushma A2
Mishra Prashant A4
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Your interpretation of clue 2 is incorrect. At first I also got confused. Hint is the statement started with "The four teachers are...."I'm reaching an impossibility.
Prashant is the only male.
Sharma is HIS last name.
So Prashant Sharma is one piece of the puzzle. (per intro and clue 2)
Tripathi, Mishra and Sushma are three different people. (per clue 1)
Sushma can't be Sharma, Mishra or Tripathi so must be Verma.
Sushma Verma doesn't teach one either end block (clue 3) and Verma doesn't teach in A3 (clue 2).
That means Sushma Verma teaches in A2... but Lalita teaches in A2.
Anyone want to tell me where I went wrong?
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The teachers Sushila, Lalita, Sushma and Prashant teach in a block of four different but adjacent class rooms numbered in order as A1, A2, A3, & A4. One of the teachers has his last name as Sharma. With the help of following clues find out last names of all the teachers and their class room numbers in which they teach:
1. Tripathi teaches so loud that her voice disturbs Mishra & Sushma, who teach in the classrooms adjacent to hers.
2. The four teachers are: Lalita, a teacher who teaches in classroom A2, Verma who does not teach in classroom A3, and only male teacher Prashant.
3. Sushma does not teach at either end of the classroom’s block.
Note: Verma, Sharma, Mishra, and Tripathi are the last names - written after first names.
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OOOPS... what a silly mistake......!
SP is correct, answer to the quiz should be 6AD.
Thank you all again....
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Thank you SP,because there was no year 0. The year after 1BC is 1AD. So, 15 years after 10BC is 6AD.
I think you will have to re-check your answer..
What I have learned is that in counting of years, 1AD starts forward from the birth of Jesus, the year 1BC starts backwards from the same instant.
Therefore when we say an event to have happened 1year after 1BC, it actually have happened in the year 2AD.
I would like to have comments of other members also.....!
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Thanks Nakulendu,5 AD. From year Michael was born till Samuel died there are 95 years. Of these 95 years, their collective lifespan covers 80. This leaves a gap of 15 years when neither lived. So Samuel would have been born 15 years after Michel dies - 5 AD
Your answer is correct.
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Some time back I came across a very interesting puzzle, though not so hard.
Samuel died 95 years after Michel was born. Their combined ages when they died amounted to 80 years. Michel died in the year 10 BC. Which year was Samuel born?
PLEASE FORGIVE IF POSTED BEFORE.
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I know a story of a farmer, who before dying made a will in which 8 horses were to be divided among his three sons, such that each son gets 1/2 nos.
But the no of horses to be distributed were only 7.
Then a wise person solved their problem by adding his own horse to the 7 horses to be distributed.
Then he gave 1/2 nos. i.e. 4 horses to the eldest, then half of remaining i.e. 2 horses to the younger, and half of remaining i.e. 1 horse to the youngest son.
One horse remained after distribution which he took back. So without losing his own horse the problem was solved
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Say you are given a heap of 50 coins, in which only one coin is real, while all other coins are fake. you can not recognize the fake or real coin by seeing or touching them.
When you grab a coin from the heap of 50 coins, there was only one out of 50 chances that you grabed right coin.
Suppose the host now removes 47 fake coins. Only two coins are left in the heap and one in your hand. Now what you will do?
Stick to the coin which you selected When there were 50 coins and there was only a very remote chance that the coin selected was real...?
OR
You will leave that and grabe another coin....?
Definitely when you do not know which coin is real, you shall like to leave the coin in hand which had more chances to have been fake, and grab a new coin which has now better chances to be real.
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Say you are given a heap of 50 coins, in which only one coin is real, while all other coins are fake. you can not recognize the fake or real coin by seeing or touching them.
When you grab a coin from the heap of 50 coins, there was only one out of 50 chances that you grabed right coin.
Suppose the host now removes 47 fake coins. Only two coins are left in the heap and one in your hand. Now what you will do?
Stick to the coin which you selected When there were 50 coins and there was only a very remote chance that the coin selected was real...?
OR
You will leave that and grabe another coin....?
Definitely when you do not know which coin is real, you shall like to leave the coin in hand which had more chances to have been fake, and grab a new coin which has now better chances to be real.
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Water level of the pond rises if it is the ice rock (Iceberg). But initially it will float on the water surface.
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There was 99 pounds of water in 100 pounds Jelly. Say X pound giant jelly is left after evaporation, which now contain 98%water. Balance 2% is jelly which will remain to be 1 pound as originally it was. So 2X/100 = 1 OR X = 50 Pounds.
Therefore Giant gelly now weighs 50 Pounds.
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Alex's first two shots scored 22 (20+2) points. Therefore, remaining four shots scored 71-22 = 49 points, so Alex had not hit the bull to score 50 points.
Davie scored 3 points on his first shot, so balance 71-3=68 points had been scored by him on remaining 5 shots. If on one shot he scored 50 points by hitting the bull than 68-50=18 points were scored by him on remaining four shots, which is possible if he scored 10+5+2+1 points on his remaining four shots.
From the figure we get the scored points on all shots by all the three:
1+1+1
2+2
3+3
5+5
10+10+10
20+20+20
25+25
50
From the above it is clear that Alex scored 20+2+ (49 on 4 shots= 25+20+3+1), to get score 71.
Similarly Davie scored 3+50+10+5+2+1, to obtain score 71.
Then Janie scored 25+20+10+10+5+1 to score 71points.
If We assume Jamie to had hit the bull than Davie could not score 3 points on any shot.
Therefore Davie hit the Bull.
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I normally do not google an answer to a puzzle, but after reading the comments I also realized that answer which I posted by applying mere common sense may be wrong and then I searched this subject on google and found there have been long discussions on this subject.I don't think that is the answer.
AC isn't a time measurement. AD is the time which stands for Anno Domini which means "The year of our lord" or something like that. If you use Before Christ and After Death then where would the years that Jesus lived fall?
So it seems worthless to me now to make further arguments, but I suggest this puzzle could be reproduced as under:
A,B,C,D are on the same level,which is longer,AD or BC ?
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Can somebody please enlighten me. Maybe I'm being stupid, but I can't understand wht this particular puzzle is about!
My answer was based on the assumtion that the statement "A,B,C are on the same level", has no concern withe the Question "which is longer,AC or BC?"
Then AC is used for the period After christ, and BC for Before Christ.
Obviously BC is longer than AC as on today.
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[1] If Ned said it, then it is not a true statement, and one child is not a girl. So probability for another child being a girl: 0%.
[2] If Red said it, then it is a true statement and one child is a girl. Then probability for other child being a girl: 50%.
[3] If Ted said it, then older child is not a boy but a girl. So probability for other child being a girl: 50%. [4] If Zed said it, and if coin selected child is taller, then the taller child is not a boy but a girl, as well as the other child.. Hence the probability for other child being a girl: 100%.
Similarly if the coin selected child is shorter, then the shorter child is not a boy, but a girl, as well as the other child. Probability for other child being a girl: 100%
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May be the barber was a male child.....!!!
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As on today BC is longer than AC.......
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1
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OP states: "They have two kids, one of them is a girl, what is the probability that the other kid is also a girl.
In my view the word 'ALSO' in OP has restricted the probability to only either the other kid (the kid other than the girl kid) is ALSO a girl, or it is Boy. So answer should be 1/2.
But It could also be like this: You are blind folded before two kids. you are told one of the kids is a Girl. NOW you are asked to touch one kid. THEN you are asked what is the probility that the OTHER kid is a girl. In this case there are three possibilities:
You touched may be a BOY, Other is a GIRL. OR
You touched may be a GIRL, Other is a GIRL. OR
You touched may be a GIRL, Other is a BOY.
But this answer will suit only when 'ALSO' word is deleted from OP.
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I converted the word file to pdf, and attached here, for those who want to solve it matematically
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Sorry I am not able to send the file
Please somebody let me how should I load a word file...?
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Sorry I am not able to send the file
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superprismatic,
Thank you very much to take so interest in this my favourite math puzzle.
It is most interesting when we work out the solution mathematically.
My solution is attached herewith in the word file. But I will be delighted if somebody will work out a better mathematical solution.
From clue [13] & [14], Table of probable numbers will be as under:
A (1-15)
B (16-30)
C (31-45)
D(46-60)
E (61-75)
J
1-15
16-30
31-45
From [13]
46,47,48,49,
56,57,58,59
From [14]
61,63,65,67,
69,71,73 & 75
K
1-15
16-30
31-45
46,47,48,49,
56,57,58,59
61,63,65,67,
69,71,73 & 75
L
1-15
16-30
EMPTY
46,47,48,49,
56,57,58,59
61,63,65,67,
69,71,73 & 75
M
1-15
16-30
31-45
46,47,48,49,
56,57,58,59
61,63,65,67,
69,71,73 & 75
N
1-15
16-30
31-45
46,47,48,49,
56,57,58,59
61,63,65,67,
69,71,73 & 75
According to clue [8], row J contains all ten numerals (0 through 9) only once, so all five spaces in this row will contain only double digit numbers and no duplicate digit numbers. So space AJ will necessarily contain numeral 1, and hence no number containing numeral 1 will be located at other spaces in row J.
For column D, 5 numbers are required from 46,47,48,49,56,57,58,59,&60, so at least one of the numbers will definitely contain numeral 5; and from clue [18], numeral ‘5’ appears in the complete table for once only. Therefore numeral ‘5’ will appear only in column D.
Then from clue [8], each numeral (0 through 9) appears in row 'J', for once only. So numeral 5 will appear only in column D & row J, i.e. at space DJ. So in all other columns numbers containing numeral 5 may be ignored.
Modified table will be as under:
A (1-15)
B (16-30)
C (31-45)
D(46-60)
E (61-75)
J
10,12,13,14
20,23,24,26,
27,28,29,30
32,34,36,37,38,
39,40,42,43
56,57,
58,59
63,67,69,
73
K
1 through 14
Except 5
16 through 30
Except 25
31 through 44
Except 35
46,47,
48,49
61,63,67,
69,71,73
L
1 through 14
Except 5
16 through 30
Except 25
EMPTY
46,47,
48,49
61,63,67,
69,71,73
M
1 through 14
Except 5
16 through 30
Except 25
31 through 44
Except 35
46,47,
48,49
61,63,67,
69,71,73
N
1 through 14
Except 5
16 through 30
Except 25
31 through 44
Except 35
46,47,
48,49
61,63,67,
69,71,73
Now if 30 appeared at space BJ, then as per clue [8], there would be no number containing numeral 3 & 0 at all other spaces in row J. Then at space CJ only 42 would be left, which in turn will mean that numerals 4 & 2 would not appear at any other space in row J; which finally would leave no choice of number at space AJ, which contradicts clue [7]. Hence 30 will not appear at space BJ. Then all probable numbers at space BJ will necessarily contain numeral 2, so at other spaces in row J, numbers containing numeral 2 will not appear.
Also as per clue [10], lowest number in each column is located in row N, so 46 will appear at space DJ.
Modified table will be as under:
A (1-15)
B (16-30)
C (31-45)
D(46-60)
E (61-75)
J
10,13,14
20,23,24,26,
27,28,29
34,36,37,38,
39,40,43
56,57,58,
59
63,67,69,
73
K
1 through 14
Except 5
16 through 30
Except 25
31 through 44
Except 35
47,48,49
61,63,67,
69,71,73
L
1 through 14
Except 5
16 through 30
Except 25
EMPTY
47,48,49
61,63,67,
69,71,73
M
1 through 14
Except 5
16 through 30
Except 25
31 through 44
Except 35
47,48,49
61,63,67,
69,71,73
N
1 through 14
Except 5
16 through 30
Except 25
31 through 44
Except 35
46
61,63,67,
69,71,73
Then if 73 or 63 appeared at EJ,
OR
If 23 appeared at BJ, then CJ would be left with only number 40, which in turn would leave no choice of number at AJ. After eliminating 63 and 73, EJ will be left with numbers 67 and 69, which contain numeral 6. So eliminate all numbers containing numeral 6 at all other spaces in row J, also eliminate number 23 at BJ.
Then according to clue [11], each row contains only one two digit prime number, therefore if 29 appeared at BJ,
OR
If 59 appeared at DJ, then EJ would contain 67, which is also a prime number, contradicting clue [11]. So eliminate 29 & 59 from row J.
Then we have modified table as under:
A (1-15)
B (16-30)
C (31-45)
D(46-60)
E (61-75)
J
10,13,14
20,24,27,28
34,37,38,
39,40,43
57,58
67,69
K
1 through 14
Except 5
16 through 30
Except 25
31 through 44
Except 35
47,48,49
61,63,67,
69,71,73
L
1 through 14
Except 5
16 through 30
Except 25
EMPTY
47,48,49
61,63,67,
69,71,73
M
1 through 14
Except 5
16 through 30
Except 25
31 through 44
Except 35
47,48,49
61,63,67,
69,71,73
N
1 through 14
Except 5
16 through 30
Except 25
31 through 44
Except 35
46
61,63,67,
69,71,73
In column D, four numbers will be 46,47,48,49 and fifth number will be either 57 OR 58.
If 57 appeared at DJ, then total of all the five numbers in column D will be 247, which is a prime number. According to clue [17], the sum of numbers in each column shares a single common prime factor, and maximum sum of all numbers in column A would be (14+13+12+11+10)=60, therefore common prime factor shared by total of all numbers in each column will be less than 60.
So DJ will not contain 57, hence DJ will contain 58, and no number containing numeral 8 will appear at all other spaces in row J.
Then total of all numbers in column D will be 248. Factorizing it, we get 31*8.
Therefore 31 is the prime factor shared by total of all numbers in each column.
Then total of all five numbers in column E would be:
63+67+69+71+73=343 (Max), and
61+63+67+69+71=331 (Min).
Neither of the above shares prime factor 31, so the total must lie in between 331 & 343.
Then only one number ‘341’ lies in between 331 & 343, which has 31 as its factor (31*10=310 which is less than minimum total of 331 required and 31*12=372 which is more than the maximum total required, so 31*11=341 must be the total).
The total of five numbers in column E is 341 which is ‘2’ less of the total of 63+67+69+71+73=343.
Therefore to have required total of 341, column E will contain 61,67,69,71,&73.
Then as per clue [10], lowest number in each column is located in row N, so 61 will appear at space EJ.
Then we have modified table as under:
A (1-15)
B (16-30)
C (31-45)
D(46-60)
E (61-75)
J
10,13,14
20,24,27
34,37,39,40,
43
58
67,69
K
1 through 14
Except 5
16 through 30
Except 25
31 through 44
Except 35
47,48,49
67,69,
71,73
L
1 through 14
Except 5
16 through 30
Except 25
EMPTY
47,48,49
67,69,
71,73
M
1 through 14
Except 5
16 through 30
Except 25
31 through 44
Except 35
47,48,49
67,69,
71,73
N
1 through 14
Except 5
16 through 30
Except 25
31 through 44
Except 35
46
61
Now it can be seen from the table above, that if EJ contained 69, then all other spaces in column E would contain a two-digit prime number. Then according to clue [11], all other rows could not contain a two-digit prime number. But in column D, two-digit prime number 47 will appear in one of the rows other than row J, which would contradict clue [11]. So 69 will not appear at EJ; therefore 67 will appear at EJ Therefore no number containing numeral 7, and no two digit prime number will appear at any other space in row J. Hence numbers 13,27,37,&43 will not appear in row J.
Again since two-digit prime numbers are contained in all spaces in column D & E, therefore no other column (A,B&C) will contain any two-digit prime number
Then numbers 20&24 will be left at BJ, so if 40 appeared at CJ, then there would be no choice of number left at space BJ. Therefore 40 will not appear in row J. So either 34 OR 39 will appear at CJ.
Now if 34 appeared at CJ, then 20 would appear at BJ, leaving no choice of number at AJ. Therefore 39 is located at CJ.
Then we have modified table as under:
A (1-15)
B (16-30)
C (31-45)
D(46-60)
E (61-75)
J
10,14
20,24
39
58
67
K
1,2,3,4,6,7,8,
9,10,12,14
16,18,20,21,22,
24,26,27,28,30
32,33,34,36,
38,40,42,44
47,48,49
69,71,73
L
1,2,3,4,6,7,8,
9,10,12,14
16,18,20,21,22,
24,26,27,28,30
EMPTY
47,48,49
69,71,73
M
1,2,3,4,6,7,8,
9,10,12,14
16,18,20,21,22,
24,26,27,28,30
32,33,34,36,
38,40,42,44
47,48,49
69,71,73
N
1,2,3,4,6,7,8,
9,10,12,14
16,18,20,21,22,
24,26,27,28,30
32,33,34,36,
38,40,42,44
46
61
Then as per clue [12], difference of lowest to highest number is 8, in column C.
Then possible group of numbers in column C, might be 36-44, OR 34-42, OR 32-40.
Consider the group 36-44, then:
Maximum possible total of all four numbers in column C will be (39+36+44)+42=161
And Min possible total of all four numbers in column C will be (39+36+44)+38=157
Note: Number at space CN has to be smaller than 39 to satisfy clue [10].
Total must have prime factor 31, and since both the above totals do not have 31 as their factor, therefore total would lie in between 157 and 161.
31*6=186 can’t be the total as it more than the required maximum value.
Also 31*5=155 could not be the total as it is lesser than the required minimum value.
So 36-44 can’t be the required number group.
Consider the group 32-40, then:
Maximum possible total of all four numbers in column C would be (39+32+40)+38=149
And Min possible total of all four numbers in column C will be (39+32+40)+33=144
Total must have prime factor 31, and since both the above totals do not have 31 as their factor and no number exist between Min and max required values also, therefore 32-40 can’t be the required group.
So 34-42 must be the right group. In this case:
Maximum possible total of all four numbers in column C will be (39+34+42)+40=155
And Min possible total of all four numbers in column C will be (39+34+42)+36=151
Here 155 (=31*5) itself has factor 31, so 34-42 is the right combination, and since it is the only possible total for this group, therefore All four numbers in column C are: 42,40,39,34.
Also from clue [10], 34 will appear at space CN.
Then modified table will be as under:
A (1-15)
B (16-30)
C (31-45)
D(46-60)
E (61-75)
J
10,14
20,24
39
58
67
K
1,2,3,4,6,7,8,
9,10,12,14
16,18,20,21,22,
24,26,27,28,30
40,42
47,48,49
69,71,73
L
1,2,3,4,6,7,8,
9,10,12,14
16,18,20,21,22,
24,26,27,28,30
EMPTY
47,48,49
69,71,73
M
1,2,3,4,6,7,8,
9,10,12,14
16,18,20,21,22,
24,26,27,28,30
40,42
47,48,49
69,71,73
N
1,2,3,4,6,7,8,
9,10,12,14
16,18,20,21,22,
24,26,27,28,30
34
46
61
Similarly maximum possible sum of five probable numbers in the column B:
24+30+28+27+22=131.
Note: To get max total we assumed 24 at space BJ, so to satisfy clue [10] we assumed a number which is largest but smaller than 24 at space BN.
Also minimum possible total of five probable numbers in column B:
20+16+18+21+22=97.
Note: To get min possible total we assumed 20 at space BJ, so to satisfy clue [10] we assumed a smallest possible number 16 at space BN.
Both of the above are not multiple of prime factor 31, therefore the total must lie in between 97&131. 31*4=124 is the only total satisfying this condition, hence required total must be 124.
But if we assume 20 at BJ & 18 at BN then we cannot get required total of five numbers as 124, even if we select largest possible numbers at other spaces.
Therefore 24 will appear at BJ, and then no number containing numerals 2 & 4 will lie at any other space in row J, then at space BN, a number smaller than 24 will appear, as per clue [10].
Then modified table will be as under:
A (1-15)
B (16-30)
C (31-45)
D(46-60)
E (61-75)
J
10
24
39
58
67
K
1,2,3,4,6,7,8,
9,10,12,14
16,18,20,21,22,
26,27,28,30
40,42
47,48,49
69,71,73
L
1,2,3,4,6,7,8,
9,10,12,14
16,18,20,21,22,
26,27,28,30
EMPTY
47,48,49
69,71,73
M
1,2,3,4,6,7,8,
9,10,12,14
16,18,20,21,22,
26,27,28,30
40,42
47,48,49
69,71,73
N
1,2,3,4,6,7,8,
9,10,12,14
16,18,20,
21,22
34
46
61
Now To get 124 as total of all five numbers in column B, we have following groups of numbers:
(a) 24+30+28+26+16=124
(b) 24+30+28+22+20=124
© 24+30+27+22+21=124
It can be seen that if the group © was correct, then and then only 27&21 would appear in the table, also 21 would appear only at space BN, to satisfy clue [10]………………………………...............[X]
Then as per clue [15], in one and only one column, numbers appear in descending order from top to bottom.
It is clear from the numbers in row J, that none of the columns B,C,D,&E can be such column.
Therefore the numbers in column A appear in descending order from top to bottom and since 10 is located at the top space AJ, so all numbers at all other spaces in column A will be single digit numbers.
As per clue [10] numbers 9,8,&7 can’t appear at space AN.
Then modified table will be as under:
A (1-15)
B (16-30)
C (31-45)
D(46-60)
E (61-75)
J
10
24
39
58
67
K
1,2,3,4,
6,7,8,9
18,20,22,26,
27,28,30
40,42
47,48,49
69,71,73
L
1,2,3,4,
6,7,8,9
18,20,22,26,
27,28,30
EMPTY
47,48,49
69,71,73
M
1,2,3,4,
6,7,8,9
18,20,22,26,
27,28,30
40,42
47,48,49
69,71,73
N
1,2,3,4,
6
16,20,
21
34
46
61
Then as per clue [16], in each column there is only one numeral which appears two times.
The numeral 1 is the only numeral which can appear two times in the column A, so number 1 will definitely appear in this column.
From clue [10] number 1 will appear at space AN.
Then modified table will be as under:
A (1-15)
B (16-30)
C (31-45)
D(46-60)
E (61-75)
J
10
24
39
58
67
K
2,3,4,
6,7,8,9
18,20,22,26,
27,28,30
40,42
47,48,49
69,71,73
L
2,3,4,
6,7,8,9
18,20,22,26,
27,28,30
EMPTY
47,48,49
69,71,73
M
2,3,4,
6,7,8,9
18,20,22,26,
27,28,30
40,42
47,48,49
69,71,73
N
1
16,20,
21
34
46
61
Then max possible sum of all five numbers in column A:
10+1+9+8+7=35
And min possible sum of all five numbers in column A:
10+1+2+3+4=20
To have 31 as factor the sum must lie in between 20 & 35. So only possible sum is 31.
The possible groups of numbers to have sum as 31 are:
(a) 10+1+9+8+3=31
(b) 10+1+9+7+4=31
As per clue [20] product of numbers at spaces AL and EL has a unit digit 2, but if we assume group (b) to be true, then this condition is not satisfied. Therefore group (b) must be true.
Then the modified table is as under:
A (1-15)
B (16-30)
C (31-45)
D(46-60)
E (61-75)
J
10
24
39
58
67
K
9
18,20,22,26,
27,28,30
40,42
47,48,49
69,71,73
L
8
18,20,22,26,
27,28,30
EMPTY
47,48,49
69,71,73
M
3
18,20,22,26,
27,28,30
40,42
47,48,49
69,71,73
N
1
16,20,
21
34
46
61
Now, to satisfy condition in clue [20], number 69 will appear at space EL, then to satisfy clue [11],
two-digit prime number 47 is located at DL.
Then as per clue [21], product of numbers at space BL & DL contains unit digit 4, which is possible only when 22 appears at space BL.
Then the table is modified as under:
A (1-15)
B (16-30)
C (31-45)
D(46-60)
E (61-75)
J
10
24
39
58
67
K
9
18,20,26,
27,28,30
40,42
48,49
71,73
L
8
22
EMPTY
47
69
M
3
18,20,26,
27,28,30
40,42
48,49
71,73
N
1
16,20,
21
34
46
61
Then as per clue [19], the sums of numbers in each diagonal are an odd number, and the two diagonals in the table contain numbers as under:
Diagonal AN to EJ: In this diagonal two numbers (1 and 67) are odd numbers; so only one of the other two numbers could be odd number to satisfy clue [19] i.e. either DK would contain 49, OR BM would contain one of the numbers 27&21.
Diagonal AJ to EN: In this diagonal one number (61) is odd, and one other number (10) is even; therefore to satisfy clue [19] either both the other two numbers should be odd numbers (i.e. 27&49) OR both the other two numbers should be even numbers. Here both the numbers can’t be odd numbers, because one of the odd numbers is necessarily required at a space in first diagonal AN to EJ; hence both the other numbers in second diagonal must be even numbers, so 48 will be located at DM, and therefore remaining number 49 will be located at remaining space DK, in column D; and therefore an even number will appear at space BK.
Then in diagonal AN to EJ, since one odd number 49 is added at space DK with the already existing two odd numbers (1&67), therefore to satisfy clue [19], remaining fourth number must be an even number at space BM. Hence number 27 (and then 21 also) will not appear in the table, and then as per [X] group (b) of numbers for column B will be true, i.e. numbers 24,30,28,22,&20 will appear in column B.
Then table will look like under:
A (1-15)
B (16-30)
C (31-45)
D(46-60)
E (61-75)
J
10
24
39
58
67
K
9
28,30
40,42
49
71,73
L
8
22
EMPTY
47
69
M
3
28,30
40,42
48
71,73
N
1
20,
34
46
61
Now as per clue [19] sum of all five numbers in row M is a perfect square.
Max possible sum: 3+30+42+48+73=196.
Min possible sum: 3+28+40+48+71=190.
Only possible sum is 196 which is a perfect square of 14. Therefore numbers 3,30,42,48&73 will appear in row M.
So finally the required table is as under:
A (1-15)
B (16-30)
C (31-45)
D(46-60)
E (61-75)
J
10
24
39
58
67
K
9
28
40
49
71
L
8
22
EMPTY
47
69
M
3
30
42
48
73
N
1
20
34
46
61
Who teaches so loudly??
in New Logic/Math Puzzles
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