A Bingo Card Puzzle (A very lengthy math puzzle)

[bhramarraj]

There is a stack of Bingo Cards. It is required to find a specific card having the number arrangement as hinted in the following clues:

[1] On the cards, there are five columns (named hereafter as A, B, C, D, & E) and five rows (named hereafter as J, K, L, M, & N).

[2] Column 'A' contains numbers 1 through 15.

[3] Column 'B' contains numbers 16 through 30.

[4] Column 'C' contains numbers 31 through 45.

[5] Column 'D' contains numbers 46 through 60.

[6] Column 'E' contains numbers 61 through 75.

[7] Center space 'CL' (a place in column C and row L) is free space i.e. contains no number.

[8] In row 'J', each numeral (0 through 9) appears only once; e.g. if there is number 43 at space DJ, then at other spaces in this row, there would be no number containing numeral 3 & 4.

[9] In row M, sum of all numbers is a perfect square.

[10] Row N contains the smallest number in each column.

[11] Each row contains only one '2-digit Prime Number'.

[12] In column C, difference of lowest to highest number is 8.

[13] In column D, each number has digit at 10th place smaller than the digit at unit place.

[14] In column E each number is an odd number.

[15] In only one column, the numbers are in descending order from top to bottom.

[16] In each column, there is only one numeral that appears two times.

[17] In each column, the sum of numbers shares a single common prime factor.

[18] On the bingo card, numeral 5 appears only once.

[19] The sum of the numbers in each diagonal is an odd number.

[20]The product of numbers at spaces AL & EL has a unit digit 2.

[21] The product of numbers at spaces BL & DL has a unit digit 4.

[MikeD]

Well that's easy (not)

Do bingo cards have a specific amount of numbers on them?

[bonanova]

I'm guessing that the puzzle is to create a 5x5 array of numbers that satisfy the clues.

I'm not sure about the reference to a stack of cards.

Nowhere does it say that at least one crd in the stack has the required array of numbers.

[bhramarraj]

I'm guessing that the puzzle is to create a 5x5 array of numbers that satisfy the clues.

I'm not sure about the reference to a stack of cards.

Nowhere does it say that at least one crd in the stack has the required array of numbers.

You are right. The puzzle is to create a 5X5 array of numbers that satisfy the given clues.

It is a lenghthy math puzzle requires lot patience.

[bhramarraj]

Well that's easy (not)

Do bingo cards have a specific amount of numbers on them?

Solution is not concerned to a bingo card, or stack. One is required to creat a 5X5 table containing numbers as per the clues given.

[MikeD]

So we need 24 numbers because 'CL' is the only blank.

[bhramarraj]

So we need 24 numbers because 'CL' is the only blank.

You are right....!

[bonanova]

Clarifications.

No two Boxes contain the same number?

8. Once and only once?

15. One and only one column?

16. Do you mean exactly two times? To prohibit having 2 sixes and 3 eights?

17. The five column sums share one and only one prime factor?

18. Once and only once?

Then, if CL is 45 then 50-59 do not appear in col D?

19. Major diagonal?

Thanks.

[bonanova]

Column D

Clues 10 and 13

Boxes DJ thru DM contain 56-59 in some order.

Box DN contains one of 46-49.

[superprismatic]

Here's what I have so far:

Column D is 58,x,y,z,46 (from J to N) with x,y, and z being 47,48,49 in some order.

Column E is p,q,r,s,61 (from J to N) with p,q,r,and s being 67,69,71,73 in some order.

Column C is either 34,38,41,42 in some order; 34,39,40,42 in some order; or 36,37,38,44 in some order.

All columns sum to a number which has 31 as a common prime factor.

[bhramarraj]

Clarifications.

No two Boxes contain the same number?

8. Once and only once?

15. One and only one column?

16. Do you mean exactly two times? To prohibit having 2 sixes and 3 eights?

17. The five column sums share one and only one prime factor?

18. Once and only once?

Then, if CL is 45 then 50-59 do not appear in col D?

19. Major diagonal?

Thanks.

Of course no two boxes contain the same number.

Clue8: Yes, once and only once.

Clue15: Yes, one and only one column.

Clue16:In each column there is one and only one numeral that appears two times. other numerals could be one or three or more times. So the condition of this clue cannot prohibit double digit numbers, at any space in the table.

Clue17: there could be more than one prime factors for the sums of different individual columns, but there is one and only one prime factor which is common to all sums.

Clue18: you are absolutely right, if CL is 45 then 50-59 do not appear in col D?

Clue19: Yes, we are talking of the major diagonals.

I think I have clarified all the doubts...?

[bhramarraj]

For them who require hint:

We know from clue [18], that numeral ‘5’ appears once only, in the complete table. So numeral ‘5’ will appear only in column D.

[superprismatic]

Here's my solution:

. A B C D E
J10 24 39 58 67
K 9 28 40 49 71
L 8 22 XX 47 69
M 3 30 42 48 73
N 1 20 34 46 61

Thanks! This was a very nice puzzle. It made me do a lot of analysis before I could get the work factor down to the point where I could write a program to complete the solution.

[bhramarraj]

Here's my solution:

. A B C D E

J10 24 39 58 67

K 9 28 40 49 71

L 8 22 XX 47 69

M 3 30 42 48 73

N 1 20 34 46 61

Thanks! This was a very nice puzzle. It made me do a lot of analysis before I could get the work factor down to the point where I could write a program to complete the solution.

Really amazing.....! How could you solve the puzzle so easily... .It took me several days to solve it. Thanks....!

[superprismatic]

In case anyone is interested in a short description of how I got the solution:

Facts 13 & 18 immediately allow one to get all but one of the numbers (not the order) of column D (which of 56,57,58,or 59 is unknow at this point). Facts 14 & 16 get column E's contents (not order) down to 2 possibilities. Factoring the four possible sums for column D and the two from column E gives the common prime factor of 31 (using Fact 17). This gives us the unordered contents of columns D & E. Furthermore, 4 of the 5 numbers in column E are 2-digit primes, and since there is one such prime in column D, we conclude (by Fact 11) that 47 in D and 69 in E are in the same row. Using Facts 11 & 17 and knowing that the common factor is 31, I wrote a program to list all possible contents of columns A, B, and C. This brought the possible number of bingo cards to almost 18 million, small enough to quickly and easily check the remaining Facts against all the possibilities with a fairly simple-to-write program. Only one possibility survived.

[bhramarraj]

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

[bhramarraj]

Sorry I am not able to send the file

Edited July 9, 2012 by bhramarraj

[bhramarraj]

Sorry I am not able to send the file

Please somebody let me how should I load a word file...?

[bhramarraj]

I converted the word file to pdf, and attached here, for those who want to solve it matematically

A very lengthy math puzzle.pdf