School-Safe Puzzle Games

## Math Puzzle- Nine Brothers and Sisters

You must find the ages of nine brothers and sisters in the same family so that:

–          There is a triplet and two pairs of twins

–          Every age is a prime number

–          The sum of the nine ages is 153

–          We can build a magic square 3×3 with the nine ages

Another fine math/logic puzzle by Bilbao!

Will give you several days to work this one out; answers to be unmasked Tuesday next week!

### 24 Comments to “Math Puzzle- Nine Brothers and Sisters”

1. Falwan | Profile

a triplet = 3 people

a pair of twins = 4 people

2 pairs of twins = 8 people

a triplet + 2 pairs of twins = 3 + 8 = 11 people…

each of them = 17 years of age !!!

Maybe something missing!!!

2. ZaKnaFeiN | Profile

Hi All, I’m a frequent visitor to the site but this is my first time posting an answer. Really fun puzzle and I had to browse Wiki and Google to relearn some prime and magic square properties.

Triplets are 17
Twins are 23 and 11
The other two are 29 and 5.

Magic square is:
23 | 17 | 11
————
5 | 17 | 29
————
23 | 17 | 11

This is how I came about the solution:
Since the sum of ages is 153, the average age is 153/9 = 17. Therefore, my hypothesis is that 17 is the central number in the square.

There’s a lemma on magic squares I found which is:
The row total of a 3×3 magic square is three times the centre value. Thus, the row total is 17*3 = 51.

Proof of lemma:
Consider the four “lines” (1 row, 1 column and 2 diagonals) that pass through the centre. Summing them gives a total of 4 times the row total, and it includes the centre value 4 times and all other values once. Adding all the values together once is the same as summing the three rows, i.e. it gives three times the row total, so the additional three centre values must sum to the row total also.

Here’s the link to the above in case anyone is interested:
http://www.dcs.gla.ac.uk/~pd/Numbers/MagicSquares/

With 51 being the row total, there are three sets of sums that can be used with 17 to make 51:
31 and 3
29 and 5
23 and 11

At this point, I used trial and error and would love to see a logical deduction of the answer. I started with 17 as the triplets so that the sets above will make the twins. Experimenting with 23 and 11 as twins in the square reveals 29 and 5 to be the remaining prime numbers available.

A few questions I have:
Can any other number besides 17 be the central number?
Would this question be easier or harder if the numbers were distinct?

Thank you.

3. suineg | PUZZLE MASTER | Profile

This puzzle was very cool, because it needs a little bit of out of the box thinking.
I think this could be the answer:
lets assume:
A= Age of the triplets
B= Age of twins 1
C= Age of twins 2
D= Age of alone birth 1
E= Age of alone birth 2

we have:

1) 3A + 2B + 2C + D + E = 153

The ages are prime numbers and form a magic square so they must add a K constant in every direction, and all the ages are different, so A,B,C,D,E are different.

For the statements above you can deduce that: you cant have combination of ages that gives you an incoherent relationship, for example: A+A+B AND A+A+D, because B and D are not equal
So first you know that the D,E should be in a common row or column, the age of the triplets should be in the same column or row and from that on, I complete the square visually.
BDB
AAA
CEC

Then you get:
2)2B+D=K
3)3A=K
4)2C+E=K
5)A+B+C=K
6)A+D+E=K

then substitute (2),(3),(4) in (1) and you get 3K =153–> K= 51
then substitute K in (3):
A= 17
then substitute A and K in (5) and (6):
5)B+C= 34–> 5)C= 34-B
6)D+E= 34–> 6)E= 34-D

then substitute K in (2) and (4):
2)2B+D= 51
3)2C+E= 51

substitute (5) y (6) in (3):
3)2(34-B)+ 34-D= 51 —> 3) 102-2B-D= 51
you get the same (2).. now what I did is try to find two primes that fulfill (2) and two that fulfill (3):
A=17 ..triplets ages
B=19 ..twins 1 ages
C=23 ..twins 2 ages
D=13 ..alone birth 1 age
E=5 …alone birth 2 age
cool man.. I think it worked, it was hard, I rated 4 bilbao, really nice puzzle!

4. Jimmy Anders | PUZZLE MASTER | Profile

I didn’t exhaust the possibilities, but the only way I found that would work even algebraically was if the triplets were all on the same row of the square. Since all of the rows sum to the same thing, and the sum of the three rows is 153, each row would add to 51. This makes the triplets each 17. After that, the rest seemed to fall into place for me with all of the ages being:

17 17 17
23 23 5
11 11 29

5. goofymushu | Profile

5,11,11,11,17,17,23,29

11 11 29

11 23 17

29 17 5

6. Bobo The Bear | PUZZLE MASTER | Profile

The ages are 5, 11, 11, 17, 17, 17, 23, 23, and 29.

The magic square looks like this (or similar)
Top row: 11, 29, 11
Middle row: 17, 17, 17
Bottom row: 23, 5, 23

7. Mashplum | PUZZLE MASTER | Profile

Can you define a magic square?

8. Hendy | Profile

05
11(
11 Twins )
17(
17
17 Triplets )
23(
23 Twins )
29

23 05 23
17 17 17
11 29 11

Sum = 51 ( across all rows, down all columns, both diagonals ) .

FYI: total being 153 average number is 17. Therefore 17 is the number to target. 17 * 3 = 51. 17 * 2 = 34. Prime numbers 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, etc. Sums of 34 ( 17+17, 23+11, 29+5, 31+3 ) . Key numbers are 3, 5, 11, 17, 23, 29, 31, but 31 and 3 didn’t fit. So had to find proper number of repeats of 5, 11, 17, 23, and 29. Answer became obvious after playing with the magic square for a while.

:P

9. Hex | PUZZLE MASTER | Profile

Hey Bilbao!

Nice to have you post yet another puzzle.

Some clarifications are needed:

Does the magic square sums include the diagonals?

Does “There is a triplet and two pairs of twins” mean that the ages of 3 siblings is the same, the age of 2 other siblings is the same, and the age of yet 2 other siblings is the same while everyone else has a different age?
In other words, I understand that one number is found 3 times only, another is found twice only, another twice only as well, and the rest are each distinct.

10. bilbao | Profile

Hi Mashplum and Hex:
in a magic square the sum of the figures of each column and line is the same. In this case the diagonals are not included.
And Hex, as you said, triplet – twins 1 – twins 2 – else 1 – else 2, are all different ages (so there are 5 different ages).

11. Shawn | PUZZLE GRANDMASTER | Profile

Well, I’m learning a bit more about the properties of magic squares! It seems to me that the magic number that all sides add up to should be 153/3, or 51. I can get it close, but there’s always a couple of 53’s and 49’s thrown in there to gum up the works.

1 is usually not considered a prime number, but if it were then this set would fit the bill:

1+7+13+13+19+19+19+31+31 = 153

magic square:

31, 01, 19
07, 31, 13
13, 19, 19

12. bizarette18 | PUZZLE MASTER | Profile

23,17,11 5,17,29 23,17,11
Do you acually know this family?

13. Hex | PUZZLE MASTER | Profile

triplets: 5
1st twins: 3
2nd twins: 41
others: 7, 43
43 3 5
3 7 41
5 41 5

14. Mashplum | PUZZLE MASTER | Profile

The average age is 17. The triplets must be 17 since there is an odd number of them. All of the other ages must make pairs that add up to 34.

17 17 17
23 05 23
11 29 11

Arranged this way, all rows and columns add up to 51.

15. Shawn | PUZZLE GRANDMASTER | Profile

Bilbao,
Have you found a combination that does not include “1” as an age?

16. bilbao | Profile

Hi Shawn, yes, you can find a solution that lacks ‘1’ as an age.

17. Shawn | PUZZLE GRANDMASTER | Profile

29, 11, (23,23,23), (17,17), (5,5)

29 17 5
17 11 23
5 23 23

18. suineg | PUZZLE MASTER | Profile

Bilbao or Anyone that can help me out, I have a question: taking into consideration that the perfect answer is the one with the diagonals included, why when you try to solve this problem by equations you get redundant equations, or a linear combination of a previous equation so that you cant solve for that values, When that happens it usually means that the solution cannot fulfill all the conditions when I saw the post that the diagonals were not included I said… ahhhh ok there you go, but it turns out that is possible, nice. Was it a mistake made by me in the substitutions?.
Cool man

19. Hex | PUZZLE MASTER | Profile

I pressed on the rating button by mistake. Thus my rating has nothing to do with me :agape

@Suineg, B+A+C=19+17+23=59 :~

@Bilbao, it seems that a solution can be found considering the diagonals as well. The logic is as follows:
Total is 153, that’s 3 equal row sums, and hence the magic number is 51.
Since the diagonals are included, 17 must be in the middle
Hence, every row/column/diagonal that includes the center cell will consist of one of the following:
31 17 3
23 17 11
17 17 17
29 17 5
This means that numbers in each of the following sets come in pairs (31,3) (23,11) and (29,5)
Consequently, 3,5,11,23,29 and 31 cannot be the triplets age. Then the triplets must be 17, and one of the rows/columns/diagonals will be 17 17 17
17 17 17 cannot be in the diagonal as if you put any pair of numbers out of the valid sets, the square will fill itself only with 3×17 and 3x the pair, which invalidates the puzzle’s given. Hence the 3×17 are in a row or column (which is equivalent).
Now we have to select each valid pair and assume it is the age of twins:
(29,5) fails even if 29 is in a corner or row/column center
(31,3) fails even if 31 is in a corner or row/column center
(23,11) succeeds if and only if the 23’s and 11’s are in the corners without any of the 2 numbers being on the same diagonal

Conclusion:
A single solution exists for the numbers:
triplets: 17
1st twins: 11
2nd twins: 23
others: 5,29

A few equivalent solutions exist as far as the arrangement within the magic square is concerned:
The 23’s and 11’s must be in the square’s corners without any of the 2 numbers being on the same diagonal
The 17’s must be in a central row or column
5 comes between the 23’s
29 comes between the 11’s

example:

23 17 11
5 17 29
23 17 11

20. Mashplum | PUZZLE MASTER | Profile

11,11,29
11,23,17
29,17,05

21. Hex | PUZZLE MASTER | Profile

@Suineg,
To start with, there are 9 unknowns and at most 8 equations (3 rows + 3 columns + 2 diagonals)
Assuming your choice for the positions of ABCDE is correct (and it is indeed):
BDB
AAA
CEC

1)2B+D=K
2)3A=K
3)2C+E=K
4)A+B+C=K
5)A+D+E=K
6)3A+2B+2C+D+E=153
from 6) -> 7) (D+A+E)+(B+A+C)+(B+A+C)=153
from 7) 5) 4) -> K+K+K=153 ie K=51
from 2) A=51/3 -> A=17
Knowing A and K, we reformulate the above equations:
1)2B+D=51
2)2C+E=51
3)B+C=34
4)D+E=34
from 1) -> 5) D=51-2B Note that B= 6) C=34-B
from 2) and 6) -> E=51-2C=51-2(34-B)=2B-17 -> 7) E=2B-17 Note that B>=9
from 4) and 5) -> E=34-D=34-(51-2B)=2B-17 which is redundant
So the system cannot be solved.
(B,C) being prime, can be: (3,31),(5,29),(11,23),(17,17) Note that B and C are interchangeable
(17,17) is ruled out as A=17 and A figures already 3 times.
As 9=<B= D=29
from 7) -> E=5
So A=17, B=11, C=23, D=29, E=5
Had we chosen B to be 23, then A=17, B=23, C=11, D=5, E=29

22. suineg | PUZZLE MASTER | Profile

I saw my mistake…. jesus christ..jajajaja, when I got K= 51 I used (2) (3) (4)so I use this value for (5) and (6) and thats it, but I CANT put the k value in (2) and (4) jajaja its a shame… the most difficult part of the puzzle solved and getting that mistake… man. In my country you said : tanto nadar para morir en la orilla jajaja, very cool puzzle Bilbao

23. bilbao | Profile

By far your solutions have exceeded mine, wow.
19 19 13
19 31 1
13 1 37
1 is not considered prime and when I solved it I didn´t even bother thinking about other possibilities.
I like the solutions that consider diagonals too (Hex’s explanations)
So, congratulatons for squeezing this puzzle!

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