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Arrange six quarters around the perimeter of a central quarter,

and twelve more around the new perimeter. That makes six rows

of 3 coins, six rows of 4 coins and three rows of 5 coins.

Number each coin with a different number from 1-19 so that

the same sum is achieved along all fifteen rows.

What is that sum?

OK now that we have that settled, find the numbers. ;)

Edited by bonanova
Add the final question
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Posted (edited) · Report post

Totaling the coins (1 - 19) = 190

In one direction you have 5 rows (with 3,4,5,4,3 coins)

The total in all these rows has to be 190. So the sum of one row is 190/5 = 38

The question would be a lot harder if you wanted to know the filled in hexagon :-)

Edited by Merrick
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Posted · Report post

Totaling the coins (1 - 19) = 190

In one direction you have 5 rows (with 3,4,5,4,3 coins)

The total in all these rows has to be 190. So the sum of one row is 190/5 = 38

The question would be a lot harder if you wanted to know the filled in hexagon :-)

Good point. Go for it. ;)

Nice analysys btw.

And, welcome to the Den

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Posted · Report post

Ok, I think I have the first answer.

Checking the distribution I realized that every coin will be on 3 and only 3 rows. Therefore if I take the 15 rows and add their sum the result would be:

(1 + 2 + 3 + ... + 19) * 3 = 570

And then since we know all the rows should sum the same then the sum on each row should be:

570 / 15 = 38

Am I on the right track? I'll keep working on the placement of the numbers.

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Posted · Report post

First post....hopefully I'm thinking about this right.

If you letter the quarters from the top row across each row you get...

abc

de fg

h i j k l

mnop

qrs

There are 15 possible equations such as a+b+c=38,

a+d+h=38...etc.

If you put those equations together as much as possible like b+c=d+h from two equations listed you end up with 19 equations. 19 equations, 19 unknowns, algebra should be enough to solve it. I'm feeling a bit lazy so I'm not solving them right now, but I'll list them all if someone else is inclined to solve them.

1. b+c=d+h

2. a+b=g+l

3. a+d=m+q

4. c+g=s+p

5. h+m=r+s

6. q+r=l+p

7. d+f+g=m+i+b

8. d+e+f=k+o+r

9. e+f+g=i+n+r

10. n+o+p=i+e+b

11. m+n+p=g+k+r

12. m+n+o=b+f+k

13. m+o+p=d+i+r

14. m+i+e=f+k+p

15. m+e+b=d+n+r

16. g+o+r=b+f+p

17. g+k+o=d+i+n

18. h+i+k+l=a+e+o+s

19. h+i+k+l=c+f+n+q

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Posted · Report post

Ok, I think I have the first answer.
Checking the distribution I realized that every coin will be on 3 and only 3 rows. Therefore if I take the 15 rows and add their sum the result would be:

(1 + 2 + 3 + ... + 19) * 3 = 570

And then since we know all the rows should sum the same then the sum on each row should be:

570 / 15 = 38

Am I on the right track? I'll keep working on the placement of the numbers.

You are correct.

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Posted · Report post

First post....hopefully I'm thinking about this right.

If you letter the quarters from the top row across each row you get...

abc

de fg

h i j k l

mnop

qrs

There are 15 possible equations such as a+b+c=38,

a+d+h=38...etc.

If you put those equations together as much as possible like b+c=d+h from two equations listed you end up with 19 equations. 19 equations, 19 unknowns, algebra should be enough to solve it. I'm feeling a bit lazy so I'm not solving them right now, but I'll list them all if someone else is inclined to solve them.

1. b+c=d+h

2. a+b=g+l

3. a+d=m+q

4. c+g=s+p

5. h+m=r+s

6. q+r=l+p

7. d+f+g=m+i+b

8. d+e+f=k+o+r

9. e+f+g=i+n+r

10. n+o+p=i+e+b

11. m+n+p=g+k+r

12. m+n+o=b+f+k

13. m+o+p=d+i+r

14. m+i+e=f+k+p

15. m+e+b=d+n+r

16. g+o+r=b+f+p

17. g+k+o=d+i+n

18. h+i+k+l=a+e+o+s

19. h+i+k+l=c+f+n+q

Good thought, but...

By having 19 equations with 19 unknowns, you assume that a unique solution exists. I doubt that one unique solution exists - there are likely many solutions (if there are any at all), similar to a magic square. And, because your equations have no way to constrain the sums to be 38 or for the individual variables to be integers from 1 to 19, that opens the door to even more solutions. Finally, I think that if you carry out the substitution process that one would normally do to solve these types of equations, you'd find that several are redundant and you'd end up with something like a = a or 0 = 0 at the end. So, I don't think algebra helps much. At least not here.

:huh:
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Posted · Report post

....16, 19, 3

...12, 2, 7,17

10, 4, 5, 1, 18

..13, 8, 6, 11

.....15,14, 9

I know the formatting wasn't perfect but I tried.

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Posted · Report post

....16, 19, 3

...12, 2, 7,17

10, 4, 5, 1, 18

..13, 8, 6, 11

.....15,14, 9

I know the formatting wasn't perfect but I tried.

Bingo. ;)

Good job.

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Posted · Report post

I'm pretty sure there is just one unique answer also, I've been trying and can't come up with another. Maybe bonanova can confirm there is only one, save rotations and mirrors.

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Posted · Report post

I'm pretty sure there is just one unique answer also, I've been trying and can't come up with another. Maybe bonanova can confirm there is only one, save rotations and mirrors.

To my knowledge the solution is unique [rotation and reflection only].

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