10 replies to this topic

[bonanova]

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, 06 March 2009 - 01:36 PM.
Add the final question

[Merrick]

Spoiler for Here is my try

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, 06 March 2009 - 12:55 PM.

[bonanova]

Spoiler for Here is my try

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

[Juan Daniel R.]

Ok, I think I have the first answer.

Spoiler for What is the sum?

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.

[kitnkat]

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

Spoiler for my thoughts

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

[bonanova]

Ok, I think I have the first answer.

Spoiler for What is the sum?

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.

[HoustonHokie]

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

Spoiler for my thoughts

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...

Spoiler for unique might matter

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.

[Orykle]

Spoiler for solution

....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.

[bonanova]

Spoiler for solution

....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.

[Orykle]

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.