When I went over to see a rare coin collection, I asked the care taker of the coins about them.
"What can you tell me about your coins?"
"In each box, the number of coins is a perfect square, and each box has a different number of coins."
"How are the coins in your 9 boxes organized?"
"I put them in chronological order by groups. The ancient coins are in boxes A, B, C; my old coins in boxes D, E, F; and my recent coins sit in boxes G, H, I."
"How else are they organized?"
"Well, within each time period (ancient, old, recent), the numbers form an arithmetic sequence, and the common difference is the same for all three time periods."
"How many coins do you have in each box?"
"I'm not sure, but I do know that a newer box would always have more coins that an older box from that time period. For example, feel box A. I have very few coins in that box, less than a dozen."
How many coins are in each box?
This solution method does involve a computer (since I used excel to create some tables) but not a computer program.
Let's focus our attention on just one set of three boxes in a time period. Let's say the three values in the boxes are (Nj)², N², and (N+k)² where N, j and k are all positive integers. The common difference is (N+k)² – N² = N² + 2kN + k² – N² = 2kN+k². The common difference is also N² – (Nj)² = N² – (N² – 2jN + j²) = 2jN–j². Since it must be that 2kN+k² = 2jN–j² we know that N = (j²+k²)/[2(jk)].
Now let's expand what we found to all three time periods. We need to find 3 sets of j, k and N = (j²+k²)/[2(jk)] such that 2jNj² and 2kN+k² is the same value across all three sets.
There are a few things we can learn from N = (j²+k²)/[2(jk)]:
First, since N, the numerator, and 2 are positive, (jk) must be positive. So j>k. We probably could have guessed this intuitively, but it's nice to have confirmation.
Second, since N is an integer and there is a 2 in the denominator, j²+k² is even. This means j² and k² are either both odd or both even, and thus j and k are both even or both odd. This means that (jk) is an even number.
Third, since (jk) is an even number and N is an integer, we know j²+k² has another factor of 2 in it. So j²+k² isn't just even (a multiple of 2), it is a multiple of 4.
I am having a hard time proving why, but I haven't found any pairs of odd j and k such that j²+k² is a multiple of 4. I think I remember something about square numbers (I guess just odd squares) can always be written as 8a+1 where a is an integer. Which means the sum of two odd squares could be written as 8a+1+8b+1 = 8(a+b)+2 = 8c+2. We know 8c will be divisible by 4, but 2 won't so the sum of two odd squares will never be a multiple of 4. I hope that's a good proof.
Anyways, so we know j and k are both even (and j²+k² is divisible by 4, but that's not really interesting since the square of an even number will always be divisible by 4… (2x)² = 2²*x² = 4*x²… so if you add two numbers that are divisible by 4, the result will also be divisible by 4.)
I made a table with j across the top and k down the side. I looked at j and k up to 200. My table showed the resulting N = (j²+k²)/[2(jk)] only if N ended up being an integer greater than j.
Then I made a second table that showed (Nj)² only if the first table had a value for N present. In this range, I only found 7 cases where (Nj)² was less than 12 (1, 4 or 9) since that is a restriction on box A. Those cases were:
j k (Nj)² N² (N+k)² Common difference
4 2 1 25 49 24
8 4 4 100 196 96
12 6 9 225 441 216
28 12 1 841 1681 840
53 24 4 3364 6724 3360
84 36 9 7569 15129 7560
168 70 1 28561 57121 28560
I made a third table that showed the common difference (2jNj²) only if the first table had a value for N present so I could find any other sets that had the same arithmetic sequence. I didn't find any other sets for 24, 96, 216 or 28560 (there might have been another set for 28560 out side of my j,k <201 range, and possibly for the others but not likely). For 840 and 7560 I found one other set each. Only for 3360 did I find two other sets. The three sets were:
j k (Nj)² N² (N+k)² Common difference
53 24 4 3364 6724 3360
16 14 9409 12769 16129 3360
28 20 2116 5476 8836 3360
The clues let us know that A<12 and A<B<C and D<E<F and G<H<I. From this we know that
A=4
B=3364
C=6724
But technically we weren't given enough info to tell which of the remaining two sets is D, E, F and G, H, I. It is implied that a more recent time period will have more coins than an older time period, but it is not actually stated that way. So I will assume
D=2116
E=5476
F=8836
G=9409
H=12769
I=16129
Edited on January 21, 2008, 7:17 pm

Posted by nikki
on 20080121 15:17:55 