| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
I had a problem published in the current issue of the Journal of Recreational Mathematics (volume 16, number 2) that I devised with the help of a computer: Problem 1292: a) For some n, partition the first n perfect squares into two sets of the same size and same sum. b) For some n, partition the first n triangular numbers into two sets of the same size and same sum. c) For some n, partition the first n perfect cubes into two sets of the same size and same sum. d) For some n, partition the first n perfect fourth powers into two sets of the same size and same sum. This problem is still "open" if you want to send your solution in to them.
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 9.1 | HARE::STAN | Wed Feb 15 1984 11:43 | 24 | ||
A more general result was published as a problem in the January issue
of the American Mathematical Monthly 91(1984)57.
Problem E3032, proposed by J. O. Shallit:
Let d and r be integers with d>1 and r>0. Let p be a polynomial
with real coefficients and deg(p)<r. Show how to partitiion the set
S = { p(0), p(1), p(2), ... , p(d^r-1) }
into d disjoint subsets whose union is S, such that the sum of each
subset is the same.
---------------------------------
Now in my problem, p(x) was (x+1)^n, so the above result says that
the first 2^3=8 squares can be partitioned into two sets with the same sum;
the first 2^4=16 cubes can be partitioned into two sets with the same sum;
the first 2^5=32 4th powers can be partitioned into two sets with the same sum;
the first 2^6=64 5th powers can be partitioned into two sets with the same sum;
Of course, Shallit's result says nothing about whether this number is
minimal. It also doesn't require the sets to each have the same cardinality.
But it's a nice result anyhow.
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| 9.2 | HARE::STAN | Wed Feb 15 1984 12:09 | 14 | ||
From: METOO::YARBROUGH 15-FEB-1984 12:06
To: HARE::STAN
Subj: 5th powers
Stan;
5 5 5 5 5 5 5 5 5 5 5 5
3 +4 +7 +8 +9 +11 +12 +13 +17 +21 +22 +23
=
5 5 5 5 5 5 5 5 5 5 5 5
1 +2 +5 +6 +10 +14 +15 +16 +18 +19 +20 +24
Lynn Yarbrough
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| 9.3 | METOO::YARBROUGH | Wed Feb 15 1984 13:53 | 21 | ||
2 2 2 2 2 2 2 2
a) 1 +4 +6 +7 = 2 +3 +5 +8
b) T1 + T2 + T6 = T2 + T4 + T5
(1 + 6 + 21 = 3 + 10 + 15)
3 3 3 3 3 3 3 3 3 3 3 3
c) 1 + 2 + 4 + 8 + 9 + 12 = 3 + 5 + 6 + 7 + 10 + 11
4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4
d) 1 +2 +3 +6 +9 +10 +14 +19 +20 = 4 +5 +7 +8 +12 +13 +15 +16 +17 +18
finally,
5 5 5 5 5 5 5 5 5 5 5 5
1 +2 +5 +6 +10 +14 +15 +16 +18 +19 +20 +24
=
5 5 5 5 5 5 5 5 5 5 5 5
3 +4 +7 +8 +9 +11 +12 +13 +17 +21 +22 +23
- Lynn Yarbrough
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| 9.4 | HARE::STAN | Wed Jan 16 1985 04:35 | 11 | ||
Kenneth M. Wilke outdid us all in his published solution
in vol 17, issue 2 of JORM:
He found one single partition of the first 32 positive integers
that simultaneously solved all 4 parts of my problem:
{1,4,6,7,10,11,13,16,18,19,21,24,25,28,30,31}
{2,3,5,8,9,12,14,15,17,20,22,23,26,27,29,32}
A neat hack!
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