THE EQUAL-SUM PROBLEM
MATH 1200 Section B — Prof. N. Madras—Sept. 21, 2015
1. Review of the original problem from the class of Sept. 14
We’ll use terminology of “set theory” to phrase the original problem. To
begin with, let’s introduce a definition (see page 134 in Liebeck).
Definition: Let S be a set. A partition of S is a collection of sets S1 , S2 , . . .
such that each Si is a subset of S, and each element of S is contained in exactly
one of the Si ’s.
For example, a partition of {1, 2, 3, 4, 5, 6} is {{1, 3, 6}, {2, 4}, {5}}. (That is:
S1 = {1, 3, 6}, S2 = {2, 4}, and S3 = {5}.) A partition of the natural numbers
is S1 being the set of odd numbers and S2 being the set of even numbers.
Also, a partition could have only one set in the collection: thus { {1, 2, 3, 4, 5, 6} }
is a partition of {1, 2, 3, 4, 5, 6} (we could call it the “trivial partition”). This is
not a very interesting partition, but it fits the definition.
Remark: A “collection” is another word for “set”. Many people prefer to say
“collection of sets” rather than “set of sets”. Since a set does not depend on
the order of its members (e.g., {A, B, C} is the same as {C, B, A}), the same is
true for a partition. For example, {{1, 3, 6}, {2, 4}, {5}} is the same partition
as {{2, 4}, {1, 3, 6}, {5}} and as {{4, 2}, {3, 1, 6}, {5}}.
Using this definition, here is a rephrased version of the original problem we
considered in the first class.
Problem:
(a) Find a partition {S1 , S2 , . . . , S6 } of {1, 2, 3, . . . , 12} such that
the sum of the elements in Si is the same for every i.
(b) Is the solution of part (a) unique?
Many people found the following solution to part (a):
{{1, 12}, {2, 11}, {3, 10}, {4, 9}, {5, 8}, {6, 7}}.
(1)
Part (b) required more thought. The answer to (b) is “Yes,” and here is a way
that the class found to prove it.
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Proof of uniqueness: The first step is to show Claim 1: In any solution to
(a), the sum for each Si must equal 13. To prove Claim 1, let ti be the sum of
the elements in Si (for i = 1, . . . , 6). Then we must have
t1 + t2 + t3 + t4 + t5 + t6 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 (2)
because each element of {1, . . . , 12} must contribute to exactly one of the sums
ti . In other words, the left-hand side of Equation (2) simply represents adding
up all of the elements of {1, . . . , 12} in the order corresponding to the partition.
Next, observe that the left-hand side of Equation (2) is 6t1 , since all of the ti ’s
are equal. And the right-hand side of Equation (2) equals 78. Therefore we
have 6t1 = 78, which leads to t1 = 78/6 = 13. This proves Claim 1.
Next we will show Claim 2: In any solution to (a), each Si must have exactly
two elements. To prove Claim 2, we can proceed as follows. Since the sum of
the elements of Si must be 13 (by Claim 1), and the largest element is 12, we
see that each Si must contain at least two elements. Since we have six subsets,
each containing at least two elements, and there are only 6 × 2 elements in the
original set {1, 2, . . . , 12}, none of the subsets can have more than two elements.
This proves Claim 2.
To complete the proof, we see from Claims 1 and 2 that in any solution to
(a), each Si must be a two-element set of the form {a, b} where a + b = 13.
Equivalently, each Si is of the form {a, 13 − a}. The only sets of this form are
the ones listed in line (1) above. Therefore the partition in line (1) is the only
solution.
Notation: Above, we used the symbol to denote the end of a proof. The
text uses a solid rectangle instead.
Remark: Here is another way to prove uniqueness. First prove Claim 1 as
above. Now, consider a partition satisfying part (a). The element 12 must be
one of the subsets, so let us call this subset S1 . Since the elements of S1 add up
to 13, clearly the only possibility for S1 is {1, 12}. Then the other subsets in
the partition are all subsets of {2, 3, . . . , 11}. Let S2 be the subset containing
11. Then S2 has to be {2, 11}. Now the remaining four subsets are all subsets
of {3, 4, . . . , 10}. Let S3 be the subset containing 10. To get the sum of 13,
the only possibility is that S3 is {3, 10}. Continuing in the same way (Question
to think about: is it okay to write “in the same way”? ), we find that we must
have S4 = {4, 9}, then S5 = {5, 8}, and finally S6 = {6, 7}. This shows that the
solution of line (1) is unique.
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2. Some generalizations
Here are some examples of further questions we could ask about the equalsum problem.
For each positive integer k, let P (k) be the statement
P (k): There exists a partition of {1, 2, . . . , 12} into k subsets, {S1 , S2 , . . . , Sk },
such that the sum of the elements in Si is the same for every i.
We have seen that P (6) is true. For what values of k is P (k) true?
We could also consider the the statement Q(k) for each k:
Q(k): There is a unique partition of {1, 2, . . . , 12} into k subsets,
{S1 , S2 , . . . , Sk }, such that the sum of the elements in Si is the
same for every i.
(A statement of the form “There is a unique object with this property” says
two things: It says that there exists an object with the property, and that no
other object has this property.)
We have seen in class that Q(6) is true. For what other values of k is Q(k)
true?
We could also ask similar questions with “12” replaced by another positive
integer. Thus, for each pair of positive integers k and N , we can consider the
statements
P (k, N ): There exists a partition of {1, 2, . . . , N } into k subsets,
{S1 , S2 , . . . , Sk }, such that the sum of the elements in Si is the
same for every i.
Q(k, N ): There is a unique partition of {1, 2, . . . , N } into k subsets,
{S1 , S2 , . . . , Sk }, such that the sum of the elements in Si is the
same for every i.
We have seen that P (6, 12) and Q(6, 12) are true. Can we make any general
statements about conditions on k and N under which the truth or falsity if
P (k, N ) or Q(k, N ) is known?
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