A Russian puzzle

Dave Richeson tweeted about a puzzle from Futility Closet (original source a Russian mathematical olympiad): can you split the integers 1, 2, …, 15 into two groups A and B, with 13 elements in A and 2 elements in B, so that the sum of the elements of A is the product of the elements of B?

Think about it for a moment. There’s of course the temptation to brute-force it, which is doable, but there’s a more elegant solution.

This got me thinking – when can you split the integers 1, 2, …, n into two groups A and B, where B has two elements, so that the sum of the elements of A is the product of the elements of B?

Say B contains x and y. Then their product is of course xy. The sum of the elements of A is 1 + 2 + ... + n - (x+y) = n(n+1)/2 - (x+y). Setting these equal and rearranging gives

n(n+1)/2 + 1 = xy + x + y + 1

where we’ve added 1 to make the factorization work out – this becomes

n(n+1)/2 + 1 = (x+1)(y+1).

So the problem is reduced to finding factorizations of n(n+1)/2 + 1, which satisfy two conditions:

  • x and y can’t be equal (for specificity we’ll say x < y), and
  • x and y are both at most n.

Since we have y ≤ n, we’re going to have x ≥ (n+1)/2 + 1/n. n must be at least 2, so we can just write x ≥ (n/2) + 1. So we’re looking for factors of n(n+1)/2 + 1 in the interval [n/2+1, n]. Here’s some brute-force Python code to find all such solutions:


import math

def solutions(n):
    out = []
    total = n*(n+1)/2+1
    xmin = int(math.ceil(n/2.0) + 1)
    xmax = n
    for x in range(xmin, xmax+1):
        if total % (x+1) == 0:
            y = total/(x+1)-1
            if x < y:
                out.append([x,y])
    return out

def all_solutions(n):
    out = []
    for i in range(2, n+1):
        sols = solutions(i)
        for sol in sols:
            sol.insert(0, i)
            out.append(sol)
    return out

solutions takes an integer n as input and returns pairs [x, y] which are solutions to the problem. For example solutions(17) returns [[10, 13]].
And all_solutions takes an integer N and returns all triples [n, x, y] with n \le N which are solutions to the problem — that is, where xy equals the sum of all the integers up to n except for x and y. The first few solutions are:

n x y
10 6 7
17 10 13
26 15 21
36 22 28
37 21 31
45 27 36
50 28 43
61 42 43
65 36 57
67 42 52
78 45 66
82 45 73
91 52 78
94 57 76
101 55 91
102 70 73
110 70 85
122 66 111
136 76 120
138 87 108

So it appears that there’s nothing particularly special about the number 15 in the initial puzzle. There are plenty of values n for which you can’t do this, and plenty for which you can. Also, there are values of n for which there are multiple solution pairs (x, y), although not surprisingly they are rare. The smallest such n is 325, for which x = 171, y = 307 and x = 175, y = 300 are both solutions. In this case $latex n(n+1)/2 + 1 = 52976 = 24 \times 7 \times 11 \times 43$, from which 52976 has (5)(2)(2)(2) = 40 factors. A typical number of this size has about log(52976) \approx 11 factors. This abundance of factors makes it more likely that 52976 would have two factorizations of the sort we’re looking for. And in fact 52976 = 172 \times 308 = 176 \times 301.

Solutions to this problem appear to have some interesting statistical properties… more on that in a future post.

Weekly links for July 1

Simulated car design using genetic algorithms.

From the arXiv:
The Supreme Court is a spin glass.
How should traffic signals be timed on two-way streets?

From the June Notices of the AMS:
Judith R. Goodstein and Donald Babbitt’s article of E. T. Bell and Caltech mathematics between the wars (of Men of Mathematics and Bell numbers fame).
Richard Hoshino and Ken-ichi Kawarabayashi, “Graph Theory and Sports Scheduling”. As you might suspect from the names of the authors, they’re Japanese; the numbers they use in their problem apply to Japanese pro baseball (NPB), and their work has been used in actual scheduling of NPB.

Bryna Kra on mathematics as a toolbox for the sciences in the Chronicle of Higher Education.

Joel Grus can analyze data and has a two-year-old daughter, so naturally he looked at the most boyish and girlish colors and eigenshirts for children’s T-shirts.

Alex Bellos at the Guardian shows us mathematical food items.

William Beaty on the physics behind traffic jams.

Tom Fawcett has a gallery of visualization of results from machine learning classifiers.

Jon McLoone asks is there any point to the 12 times table?

Rafe Kinsey, at the University of Michigan, is teaching a freshman writing course on math, writing, and the world in the fall of 2013.

The boy who loved math: the improbable life of Paul Erdos is an illustrated children’s book.

John Cook on statistical evidence vs. legal evidence.

Gurmeet Manku has a collection of “75 combinatorial puzzles for mathematicians and computer scientists.”.

Celebrities die e at a time. (via Reddit)

From Nautilus magazine: how to insure against a rainy day and taming the unfriendly skies.