- LUCI GUTIÉRREZ
Now the team is prepping for one of its own biggest challenges, the annual Putnam Competition, a prestigious and famously challenging contest for undergraduate college students. Here are some sample practice problems.
Midpoint Well Taken | Varsity Relay
S is a set of points in the plane. The exact number of distinct elements of S is equal to the answer to Gothic Arc from last week. M is the set of midpoints of every line segment both of whose endpoints are in S.
What is the minimum possible number of elements in M?
Closing Times | Varsity
A set is called closed under multiplication if the product of any two elements of the set is always a member of the set. For example, the integers are closed under multiplication. Suppose that every real number is colored either green or blue such that the product of any three green numbers is green and of any three blue numbers is blue. Let G be the set of green numbers and B be the set of blue numbers.
Which of the following is closed under multiplication: neither G nor B; at least one of G or B; or both G and B?
See answers to last week’s puzzles below. Come back next week for answers and more puzzles. Note: Comments may include spoilers.
The National Museum of Mathematics (MoMath) is North America’s only museum devoted to mathematics. Learn more at momath.org.
In Fancy Dice there are 48 distinct ways to construct a truncated tetrahedron. Look for the answer to Gothic Arc and the rest of the relay questions next week.
JK,
That proof was mine, as I clarified above, but thanks. (If you forget to type in your name, the posting system automatically dubs you "anonymous," which can be confusing.)
Both examples that I gave for the second set (one closed under multiplication and one open) proved on reflection to be impossible, so I've been trying to think of two possible sets that are consistent with the conditions that the product of any three elements of a set be in the same set, and that all real numbers are in one set or the other but not both.
It seems to me that one set could consist of all algebraic numbers and the other of all transcendental numbers. (Both would be closed.) That's kind of a natural division, because the product of algebraic numbers can never be transcendental, and vice versa. That's the only answer I could think of.
The other possible answer that's been suggested was positive numbers in one set and negative numbers in the other, with zero going somewhere. But that one gives me pause. Where do you put numbers that are simultaneously both positive and negative, like square roots, 4th roots, 6th roots, etc? I suppose you could somehow split them all in two, but that feels like cheating.
Anonymous provides a nice proof that at least one of the sets (the one containing '1') must be closed under multiplication. And it is possible for exactly one of the sets to be closed under multiplication: an example is given by the sets A = {x | x \geq 0} and B = {x | x < 0}. It is also possible for both sets to be closed under multiplication: an example is given by the sets A = {0} and B = {x | x \neq 0}.
chas: Hmm, I need to amend my example a bit. Thinking along integer lines, not real numbers. There are real numbers between -1 and 1 other than 0, but none of these can multiply to -1, 0 or 1. The remainder of my logic stands.
Oops again. {0,1,n} is shown above to meet the simple product criterion. It also obviously meets the triple-product criterion as 0 x 1 x n =0.
I made a mistake above when I wrote that the second set, the one which does not contain 1, "might consist solely of all the even integers, in which case it's closed under multiplication." In fact such a set can't exist if the first set contains all other numbers All other numbers would include, for example, the square roots of 2, 3 and 6, whose product is 6, an even integer. I haven't thought of a different example of a second closed set that would work.