“Potatoes.”

1. Qh6 and the queen or knight will mate.

One in four. There are three intersections, so there are eight possible ways the lace can cross itself. Of these, only two (over/under/over and under/over/under) will produce a knot.

1. Qa6 and White will mate by 2. Qh6, 2. Nd3, or 2. Qa1.

Consider first the endpoints, A and K: The sum of the distances to these two points will be the same for any point between them. Similarly, any point between B and J will have the same total distance to points A, K, B, and J. Proceeding in this way and taking the points in pairs, we find that every point between E and G produces the same total distance to the outermost 10 labeled points. That leaves only F. So the answer is that point F has the minimum total distance to the labeled points.

Strangely, this means that we can shift the points around without affecting this result. If points A through E crowded F closely on the left, and point K were 90 light-years away on the right, point F would still yield the smallest total sum, so long as the points appeared in the order shown above.

1. Ng6 and Black must lose by 2. Qd3#, 2. Qe5#, or 2. Qd6#.

We can solve this problem by an application of the pigeonhole principle. Get a calendar for the month of June and on each day record the running total number of pills I’ve taken so far in the month. So June 1 might read “1,” June 2 “5,” and so on. The entry for June 30 will read “48.” Write these numbers in red.

Now revisit each date, add 11 to the red number, and enter the result in blue. In the example above June 1 now reads “1 12,” June 2 “5 16,” and so on. June 30 reads “48 59.”

Now we’ve filled the calendar with 60 numbers, all of which fall in the range 1-59. This means that at least one number falls on two different days. In the interval between these two days I took 11 pills.

By similar arguments it can be shown that intervals must exist during which I took precisely 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 pills.

Adapted from Kenneth R. Rebman, “The Pigeonhole Principle (What It Is, How It Works, and How It Applies to Map Coloring),” The Two-Year College Mathematics Journal, January 1979.

Yes. Statement 1 says that there are Futility Closet readers who are not U.S. residents. Let A be one of them. We’re claiming that A is not a walrus veterinarian, which would imply that statement 3 is true.

Suppose the contrary, that A is a walrus veterinarian. Since he’s not a U.S. resident, statement 2 implies that he’s not a Futility Closet reader, which isn’t true.

Adapted from Svetoslav Savchev, Mathematical Miniatures, 2003.

To list the numbers that don’t contain a 1, imagine six spaces and fill each with the digits 0, 2, 3, 4, 5, 6, 7, 8, or 9. The number of ways of doing this is 9⁶. There’s one exception: 000000 doesn’t fall between 1 and 1,000,000. So the number of integers without 1 is 9⁶ – 1 = 531,440, and the number with 1 is 468,560.