These are the answers to yesterday’s math questions. If you want to look at the questions before the answers, scroll down to the previous post quickly, or click here.

To space the answers further down the page, I will insert some of my favorite spoonerisms. Reverend William Archibald Spooner (1844-1930) was a dignified clergyman who had an unfortunate mental quirk. He would unintentionally switch the initial sounds of two words in his sentences, often with ludicrous results. It is said that he once proposed a toast to “our queer old dean” (dear old queen). Many of the quotes from him are probably apocryphal.

I had an uncle who loved to spoonerize. His wife, my aunt, would sometimes react to a sentimental story with “Doesn’t that harm your wart” (warm your heart). An annual fund-raiser at our church was called “One great hour of sharing,” which I liked to speak of as “One great shower of herring.”

1. Walter must have said that the trip began on a Tuesday and ended on a Monday, and the Tuesday must have been February 21, 1888. Given that the trip took place after the start of the American Civil War (April 12, 1861) and before Lindbergh’s solo flight (May 20, 1927), if Walter had said any other combination of days, the answer would have been either ambiguous or impossible.

The reasoning used to solve this problem hinges on recognizing that Walter’s information would have allowed Everett to deduce the length of the month (28, 29, 30, or 31 days). You see, if Walter had said it took from Monday the 21^{st} to Tuesday the 19^{th}, from a Monday to a Tuesday is some number of weeks and one day, and from the 21^{st} of one month to the 19^{th} of the next month is two days less than the length of the first month. For example, August has 31 days, so from Aug. 21 to Sept. 19 is 29 days. That would work, because 29 days is four weeks and one day. But it could be August of any year from 1861 to 1926, as long as Aug. 21 was a Monday. There were 66 Augusts in that period, and the 21^{st} fell on a Monday in 10 of them. Everett would not have been able to get the year right with certainty.

In the period 1861-1927, there would be too many 28-day, 30-day, or 31-day months to determine which of them was the right one, but there were only 15 29-day months (the leap-year Februarys). Looking up the day on which the 21st of each of these Februarys fell, we find that only one of the days (Tuesday) was unique, permitting Everett to solve the problem.

Note that if you knew that the Civil War started between 2/22/1860 and 2/18/1864, and that Lindbergh flew between 3/20/1924 and 3/18/1928, you could have answered the question without an almanac. You didn’t need to deduce that the days of the week were Tuesday and Monday. Just the facts that a normal year has one day more than 52 weeks, a leap year has two days more than 52 weeks, and 1900 was not a leap year, would have enabled you to construct a table showing which leap February started on a unique day of the week for that period.

You might have hit on this answer faster if you regularly used my trick for calculating day of the week in the near future. Today is Saturday, January 23. Someone in my family has a birthday coming up on March 6. What day of the week will that be? Well, March 6 is the same as February 29+6=35, because this is a leap year and February has 29 days. And that’s the same as January 31+35=66. January 66 is 66-23=43 days from now. Forty-three days is six weeks and one day, so March 6 will be one day later in the week than Saturday, i.e. Sunday.

2. The first answer is yes. The second is no. Basically, it’s because kilometers are smaller than miles, so a distance given in kilometers conveys more information. If you convert kilometers to miles by dividing by 1.609344 and rounding to the nearest mile, either 6 km. or 7 km. will give 4 miles, and there’s no single conversion method that can transform 4 miles into both 6 km. and 7 km., as needed. Multiplying 4 miles by 1.609344 and rounding gives 6 km.

To prove that converting miles to kilometers and back will always produce the number you started with, let the conversion factor be called f (in this case, f = 1.609344, but all we need to know is that f > 1). If m is the number of miles, multiplying by f and rounding to the nearest whole number produces the value k = [ fm + .5 ], where brackets represent the greatest integer function. This means that k is an integer such that fm – .5 < k <= fm + .5. Doing the reverse conversion, we calculate M = [ k/f + .5 ]. That means that k/f – .5 < M <= k/f + .5.

Is it possible that m < M? If so, m < M <= k/f + .5 <= (fm + .5)/f + .5 = m + .5(1/f + 1) < m + 1, since f > 1. Ignore the intermediate inequalities and look at m < M < m+1. Since m and m+1 are consecutive integers, it’s impossible for M to be an integer that falls in between them. This disproves the original assumption that m < M. In much the same way, we can disprove that M < m, and it follows that M = m, as desired.

3. Just three farthings. A farthing was a quarter of a penny. In those days, they wrote a money amount with a pound sign (£) first, then the number of pounds, then a space, then the number of shillings, then a slash (called a solidus, Latin for shilling), then the number of pence, then d (for denarii). If there were no pence, there would be a hyphen after the slash. The cashier would probably have figured it out about like this: 39 * 2/6¾d = 78/- + 234d + 117*¼d = 78/263¼d. There were 12 pence in a shilling, so 263¼d is the same as 21/11¼d. Adding the 78/- makes the total price 99/11¼d. There were 20 shillings in a pound, so £5 = 100/-. The change would be 100/- minus 99/11¼d. Cancelling out the 99 shillings, the change is 1/- minus 11¼d. Remembering that 1/ is 12d, that’s 12d – 11¼d, which leaves just ¾d, which is three farthings.

British cashiers these days just don’t know how good they have it.

The United Kingdom converted to decimal currency in mid-February, 1971. I was touring in Great Britain for four weeks in January, 1971. My travel journal reports, “About 5 times I’ve been given incorrect change, always in my favor.” I suppose I took some glee in correcting the pros in their work.

4. No. If you learned in school about casting out the nines, you can see why not. Take a number like 2076. If you add its digits (2+0+7+6 = 15), the result has the same remainder on division by 9 as the number you started with (2076/9 = 230 r. 6; 15/9 = 1 r. 6). This fact is used, among other things, to test whether a number is divisible by 9. If a ten-digit number contains every digit from 0 to 9 once, then the sum of its digits is 0+1+2+…+9 = 45. Since 45 is divisible by 9 with no remainder, the ten-digit number must also be divisible by 9, so it can’t be a prime.

For the general case, let b be the base of notation (b is an integer ≥ 2). A number in base b has the same remainder as the sum of its digits when they are both divided by b-1; when b=10 and b-1=9, this is the same thing as casting out the nines. Given a b-digit number in base b containing every digit from 0 to b-1, the sum of its digits is (b-1)b/2. For example, I asked about base 5. The sum of the digits in that system is 0+1+2+3+4 = 10, so any number formed with those five digits has the same remainder when divided by 4 as 10 does. That remainder is 2. So any such number is even. In the general case, if b is even, then (b-1)b/2 is b-1 times b/2, and b/2 is an integer; therefore the sum of the digits is divisible by b-1; therefore the number itself is divisible by b-1. If b = 2, that doesn’t help, because a prime can be divisible by 1; but if b > 2 (and b is still even), then b must be at least 4, b-1 ≥ 3, and if the number is divisible by b-1, it can’t be a prime. If b is odd, on the other hand, then (b-1)/2 is an integer. Say d = (b-1)/2. The sum of the digits is divisible by d. But b-1 is also divisible by d (two times). It follows that the original number is divisible by d. If b = 3, then d = 1, and the number could still be prime, but if b > 3, then d ≥ 2, and the number can’t be a prime.

So far, we’ve seen that if b is an even number > 2 or an odd number > 3, the number in base b formed with every digit from 0 to b-1 can’t be a prime. Since 1 is not a possible base, that leaves only 2 and 3 as possibilities. It turns out that they actually work. In base 2, the number 10 uses all the digits and is prime. In base 3, the numbers 012, 021, 102, and 201 use all the digits and are prime (in ordinary base-10 arithmetic, they are equal to 5, 7, 11, and 19, respectively).

5. Surprisingly, no. The two Archimedean solids shown below are not mirror-symmetric. Their shaded faces are twisted to the right; if they had been twisted to the left instead, that would produce their mirror images. (L. to r.: snub cube, snub dodecahedron)

6. Eight. There are deltahedra with 4, 6, 8, 10, 12, 14, 16, and 20 sides. The ones with 4, 8, and 20 sides are the Platonic solids, tetrahedron, octahedron, and icosahedron. The one with 6 sides looks like two tetrahedra glued together; the one with 10 sides looks like two pentagonal pyramids with their pentagonal sides glued together; and the octahedron is equivalent to two square pyramids with their square sides glued together. The one with 16 sides can be described as a square anti-prism with a square pyramid glued to each of its square sides. The one with 14 sides is a triangular prism with square pyramids glued to each of its square sides. The one with 12 sides isn’t so easy to describe.

You might ask, how about two hexagonal pyramids glued together? Well, a hexagonal pyramid is flat. If you erect six equilateral triangles on a hexagonal base, they will only meet if they lie in the same plane as the hexagon. How about three tetrahedra glued together? No, the resulting figure is not convex. It will have one edge common to all three tetrahedra, and at that edge the two triangles will form an angle of more than 180°. How about an octahedron with a tetrahedron glued to one side? If you glue a tetrahedron to an octahedron, their two adjacent faces will lie in the same plane, forming a single rhombus-shaped face.

7. You may have thought it was a cryptarithm, where you have to replace each letter by a digit (the same digit for the same letter throughout). If so, surprise! It’s just a straight long division problem in hexadecimal notation (base-16 arithmetic, where the digits are 0 1 2 3 4 5 6 7 8 9 A B C D E F). I replaced the digit 0 with the letter O, because it was a giveaway that the digit 0 is slashed in this font.

8. Real! Euler’s well-known formula says e^{iπ} = -1, and is often used to define powers of numbers when the exponent is not an integer. Take the square root of both sides: e^{iπ/2} = √-1 = i; raise both sides to the ith power: i^{i} = (e^{iπ/2})^{i} = e^{iiπ/2} = e^{-π/2}. Since e > 1 and e and π are both real, the answer is real. Its value is about 0.20788.