»  Solutions to puzzles in my National Review Online Diary

  July 2009


In my July diary I posed the following two brain-teasers:

(1)   Prove that the following pattern continues:

          1 + 2 = 3
          1×2 + 2×3 + 3×4 = 4×5
          1×2×3 + 2×3×4 + 3×4×5 + 4×5×6 = 5×6×7
(2)   Tell me the exact significance of all the following seven numbers. (Note: I have rigged it so that you almost certainly can't, but you should be able to get close).




(1)   Several quick-witted readers spotted that we are in perms'n'coms territory here — permutations and combinations, that is. The key symbol here is C(np), the number of ways of choosing p from n. C(5,2) is 10: if the five objects are A, B, C, D, and E, I can choose AB, AC, AD, AE, BC, BD, BE, CD, CE, or DE. (The order of choice doesn't matter, and you are choosing without replacement.)
The main thing you need to know is that C(np)  =  n! ÷ (p! × (np)!). We remember this in a simpler form as "factorial on the bottom, same number of terms on top." So C(5,2) is 5×4 divided by 2×1. It also follows from the formula that C(5,2) = C(5,3);  of course! — 5×4 divided by 2×1 is equal to 5×4×3 divided by 3×2×1.

When you've messed with perms'n'coms a bit, this simple form becomes so ingrained that on seeing an expression like

          1×2×3 + 2×3×4 + 3×4×5 + 4×5×6 = 5×6×7

you immediately want to divide through by the first tem. This reduces the whole expression to coms:

1 + C(4,3) + C(5,3) + C(6,3) = C(7,3)

Rewrite the right-hand side as C(7,4). How many ways can I choose 4 items from 7? If I insist on including some particular element, there are C(6,3) ways of picking the rest; if I insist on leaving that out but insist on including some other particular one, that leaves C(5,3) ways of choosing the rest; if I insist on leaving those two out but including some other one, there are C(4,3) ways to pick the rest; and if I insist on leaving out some particular three but including a fourth, I have only one way to choose the rest. Q.E.D.

(2)   Oh dear, I'm afraid I messed up at least one of these. Several readers none the less got my drift, though of course nobody nailed them all.

Write down the English words for the ten digits:  zero, one, two, three, four, five, six, seven, eight, nine. Rearrange them in alphabetic order:  eight, five, four, nine, one, seven, six, three, two, zero. That's your first number. For the others, just do the same thing in French, German, Italian, Russian (on the Cyrillic alphabet, of course!), Chinese (pinyin), and — just for the heck of it — Hungarian.

The German I know I screwed up, "hearing" vier with an initial "f" instead of "v."  (I've corrected it here.)  Someone tells me I got the Italian wrong, too, though I don't see it. (But he said it works for Portuguese …)

Curious that the two closest numbers in the list are generated by Russian and Chinese, two utterly unrelated languages.

Martin Gardner's latest book has some interesting facts about the first number, the one generated from English. If you divide it by five, for example, you get 1,709,835,264, which also has each of the ten digits once! Then if you divide by five again …