Five Points, Ten Distances

Five points, A, B, C, D, and E, lie on a line. The distances between them in ascending order are: 2, 5, 6, 8, 9, X, 15, 17, 20, and 22. What is X?

We assume that the points are ordered A to E from left to right. We have AE = 22 and either AD = 20, BE = 17, or AD = 17, BE = 20. Without loss of generality AD = 20, BE = 17, and therefore AB = 5, BD = 15, DE = 2. The distance of 6 is associated with either BC or CD, and therefore the points are arranged in one of these two ways:

  1. AB = 5, BC = 6, CD = 9, DE = 2
  2. AB = 5, BC = 9, CD = 6, DE = 2

If it is the latter, we get the sequence of distances: 2, 5, 6, 9, 11, 14, 15, 17, 20, 22, which does not fit the provided sequence.

If it is the former, we get the sequence of distances: 2, 5, 6, 8, 9, 14, 15, 17, 20, 22, and therefore X = 14.

Wacky Wordies 2

Can you figure out what phrases and sayings are represented in the Wacky Wordies below?

The answers are:

  1. Leave no stone unturned
  2. Foot in the door
  3. Go on a double-date
  4. Green with envy
  5. Look square in the eye
  6. Broken promise
  7. Pull up alongside the curb
  8. Excuse me
  9. High-grade performance
  10. Take on a big job
  11. Split the difference
  12. He came out of nowhere
  13. Wait on hand and foot
  14. Suit to a T
  15. Know it forward and back
  16. A period in history
  17. Crooked lawyer
  18. Get up on the wrong side of bed
  19. Sign on the dotted line
  20. Disorderly conduct

Perfect Square Sums

Find all possible arrangements of the numbers 1 to 15 in a sequence, where the sum of any two consecutive numbers is a perfect square.

Note that the number 8 can be neighboring only with the number 1, so it must be at one end of the sequence. The number 15 can be neighboring only with the numbers 1 and 10, so it either needs to be next to 1 or at the other end of the sequence. In either case, 10 should be next to it. The only other number that can be neighboring 10 is 6. Then 3 should follow, then 13 (since 1 is already taken), then 12, then 4, then 5, then 11, then 14, then 2, then 7, then 9. Since 1+9=10 is not a perfect square, we find that the only solutions are

8-1-15-10-6-3-13-12-4-5-11-14-2-7-9

and its reverse.