Skip to content

Sunday Times Teaser 3117 – Save Stop Me If You’ve Heard This One

by Brian Gladman on June 17, 2022

by Colin Vout

Published Sunday June 19 2022 (link)

A square, a triangle and a circle went into a bar. The barman said, “Are you numbers over 18?” They replied, “Yes, but we’re under a million.” The square boasted, “I’m interesting, because I’m the square of a certain integer.” The triangle said, “I’m more interesting; I’m a triangular number, the sum of all the integers up to that same integer.” The circle said, “I’m most interesting; I’m the sum of you other two.” “Well, are you actually a circular number?” “Certainly, in base 1501, because there my square ends in my number exactly. Now, shall we get the drinks in?” The square considered a while, and said, “All right, then. You(’)r(e) round!”

In base 10, what is the circular number?

From → Uncategorized

9 Comments Leave one →
  1. Brian Gladman permalink

    Here is a straightforward but slow solution:

    Here are two alternative fast solutions, both of which use my number_theory library.

    This version uses the Chinese Remainder Theorem:

    and a second using modular square roots of unity:

    • Brian Gladman permalink


      This is not an assumption. Since C is the last digit of C^2 when C^2 is expressed
      as a number in base 1501, C must be less than 1501 (all digits in a base B integer
      are less than B).

      Note that I have just edited the first version.


      • Frits permalink

        If the base is 209 you seem to miss C’s 21660 and 22022.

        • Brian Gladman permalink

          I wouldn’t expect them to see these solutions since both
          programs use a short cut to solve the specific case set
          by this teaser.

    • Frits permalink

      @Brian, your alternative program doesn’t print a solution.

      C being less than 1501 seems a bit strange as all three numbers are reported to be under a million.

      • Brian Gladman permalink


        Thanks, I have corrected the alternative version now.

        Yes, its a bit surprising that the solution is so low compared to the
        range given in the teaser. But the teasers are intended for manual
        solution so including a clue that C is less than 1501 makes sense
        in my view.

    • Frits permalink

      Easier for me to understand (modular square roots):

      (C^2 – C) % 1501 = 0
      (4C^2 – 4C) % 1501 = 0
      (4C^2 – 4C + 1) % 1501 = 1
      (2C – 1)^2 % 1501 = 1

      • Brian Gladman permalink

        Modular square roots are tricky to compute in general but the mod
        value in this case has been chosen to make this easy since 1501 is
        the product of two primes – 19 and 79 .

        The extended GCD algorithm can be used to express any number as a
        linear combination of multiples of 19 and 79. Finding the square
        roots of 1 mod 19 and 79 is easy enough to do manually, after which
        the Chinese Remainder Theorem can be used to combine them to find
        square roots of 1 mod 1501.

  2. Arthur Vause permalink

    I also found it disappointing that the solution was so small

Leave a Reply

Note: HTML is allowed. Your email address will not be published.

Subscribe to this comment feed via RSS