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Re: [MATH for FUN] Old fashioned arithmetic fun.

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  • bogaduck
    I copied the question from a maths competition paper. I did it the same way as you, though didn t bother with squares like 169, since 196 is the same digits
    Message 1 of 4 , Sep 1, 2011
      I copied the question from a "maths competition" paper.
      I did it the same way as you, though didn't bother with squares like 169, since 196 is the same digits so would lead to two final answers. That method eliminated a lot of other.

      I do like the way they ask for the last 2 digits rather than all of them - a property I ignored until the end.

      I would like to know if there is some subtle way of establishing this answer, but enjoyed playing with the numbers rather than working out which end the string to light.

      And I think the insight, and understanding of numbers, you showed is probably all that is required.

      Hope you had fun.

      Peter

      --- In mathforfun@yahoogroups.com, MorphemeAddict <lytlesw@...> wrote:
      >
      > I don't understand why you specify only the last two digits of the sum of
      > the squares, but the sum is 1674, so it's last two digits are 74.
      > I made a list of all the squares between 100 and 999 that contain no
      > repeating digits, then made a table with 9 rows, and in each row entered all
      > the squares that contain that digit. Since only two squares (324, 361)
      > contain a 3, I tried 324 first. Of the digits not contained in 324 (156789),
      > the only squares that contain a 1 are 169 and 196. No square contains only
      > the remaining digits (578). So 324 cannot be the number containing 3, which
      > means that 361 is that number. Of the digits not in 361 (245789), only three
      > contain a 4 (324, 784, 841). Both 324 and 841 contain digits already used,
      > leaving only the digits 259, which are in the square 529. So the three
      > squares are 361, 529, 784.
      >
      > I suspect there is a much simpler way to get the answer, though, although it
      > requires more insight than I have.
      >
      > stevo
      >
      > On Wed, Aug 31, 2011 at 8:46 AM, bogaduck <pmaxotzen@...> wrote:
      >
      > > **
      > >
      > >
      > > The digits 1 to 9 can be separated into 3 disjoint sets of 3 digits each so
      > > that the digits in each set can be arranged to form a 3-digit perfect
      > > square. Find the last two digits of the sum of these three perfect squares.
      > >
      > > Peter
      > >
      > >
      > >
      >
      >
      > [Non-text portions of this message have been removed]
      >
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