A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 4 years ago

Fool's cryptarithm of the day, Here is just another nice problem from an old issue of mathematics exacalibur ( http://www.math.ust.hk/excalibur/v1_n3.pdf ) check out "Qui Trouve Ceci" one. Good luck!

  • This Question is Closed
  1. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    FORTY TEN TEN ----- SIXTY to get Y + 2N = Y N must be either 0 or 5. if N = 0, no 1 is carried to 2nd column and E must = 5 and 1 is carried to 3rd. if N = 5, 1 is carried to 2nd column and 2E + 1 + T must = T or T + 10. that is 2E = -1 or 2E + 1 = 9 which is not possible so N must be = 0 and E = 5

  2. asnaseer
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I think ffm was referring to the problem after this one.

  3. asnaseer
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    the long division involving Qui Trouve Ceci

  4. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh yes

  5. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i did manage the above one though - after some sweating!! 29786 850 850 ----- 31486

  6. asnaseer
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    The one ffm wanted us to tackle is far more devious (as usual for an ffm problem!)

  7. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yea - i just looked at it!!!

  8. asnaseer
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    After lots and lots of scribblings and scratching of the head, I have found these sets so far: Q = 1 E = 4, 6 or 8 V = 0, 2 or 4 I = 2, 4, 6 or 8 C = 2, 3, 4, 6, 7 or 8 maybe this will help others progress further but I have hit a brick wall right now. I'll back to this tomorrow with a fresh mind. FFM: Please let me know if I am way off the mark in my findings so far.

  9. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Good work jimmyrep :) Here is the solution for that one : http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/ALPHAMETIC/forty+ten+ten=sixty More could be found here: http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/ALPHAMETIC/f @asnaseer: Seems okay to me :)

  10. asnaseer
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Ok, I've got further. These are my current findings: Q = 1 U = 3 I = 2 C = 6 or 7 E = 4 T = 5, 6, 7, 8 or 9 R = 5, 6, 7, 8 or 9 O = 5, 6, 7, 8 or 9 V = 0

  11. asnaseer
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    This was a real messy problem - lots of head scratching, trial and error, and luck, but I think I finally got it: Q = 1 U = 3 I = 2 C = 7 E = 4 T = 9 R = 8 O = 6 V = 0

  12. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.