Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

sritama

  • 2 years ago

Let a secret three digit number be cba. If the sum of cab + bac + bca + abc + abc = 2536, what is cba ?

  • This Question is Closed
  1. Arnab09
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    abc+abc??

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

    it was in the question..

  3. Ishaan94
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    We should let the number be secret.

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

    nice idea

  5. Arnab09
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    no, open the secret :D

  6. vishal_kothari
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 5

    The sum of all six numbers is 222c + 222b + 222a If u omit cba = (100c + 10b + a), what is left is N = 122c + 212b + 221a which is what we are given. Observing 221 = 13*17, we have N = 5c + 4b mod 13 and N = 3c + 8b mod 17 (u could have used factors of 122 or 212, but 2*61 and 2*2*53 are not nearly so convenient as 13 and 17.) For N = 2536, we have N mod 13 = 1 and N mod 17 = 3 So u have 1 = 5c + 4b mod 13 and 3 = 3c + 8b mod 17 Since b and c are digits in range 0-9, u can try each value of b for 0 through 9 . Compute 4 x b and 4 x b mod 13. Subtract that from 1 to (5c mod 13). Then find a suitable value for c which satisfies that. For example 5c = 1 mod 13 -> 5c = 40, c = 8. compute 3c+8b (mod 17) and find the one which equals 3.. we see it is the row where b = 7 and c = 5. So we know that the original number was 57a. To determine a, we take 2536 - 122*5 - 212*7 = 2536 - 610 - 1484 = 442 Dividing by 221, we get a = 2, so the original number was 572. We can verify that 527+725+752+257+275 = 2536. ..

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

    fun will be gone

  8. Arnab09
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    hm.. i guess.. this is easier method: 2536 +/-multiple of 99=multiple of 111

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

    vishal is a fraud , he finds answers online

  10. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    here is the banana problem, compare http://www.gottfriedville.net/mathprob/misc-bananas.html

  11. Arnab09
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    HAHA, lol

  12. Arnab09
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    copy, paste?

  13. Arnab09
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    or typed entirely?

  14. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    he copied and pasted the whole thing. he is a fraud

  15. Arnab09
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    ok, lets banish him xD

  16. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    whats the challenge in googling?

  17. Arnab09
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    no challenge, probably the question-poster is making a fun of us :/

  18. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    well the question is fine, i dont like people who just post the whole answer. i want to work on it

  19. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    well, maybe its not a big deal. ok , forget about it

  20. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    better have the answer than not, right?

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

    i agree ...

  22. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok im angry that he gets a medal for it

  23. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    a medal for using google?

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

    it's all right ... medal's just nothing

  25. Arnab09
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    but i gave him a medal, can i take it back? xD

  26. sritama
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i became his fan..

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

    still ... i got to see a nice question and answer.

  28. Arnab09
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    un-fan :D

  29. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hehe

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

    lol ... seriously i don't like the word fan ... i believe we all have equal abilities .. some use and some don't use ... but i think it's fun

  31. Arnab09
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    but he can do it himself too..

  32. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    this one is do-able

  33. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    he says, i have all the answers on a file. yeah its called GOOGLE!!!

  34. sritama
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    :D

  35. myko
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    cab + bac + bca + abc + abc = 2536 100c+10a+b+100c+10a+c+100b+10c+a+2(100a+10b+c) = 2536 213c +221b + 221a =2536 maybe this is good start?

  36. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i still dont understand the gorilla problem

  37. Arnab09
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    ok, if u would replace last abc by cba, the result would be different, the sum would b divisible by 222 and the difference between them is divisible by 99 then, i guess hit and trial method

  38. Arnab09
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    the gorilla cant understand how he can solve that too xD

  39. myko
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    213c +(a+b)221 = 2536 but this gives more then one solution if 221 is more than once

  40. myko
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and it does

  41. myko
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    cab + bac + bca + abc + abc = 2536 100c+10a+b+100c+10a+c+100b+10c+a+2(100a+10b+c) = 2536 213c +221b + 221a =2536 maybe i made a mistake here somehere...

  42. Ishaan94
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[cab+bac+bca+2abc=2536\]\[100∗(2a+2b+c)+10∗(2a+2b+c)+(a+b+3c)=2536\]\[a+b+3c=10x+3\tag1\]\[x+2a+2b+c=10y+6\tag2\]\[y+2a+2b+c=25\tag3\] From (1), (2) and (3), \[x−11y=−22\] I think it would be safe to assume x=0 and y=2.\[2a+2b+c=23,a+b+3c=6\]We all know 0<a,b,c<9. So, it wouldn't be that hard to guess all the numbers.

  43. Ishaan94
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    No :/ this isn't right

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

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

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.