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sritama Group Title

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

  • 2 years ago
  • 2 years ago

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  1. Arnab09 Group Title
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    abc+abc??

    • 2 years ago
  2. sritama Group Title
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    it was in the question..

    • 2 years ago
  3. Ishaan94 Group Title
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    We should let the number be secret.

    • 2 years ago
  4. experimentX Group Title
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    nice idea

    • 2 years ago
  5. Arnab09 Group Title
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    no, open the secret :D

    • 2 years ago
  6. vishal_kothari Group Title
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    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. ..

    • 2 years ago
  7. experimentX Group Title
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    fun will be gone

    • 2 years ago
  8. Arnab09 Group Title
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    hm.. i guess.. this is easier method: 2536 +/-multiple of 99=multiple of 111

    • 2 years ago
  9. perl Group Title
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    vishal is a fraud , he finds answers online

    • 2 years ago
  10. perl Group Title
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    here is the banana problem, compare http://www.gottfriedville.net/mathprob/misc-bananas.html

    • 2 years ago
  11. Arnab09 Group Title
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    HAHA, lol

    • 2 years ago
  12. Arnab09 Group Title
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    copy, paste?

    • 2 years ago
  13. Arnab09 Group Title
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    or typed entirely?

    • 2 years ago
  14. perl Group Title
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    he copied and pasted the whole thing. he is a fraud

    • 2 years ago
  15. Arnab09 Group Title
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    ok, lets banish him xD

    • 2 years ago
  16. perl Group Title
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    whats the challenge in googling?

    • 2 years ago
  17. Arnab09 Group Title
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    no challenge, probably the question-poster is making a fun of us :/

    • 2 years ago
  18. perl Group Title
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    well the question is fine, i dont like people who just post the whole answer. i want to work on it

    • 2 years ago
  19. perl Group Title
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    well, maybe its not a big deal. ok , forget about it

    • 2 years ago
  20. perl Group Title
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    better have the answer than not, right?

    • 2 years ago
  21. experimentX Group Title
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    i agree ...

    • 2 years ago
  22. perl Group Title
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    ok im angry that he gets a medal for it

    • 2 years ago
  23. perl Group Title
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    a medal for using google?

    • 2 years ago
  24. experimentX Group Title
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    it's all right ... medal's just nothing

    • 2 years ago
  25. Arnab09 Group Title
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    but i gave him a medal, can i take it back? xD

    • 2 years ago
  26. sritama Group Title
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    i became his fan..

    • 2 years ago
  27. experimentX Group Title
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    still ... i got to see a nice question and answer.

    • 2 years ago
  28. Arnab09 Group Title
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    un-fan :D

    • 2 years ago
  29. perl Group Title
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    hehe

    • 2 years ago
  30. experimentX Group Title
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    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

    • 2 years ago
  31. Arnab09 Group Title
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    but he can do it himself too..

    • 2 years ago
  32. perl Group Title
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    this one is do-able

    • 2 years ago
  33. perl Group Title
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    he says, i have all the answers on a file. yeah its called GOOGLE!!!

    • 2 years ago
  34. sritama Group Title
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    :D

    • 2 years ago
  35. myko Group Title
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    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?

    • 2 years ago
  36. perl Group Title
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    i still dont understand the gorilla problem

    • 2 years ago
  37. Arnab09 Group Title
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    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

    • 2 years ago
  38. Arnab09 Group Title
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    the gorilla cant understand how he can solve that too xD

    • 2 years ago
  39. myko Group Title
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    213c +(a+b)221 = 2536 but this gives more then one solution if 221 is more than once

    • 2 years ago
  40. myko Group Title
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    and it does

    • 2 years ago
  41. myko Group Title
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    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...

    • 2 years ago
  42. Ishaan94 Group Title
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    \[cab+bac+bca+2abc=2536\]\[100∗(2a+2b+c)+10∗(2a+2b+c)+(a+b+3c)=2536\]\[a+b+3c=10x+3\tag1\]\[x+2a+2b+c=10y+6\tag2\]\[y+2a+2b+c=25\tag3\] 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.

    • 2 years ago
  43. Ishaan94 Group Title
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    No :/ this isn't right

    • 2 years ago
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