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sritama

  • 4 years ago

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

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  1. Arnab09
    • 4 years ago
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    abc+abc??

  2. sritama
    • 4 years ago
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    it was in the question..

  3. Ishaan94
    • 4 years ago
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    We should let the number be secret.

  4. experimentX
    • 4 years ago
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    nice idea

  5. Arnab09
    • 4 years ago
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    no, open the secret :D

  6. vishal_kothari
    • 4 years ago
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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. ..

  7. experimentX
    • 4 years ago
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    fun will be gone

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

  9. perl
    • 4 years ago
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    vishal is a fraud , he finds answers online

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

  11. Arnab09
    • 4 years ago
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    HAHA, lol

  12. Arnab09
    • 4 years ago
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    copy, paste?

  13. Arnab09
    • 4 years ago
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    or typed entirely?

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

  15. Arnab09
    • 4 years ago
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    ok, lets banish him xD

  16. perl
    • 4 years ago
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    whats the challenge in googling?

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

  18. perl
    • 4 years ago
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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

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

  20. perl
    • 4 years ago
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    better have the answer than not, right?

  21. experimentX
    • 4 years ago
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    i agree ...

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

  23. perl
    • 4 years ago
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    a medal for using google?

  24. experimentX
    • 4 years ago
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    it's all right ... medal's just nothing

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

  26. sritama
    • 4 years ago
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    i became his fan..

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

  28. Arnab09
    • 4 years ago
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    un-fan :D

  29. perl
    • 4 years ago
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    hehe

  30. experimentX
    • 4 years ago
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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

  31. Arnab09
    • 4 years ago
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    but he can do it himself too..

  32. perl
    • 4 years ago
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    this one is do-able

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

  34. sritama
    • 4 years ago
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    :D

  35. myko
    • 4 years ago
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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?

  36. perl
    • 4 years ago
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    i still dont understand the gorilla problem

  37. Arnab09
    • 4 years ago
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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

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

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

  40. myko
    • 4 years ago
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    and it does

  41. myko
    • 4 years ago
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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...

  42. Ishaan94
    • 4 years ago
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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.

  43. Ishaan94
    • 4 years ago
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    No :/ this isn't right

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