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klimenkov

Find all 3-digit numbers such that the sum of the digits is 11 times less than the number.

  • one year ago
  • one year ago

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  1. Azteck
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    that are more than the sum of its digits in 11 <===that part of the question is confusing. Did you write the question properly, because it doesn't make sense to me?

    • one year ago
  2. Azteck
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    I don't know how to correct that part. Is this a question from a worksheet or textbook?

    • one year ago
  3. klimenkov
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    This is my translation. You have to find all 3-digit numbers, the sum of digits of which is 11 times less than the number. Can you write this correctly in English?

    • one year ago
  4. ParthKohli
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    Find all 3-digit numbers such that the sum of the digits is 11 times less than the number.

    • one year ago
  5. klimenkov
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    Actually, I used Google Translate, because I did not know the grammar.

    • one year ago
  6. Azteck
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    Ah okay, no worries mate.

    • one year ago
  7. klimenkov
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    Thank you @ParthKohli . Now solve this.

    • one year ago
  8. ParthKohli
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    So translating your question into an equation:\[\dfrac{100a + 10b + c}{11} = a + b + c\]Where the digits are \(a,b,c\).

    • one year ago
  9. ParthKohli
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    Another way to think is to see the multiples of \(11\), because we are sure that our three-digit number is a multiple of \(11\): 110 - no. 121 - no. 132 - no. 143 - no. 154 - no. 165 - no. Too long.

    • one year ago
  10. whpalmer4
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    What about 198?

    • one year ago
  11. ParthKohli
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    oh myyy...

    • one year ago
  12. klimenkov
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    You don't find easy ways...

    • one year ago
  13. whpalmer4
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    Also 0 :-)

    • one year ago
  14. klimenkov
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    0 is not a 3-digit number @whpalmer4 .

    • one year ago
  15. ParthKohli
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    \[11a + 11b + 11c = 100a + 10b + c \iff -89a + b + 10c = 0\]

    • one year ago
  16. klimenkov
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    Yes. It is easier to find.

    • one year ago
  17. ParthKohli
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    Yeah, let's try \(a = 1\). Then \(b = 9\) and \(c = 8\) which is exactly what @whpalmer4 said

    • one year ago
  18. ParthKohli
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    If \(a = 2\), then \(b = 8\) but \(c > 9\).

    • one year ago
  19. ParthKohli
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    So I guess \(198\) is the only number.

    • one year ago
  20. whpalmer4
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    for (i = 0; i < 10; i++) for (j = 0; j < 10; j++) for (k = 0; k < 10; k++) if ((i+j+k)*11 == (i*100+j*10+k)) then printf("%d%d%d is solution", i, j, k) works too

    • one year ago
  21. whpalmer4
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    (though it will toss out 000 as a solution)

    • one year ago
  22. klimenkov
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    I would give you both medals if i could.

    • one year ago
  23. ParthKohli
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    Don't worry, whpalmer has mine.

    • one year ago
  24. ParthKohli
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    @whpalmer4 Don't worry with the 000. We can figure out trivial solutions in the end. :-D

    • one year ago
  25. whpalmer4
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    WolframAlpha writes the solution as \[b=9a + 10n, c = 8a-n, n \in Z\] which makes it clear that 198 is the only solution if a b and c all have to be single digits

    • one year ago
  26. whpalmer4
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    Fun little late-night problem :-)

    • one year ago
  27. ParthKohli
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    Holy Wolfram|Alpha!

    • one year ago
  28. whpalmer4
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    So, here's my parting shot — what about a similar problem, except in base 16, and the multiple is 17 * the sum of the digits? :-)

    • one year ago
  29. ParthKohli
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    \[17a + 17b + 17 c= 256a + 16b + c\]

    • one year ago
  30. ParthKohli
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    That?

    • one year ago
  31. whpalmer4
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    Yes, that's it, I think.

    • one year ago
  32. ParthKohli
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    And the solutions can't exceed \(15\).

    • one year ago
  33. klimenkov
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    Do[If[11*(IntegerDigits[i][[1]] + IntegerDigits[i][[2]] + IntegerDigits[i][[3]]) == i, Print[i]], {i, 100, 999}] 198

    • one year ago
  34. whpalmer4
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    Here's a question: what effect does the multiplier have on the number of solutions? in base 16, a multiplier of 11 gets all sorts of solutions. a multiplier of 17 gets 1, a multiplier of 19 gets a small handful.

    • one year ago
  35. ParthKohli
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    What's that? Mathematica?

    • one year ago
  36. klimenkov
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    Yes.

    • one year ago
  37. ParthKohli
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    So does that mean this: If we are looking at base \(n\), and the multiplier is \(n + 1\), then there's one solution.

    • one year ago
  38. whpalmer4
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    For no good reason, I thought that a prime number as the multiplier would lead to few solutions, and a composite number many, but that doesn't seem to be true

    • one year ago
  39. whpalmer4
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    I'll check back in the morning to see what you guys have figured out :-)

    • one year ago
  40. ParthKohli
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    Nah, I am pretty bad at mathematics, I'd just leave this thread and we'd on this tomorrow. =)

    • one year ago
  41. ParthKohli
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    Work on this*

    • one year ago
  42. whpalmer4
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    I'm confident that our friend @klimenkov will have written a program to analyze it before long :-) With that, I'm out of here for the night!

    • one year ago
  43. klimenkov
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    Ok. I'll try. Good night.

    • one year ago
  44. LOOSEr
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    11(a+b+c)<100a+10b+c b+10c<89a Always true when a>=2 When a=1 always true for 0<=c<=7 and 0<=b<=9 always true when c=8 and 0<=b<=8

    • one year ago
  45. LOOSEr
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    So, 9*10*10+1*10*8+1*9*1

    • one year ago
  46. whpalmer4
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    @LOOSEr the problem is that 9*10*10+1*10*8+1*9*1 = 989 which is not divisible by 11, and 9+8+9 = 26 * 11 = 286 so it fails to meet the requirements. Because the maximum sum of the digits is 9+9+9 = 27, and 27*11 = 297, any solution to this problem must be <= 297...

    • one year ago
  47. LOOSEr
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    No.... what I mean is Number of 3-digit numbers such that the sum of the digits is 11 times less than the number = 989

    • one year ago
  48. LOOSEr
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    Sorry, it is 8*10*10+1*10*8+1*9*1=889

    • one year ago
  49. whpalmer4
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    Ah, okay, I agree with that, as I find 110 numbers where the sum of the digits * 11 is >= the number, and we aren't counting 0, so there are 999 candidates in all. Sorry for misunderstanding your point earlier.

    • one year ago
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