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hartnn Group Title

If a,b are integers, what is the probability that \[a^2+b^2 \] is divisible by 10 ?

  • 2 years ago
  • 2 years ago

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  1. hartnn Group Title
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    \((a^2+b^2)/10\) is integer

    • 2 years ago
  2. mahmit2012 Group Title
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    |dw:1351968021756:dw|

    • 2 years ago
  3. hartnn Group Title
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    [0] means ?

    • 2 years ago
  4. mahmit2012 Group Title
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    |dw:1351968165513:dw|

    • 2 years ago
  5. mahmit2012 Group Title
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    |dw:1351968262315:dw|

    • 2 years ago
  6. mahmit2012 Group Title
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    Is it clear?

    • 2 years ago
  7. hartnn Group Title
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    trying to understand...seeing this type of approach for first time...

    • 2 years ago
  8. hartnn Group Title
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    how did u get a,b belongs t0 {......}mod 5 ?

    • 2 years ago
  9. mahmit2012 Group Title
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    |dw:1351968633462:dw|

    • 2 years ago
  10. mahmit2012 Group Title
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    |dw:1351968728104:dw|

    • 2 years ago
  11. mahmit2012 Group Title
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    |dw:1351968805888:dw|

    • 2 years ago
  12. hartnn Group Title
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    i think i am getting this a bit.....need to think

    • 2 years ago
  13. mahmit2012 Group Title
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    |dw:1351968902151:dw|

    • 2 years ago
  14. hartnn Group Title
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    can u expalin more on why there is just one option ?

    • 2 years ago
  15. mahmit2012 Group Title
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    a and b are independent so P(a,b)=P(a).P(b)

    • 2 years ago
  16. hartnn Group Title
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    i think i got other parts, except that there is only 1 option

    • 2 years ago
  17. mahmit2012 Group Title
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    |dw:1351969216109:dw|

    • 2 years ago
  18. hartnn Group Title
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    ohh....right...i'll go through this again. thanks!

    • 2 years ago
  19. mahmit2012 Group Title
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    |dw:1351969399186:dw|

    • 2 years ago
  20. hartnn Group Title
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    Also, any alternative approach will be appreciated....

    • 2 years ago
  21. hartnn Group Title
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    @mukushla any other approach you can think of?

    • 2 years ago
  22. sirm3d Group Title
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    for some reason, i only have p = 0.10

    • 2 years ago
  23. hartnn Group Title
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    me too, but thats incorrect.

    • 2 years ago
  24. sirm3d Group Title
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    i found my error. it is as what mahmit gave. but i have a different proof.

    • 2 years ago
  25. hartnn Group Title
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    plz share.

    • 2 years ago
  26. sirm3d Group Title
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    break the parity of a and b into odd-odd, odd-even, even-odd and even-even. clearly, p(odd-even) and p(even-odd) = 0 since the sum of squares is odd, not divisible by 2.

    • 2 years ago
  27. hartnn Group Title
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    sorry,didn't get u...

    • 2 years ago
  28. sirm3d Group Title
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    still polishing my argument.

    • 2 years ago
  29. klimenkov Group Title
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    Use this rule. If \(a\equiv x \mod 10 \) then \(a^2\equiv x^2 \mod 10 \) All different values for a can be 0,1,2,3,4,5,6,7,8,9,10. The probability that a is divisible by 10 is 1/10. Now find the values for \(a^2\) in mod 10.

    • 2 years ago
  30. hartnn Group Title
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    1/6

    • 2 years ago
  31. sirm3d Group Title
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    oops, i got 0.18 this time. hahaha. do you have the correct answer?

    • 2 years ago
  32. klimenkov Group Title
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    |dw:1352033013459:dw|

    • 2 years ago
  33. hartnn Group Title
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    no, i don't have correct answer

    • 2 years ago
  34. hartnn Group Title
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    for a^2, probability is 1/6, i get that what about a^2+b^2

    • 2 years ago
  35. sirm3d Group Title
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    i got it, 0.18

    • 2 years ago
  36. klimenkov Group Title
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    Now what is probability:|dw:1352033286386:dw|

    • 2 years ago
  37. klimenkov Group Title
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    The same table is for b. Now you have to write a table for \(a^2+b^2\) and watch what is for 0. Got it?

    • 2 years ago
  38. sirm3d Group Title
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    |dw:1352033270108:dw|

    • 2 years ago
  39. hartnn Group Title
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    are both of you doing same thing ?

    • 2 years ago
  40. klimenkov Group Title
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    Yes.

    • 2 years ago
  41. klimenkov Group Title
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    Can you write a table for \(a^2+b^2\) ? With probabilities?

    • 2 years ago
  42. hartnn Group Title
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    it will also contain 10 columns,right ? trying....

    • 2 years ago
  43. klimenkov Group Title
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    Right.

    • 2 years ago
  44. sirm3d Group Title
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    just count pairs that add up to 10 in my table. there should be 18 of them. (0,0), (1,3), (1,7), to list a few

    • 2 years ago
  45. klimenkov Group Title
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    If you want to get it, better to draw the whole table for \(a^2+b^2\).

    • 2 years ago
  46. hartnn Group Title
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    why is the answer drastically different from 2% ?

    • 2 years ago
  47. Zarkon Group Title
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    2 5 0 7 6 7 0 5 2 1 5 8 3 0 9 0 3 8 5 4 0 3 8 5 4 5 8 3 0 9 7 0 5 2 1 2 5 0 7 6 6 9 4 1 0 1 4 9 6 5 7 0 5 2 1 2 5 0 7 6 0 3 8 5 4 5 8 3 0 9 5 8 3 0 9 0 3 8 5 4 2 5 0 7 6 7 0 5 2 1 1 4 9 6 5 6 9 4 1 0

    • 2 years ago
  48. sirm3d Group Title
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    here's the first half of (a,b) mod 10 (0,0), (1,3), (1,7), (2,4), (2,6), (3,1), (3,9), (4,2),(4,8)

    • 2 years ago
  49. klimenkov Group Title
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    Answer is \(\frac{14}{100}=0.14\).

    • 2 years ago
  50. sirm3d Group Title
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    i say it's 18/100, same as @Zarkon

    • 2 years ago
  51. hartnn Group Title
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    why is the answer drastically different from 2% ? how is that table formed ? @Zarkon ?

    • 2 years ago
  52. sirm3d Group Title
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    the 2% only counted 5,5 and 0,0.

    • 2 years ago
  53. klimenkov Group Title
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    @sirm3d please write all the pairs.

    • 2 years ago
  54. Zarkon Group Title
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    I used Octave

    • 2 years ago
  55. Zarkon Group Title
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    for i=1:10;for j=1:10;A(i,j)=mod(i^2+j^2,10);endfor;endfor

    • 2 years ago
  56. Zarkon Group Title
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    >> A A = 2 5 0 7 6 7 0 5 2 1 5 8 3 0 9 0 3 8 5 4 0 3 8 5 4 5 8 3 0 9 7 0 5 2 1 2 5 0 7 6 6 9 4 1 0 1 4 9 6 5 7 0 5 2 1 2 5 0 7 6 0 3 8 5 4 5 8 3 0 9 5 8 3 0 9 0 3 8 5 4 2 5 0 7 6 7 0 5 2 1 1 4 9 6 5 6 9 4 1 0 >> length(find(A==0)) ans = 18

    • 2 years ago
  57. sirm3d Group Title
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    here's the other nine (5,5) (6,2), (6,8) (7,1), (7,9) (8,4), (8,6) (9,3), (9,7)

    • 2 years ago
  58. hartnn Group Title
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    also why are we only considering numbers till 100, will it make any difference, if we consider 1000 numbers ?

    • 2 years ago
  59. klimenkov Group Title
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    Sorry, my fault. Really - \(\frac{18}{100}\)

    • 2 years ago
  60. hartnn Group Title
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    i got now, how we got 18 pairs for 100....

    • 2 years ago
  61. sirm3d Group Title
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    finally, we are all in agreement. hurrah.

    • 2 years ago
  62. hartnn Group Title
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    why consider only 100? will it make any difference, if we consider 1000 numbers ?

    • 2 years ago
  63. hartnn Group Title
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    because if we consider only 10 numbers, we get different answer....

    • 2 years ago
  64. hartnn Group Title
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    0.2

    • 2 years ago
  65. sirm3d Group Title
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    5^2 + 15^2 is divisible by 10, so is 5^2 + (10k+5)^2, so why bother bigger numbers, when the last digit suffices?

    • 2 years ago
  66. Zarkon Group Title
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    you are doing division by 10 (mod 10) there are only 10 numbers in that system. So we only need to consider the 10x10 possibilities to get the proportion

    • 2 years ago
  67. hartnn Group Title
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    aah! okk, got that....

    • 2 years ago
  68. mahmit2012 Group Title
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    |dw:1352043506996:dw|

    • 2 years ago
  69. mahmit2012 Group Title
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    |dw:1352043883988:dw|

    • 2 years ago
  70. mahmit2012 Group Title
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    0.18 is not true. 0.02 is true. Be sure.

    • 2 years ago
  71. hartnn Group Title
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    where's the fault in our discussion ?

    • 2 years ago
  72. sirm3d Group Title
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    @mahmit2012 what about the pairs 1,7 and 1,3? are they not solutions too? they were not counted in your argument.

    • 2 years ago
  73. sirm3d Group Title
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    @mahmit2012 you only saw 5,5 and 0,0 in your proof. that explains your 2/100=0.02. what about 1,7 pair?

    • 2 years ago
  74. mahmit2012 Group Title
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    |dw:1352045197951:dw|

    • 2 years ago
  75. mahmit2012 Group Title
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    |dw:1352045432135:dw|

    • 2 years ago
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