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hartnn

  • 2 years ago

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

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  1. hartnn
    • 2 years ago
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    \((a^2+b^2)/10\) is integer

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

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

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

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

  6. mahmit2012
    • 2 years ago
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    Is it clear?

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

  25. hartnn
    • 2 years ago
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    plz share.

  26. sirm3d
    • 2 years ago
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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.

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

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

  29. klimenkov
    • 2 years ago
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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.

  30. hartnn
    • 2 years ago
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    1/6

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

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

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

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

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

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

  37. klimenkov
    • 2 years ago
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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?

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

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

  40. klimenkov
    • 2 years ago
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    Yes.

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

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

  43. klimenkov
    • 2 years ago
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    Right.

  44. sirm3d
    • 2 years ago
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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

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

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

  47. Zarkon
    • 2 years ago
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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

  48. sirm3d
    • 2 years ago
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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)

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

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

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

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

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

  54. Zarkon
    • 2 years ago
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    I used Octave

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

  56. Zarkon
    • 2 years ago
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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

  57. sirm3d
    • 2 years ago
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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)

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

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

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

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

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

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

  64. hartnn
    • 2 years ago
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    0.2

  65. sirm3d
    • 2 years ago
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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?

  66. Zarkon
    • 2 years ago
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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

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

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

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

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

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

  72. sirm3d
    • 2 years ago
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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.

  73. sirm3d
    • 2 years ago
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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?

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

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

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