## hartnn Group Title If a,b are integers, what is the probability that $a^2+b^2$ is divisible by 10 ? one year ago one year ago

1. hartnn Group Title

$$(a^2+b^2)/10$$ is integer

2. mahmit2012 Group Title

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3. hartnn Group Title

[0] means ?

4. mahmit2012 Group Title

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5. mahmit2012 Group Title

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6. mahmit2012 Group Title

Is it clear?

7. hartnn Group Title

trying to understand...seeing this type of approach for first time...

8. hartnn Group Title

how did u get a,b belongs t0 {......}mod 5 ?

9. mahmit2012 Group Title

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10. mahmit2012 Group Title

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11. mahmit2012 Group Title

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12. hartnn Group Title

i think i am getting this a bit.....need to think

13. mahmit2012 Group Title

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14. hartnn Group Title

can u expalin more on why there is just one option ?

15. mahmit2012 Group Title

a and b are independent so P(a,b)=P(a).P(b)

16. hartnn Group Title

i think i got other parts, except that there is only 1 option

17. mahmit2012 Group Title

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18. hartnn Group Title

ohh....right...i'll go through this again. thanks!

19. mahmit2012 Group Title

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20. hartnn Group Title

Also, any alternative approach will be appreciated....

21. hartnn Group Title

@mukushla any other approach you can think of?

22. sirm3d Group Title

for some reason, i only have p = 0.10

23. hartnn Group Title

me too, but thats incorrect.

24. sirm3d Group Title

i found my error. it is as what mahmit gave. but i have a different proof.

25. hartnn Group Title

plz share.

26. sirm3d Group Title

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 Group Title

sorry,didn't get u...

28. sirm3d Group Title

still polishing my argument.

29. klimenkov Group Title

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 Group Title

1/6

31. sirm3d Group Title

oops, i got 0.18 this time. hahaha. do you have the correct answer?

32. klimenkov Group Title

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33. hartnn Group Title

no, i don't have correct answer

34. hartnn Group Title

for a^2, probability is 1/6, i get that what about a^2+b^2

35. sirm3d Group Title

i got it, 0.18

36. klimenkov Group Title

Now what is probability:|dw:1352033286386:dw|

37. klimenkov Group Title

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 Group Title

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39. hartnn Group Title

are both of you doing same thing ?

40. klimenkov Group Title

Yes.

41. klimenkov Group Title

Can you write a table for $$a^2+b^2$$ ? With probabilities?

42. hartnn Group Title

it will also contain 10 columns,right ? trying....

43. klimenkov Group Title

Right.

44. sirm3d Group Title

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 Group Title

If you want to get it, better to draw the whole table for $$a^2+b^2$$.

46. hartnn Group Title

why is the answer drastically different from 2% ?

47. Zarkon Group Title

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 Group Title

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 Group Title

Answer is $$\frac{14}{100}=0.14$$.

50. sirm3d Group Title

i say it's 18/100, same as @Zarkon

51. hartnn Group Title

why is the answer drastically different from 2% ? how is that table formed ? @Zarkon ?

52. sirm3d Group Title

the 2% only counted 5,5 and 0,0.

53. klimenkov Group Title

@sirm3d please write all the pairs.

54. Zarkon Group Title

I used Octave

55. Zarkon Group Title

for i=1:10;for j=1:10;A(i,j)=mod(i^2+j^2,10);endfor;endfor

56. Zarkon Group Title

>> 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 Group Title

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 Group Title

also why are we only considering numbers till 100, will it make any difference, if we consider 1000 numbers ?

59. klimenkov Group Title

Sorry, my fault. Really - $$\frac{18}{100}$$

60. hartnn Group Title

i got now, how we got 18 pairs for 100....

61. sirm3d Group Title

finally, we are all in agreement. hurrah.

62. hartnn Group Title

why consider only 100? will it make any difference, if we consider 1000 numbers ?

63. hartnn Group Title

because if we consider only 10 numbers, we get different answer....

64. hartnn Group Title

0.2

65. sirm3d Group Title

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 Group Title

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 Group Title

aah! okk, got that....

68. mahmit2012 Group Title

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69. mahmit2012 Group Title

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70. mahmit2012 Group Title

0.18 is not true. 0.02 is true. Be sure.

71. hartnn Group Title

where's the fault in our discussion ?

72. sirm3d Group Title

@mahmit2012 what about the pairs 1,7 and 1,3? are they not solutions too? they were not counted in your argument.

73. sirm3d Group Title

@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 Group Title

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75. mahmit2012 Group Title

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