hartnn
  • hartnn
If a,b are integers, what is the probability that \[a^2+b^2 \] is divisible by 10 ?
Probability
jamiebookeater
  • jamiebookeater
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hartnn
  • hartnn
\((a^2+b^2)/10\) is integer
anonymous
  • anonymous
|dw:1351968021756:dw|
hartnn
  • hartnn
[0] means ?

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anonymous
  • anonymous
|dw:1351968165513:dw|
anonymous
  • anonymous
|dw:1351968262315:dw|
anonymous
  • anonymous
Is it clear?
hartnn
  • hartnn
trying to understand...seeing this type of approach for first time...
hartnn
  • hartnn
how did u get a,b belongs t0 {......}mod 5 ?
anonymous
  • anonymous
|dw:1351968633462:dw|
anonymous
  • anonymous
|dw:1351968728104:dw|
anonymous
  • anonymous
|dw:1351968805888:dw|
hartnn
  • hartnn
i think i am getting this a bit.....need to think
anonymous
  • anonymous
|dw:1351968902151:dw|
hartnn
  • hartnn
can u expalin more on why there is just one option ?
anonymous
  • anonymous
a and b are independent so P(a,b)=P(a).P(b)
hartnn
  • hartnn
i think i got other parts, except that there is only 1 option
anonymous
  • anonymous
|dw:1351969216109:dw|
hartnn
  • hartnn
ohh....right...i'll go through this again. thanks!
anonymous
  • anonymous
|dw:1351969399186:dw|
hartnn
  • hartnn
Also, any alternative approach will be appreciated....
hartnn
  • hartnn
@mukushla any other approach you can think of?
sirm3d
  • sirm3d
for some reason, i only have p = 0.10
hartnn
  • hartnn
me too, but thats incorrect.
sirm3d
  • sirm3d
i found my error. it is as what mahmit gave. but i have a different proof.
hartnn
  • hartnn
plz share.
sirm3d
  • sirm3d
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.
hartnn
  • hartnn
sorry,didn't get u...
sirm3d
  • sirm3d
still polishing my argument.
klimenkov
  • klimenkov
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.
hartnn
  • hartnn
1/6
sirm3d
  • sirm3d
oops, i got 0.18 this time. hahaha. do you have the correct answer?
klimenkov
  • klimenkov
|dw:1352033013459:dw|
hartnn
  • hartnn
no, i don't have correct answer
hartnn
  • hartnn
for a^2, probability is 1/6, i get that what about a^2+b^2
sirm3d
  • sirm3d
i got it, 0.18
klimenkov
  • klimenkov
Now what is probability:|dw:1352033286386:dw|
klimenkov
  • klimenkov
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?
sirm3d
  • sirm3d
|dw:1352033270108:dw|
hartnn
  • hartnn
are both of you doing same thing ?
klimenkov
  • klimenkov
Yes.
klimenkov
  • klimenkov
Can you write a table for \(a^2+b^2\) ? With probabilities?
hartnn
  • hartnn
it will also contain 10 columns,right ? trying....
klimenkov
  • klimenkov
Right.
sirm3d
  • sirm3d
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
klimenkov
  • klimenkov
If you want to get it, better to draw the whole table for \(a^2+b^2\).
hartnn
  • hartnn
why is the answer drastically different from 2% ?
Zarkon
  • Zarkon
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
sirm3d
  • sirm3d
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)
klimenkov
  • klimenkov
Answer is \(\frac{14}{100}=0.14\).
sirm3d
  • sirm3d
i say it's 18/100, same as @Zarkon
hartnn
  • hartnn
why is the answer drastically different from 2% ? how is that table formed ? @Zarkon ?
sirm3d
  • sirm3d
the 2% only counted 5,5 and 0,0.
klimenkov
  • klimenkov
@sirm3d please write all the pairs.
Zarkon
  • Zarkon
I used Octave
Zarkon
  • Zarkon
for i=1:10;for j=1:10;A(i,j)=mod(i^2+j^2,10);endfor;endfor
Zarkon
  • Zarkon
>> 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
sirm3d
  • sirm3d
here's the other nine (5,5) (6,2), (6,8) (7,1), (7,9) (8,4), (8,6) (9,3), (9,7)
hartnn
  • hartnn
also why are we only considering numbers till 100, will it make any difference, if we consider 1000 numbers ?
klimenkov
  • klimenkov
Sorry, my fault. Really - \(\frac{18}{100}\)
hartnn
  • hartnn
i got now, how we got 18 pairs for 100....
sirm3d
  • sirm3d
finally, we are all in agreement. hurrah.
hartnn
  • hartnn
why consider only 100? will it make any difference, if we consider 1000 numbers ?
hartnn
  • hartnn
because if we consider only 10 numbers, we get different answer....
hartnn
  • hartnn
0.2
sirm3d
  • sirm3d
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?
Zarkon
  • Zarkon
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
hartnn
  • hartnn
aah! okk, got that....
anonymous
  • anonymous
|dw:1352043506996:dw|
anonymous
  • anonymous
|dw:1352043883988:dw|
anonymous
  • anonymous
0.18 is not true. 0.02 is true. Be sure.
hartnn
  • hartnn
where's the fault in our discussion ?
sirm3d
  • sirm3d
@mahmit2012 what about the pairs 1,7 and 1,3? are they not solutions too? they were not counted in your argument.
sirm3d
  • sirm3d
@mahmit2012 you only saw 5,5 and 0,0 in your proof. that explains your 2/100=0.02. what about 1,7 pair?
anonymous
  • anonymous
|dw:1352045197951:dw|
anonymous
  • anonymous
|dw:1352045432135:dw|

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