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