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

- hartnn

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

- schrodinger

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- hartnn

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

- anonymous

|dw:1351968021756:dw|

- hartnn

[0] means ?

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## More answers

- anonymous

|dw:1351968165513:dw|

- anonymous

|dw:1351968262315:dw|

- anonymous

Is it clear?

- hartnn

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

- hartnn

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

- anonymous

|dw:1351968633462:dw|

- anonymous

|dw:1351968728104:dw|

- anonymous

|dw:1351968805888:dw|

- hartnn

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

- anonymous

|dw:1351968902151:dw|

- hartnn

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

- anonymous

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

- hartnn

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

- anonymous

|dw:1351969216109:dw|

- hartnn

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

- anonymous

|dw:1351969399186:dw|

- hartnn

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

- hartnn

@mukushla any other approach you can think of?

- sirm3d

for some reason, i only have p = 0.10

- hartnn

me too, but thats incorrect.

- sirm3d

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

- hartnn

plz share.

- 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

sorry,didn't get u...

- sirm3d

still polishing my argument.

- 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

1/6

- sirm3d

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

- klimenkov

|dw:1352033013459:dw|

- hartnn

no, i don't have correct answer

- hartnn

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

- sirm3d

i got it, 0.18

- klimenkov

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

- 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

|dw:1352033270108:dw|

- hartnn

are both of you doing same thing ?

- klimenkov

Yes.

- klimenkov

Can you write a table for \(a^2+b^2\) ? With probabilities?

- hartnn

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

- klimenkov

Right.

- 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

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

- hartnn

why is the answer drastically different from 2% ?

- 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

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

Answer is \(\frac{14}{100}=0.14\).

- sirm3d

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

- hartnn

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

- sirm3d

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

- klimenkov

@sirm3d please write all the pairs.

- Zarkon

I used Octave

- Zarkon

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

- 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

here's the other nine
(5,5)
(6,2), (6,8)
(7,1), (7,9)
(8,4), (8,6)
(9,3), (9,7)

- hartnn

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

- klimenkov

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

- hartnn

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

- sirm3d

finally, we are all in agreement. hurrah.

- hartnn

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

- hartnn

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

- hartnn

0.2

- 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

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

aah! okk, got that....

- anonymous

|dw:1352043506996:dw|

- anonymous

|dw:1352043883988:dw|

- anonymous

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

- hartnn

where's the fault in our discussion ?

- sirm3d

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

- 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

|dw:1352045197951:dw|

- anonymous

|dw:1352045432135:dw|

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