x , y and z are integers such that − 1 0 ≤ x , y , z ≤ 1 0 . How many ordered triplets ( x , y , z ) satisfy x 3 + y 3 + z 3 = 3 x y z ?
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This problem was done quite badly. Without the factorization, students were unable to explain that there were only these 2 cases. Most complained that there were 21 solutions arising from x = y = z . Students also had problems applying the Principle of Inclusion and Exclusion , and several students didn't account for the overlapped solution in Case 1 and Case 2.
Finally, Mahbubul shows a straightforward way to calculate solutions to x + y + z = 0 under the restriction that − 1 0 ≤ x , y , z ≤ 1 0 . This method was used in a similar problem in the past.
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Sir this should be under combinatorics.
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Sometimes, you just can't categorize great problems.
Ya, I agee with you, and I think the problem I is little i er rated for having 24% solvers.
We know that x 3 + y 3 + z 3 − 3 x y z = ( x + y + z ) ( x 2 + y 2 + z 2 − x y − y z − z x ) So x 3 + y 3 + z 2 = 3 x y z is possible when:
i) x + y + z = 0 . In this case z = − ( x + y ) . Hence z is uniquely determined for each ordered pair ( x , y ) . Since − 1 0 ≤ z ≤ 1 0 , we get − 1 0 ≤ x + y ≤ 1 0 . If we choose x = − 1 0 then we have 11 choices for y i.e., 0 , 1 , 2 . . . , 1 0 . Similarly, for x = − 9 , we have 12 choices for y i.e., − 1 , 0 , 1 , 2 . . . , 1 0 . The number of choices for y goes on increasing by 1 from 1 1 ( when x = − 1 0 ) ) to 2 0 (when x = − 1 ). When − 1 0 ≤ x ≤ − 1 , the number of ordered pairs ( x , y ) is 1 1 + 1 2 + . . . + 2 0 = 1 5 5 . Same is the number of pairs when 1 ≤ x ≤ 1 0 and can be verified easily. When x = 0 , we have all the 2 1 choices for y . So total number of pairs in this case is 2 ⋅ 1 5 5 + 2 1 = 3 3 1 .
ii) x 2 + y 2 + z 2 − x y − y z − z x = 0 ⇒ 2 1 [ ( x − y ) 2 + ( y − z ) 2 + ( z − x ) 2 ] = 0 ⇒ ( x − y ) 2 + ( y − z ) 2 + ( z − x ) 2 = 0 ⇒ ( x − y ) 2 = ( y − z ) 2 = ( z − x ) 2 = 0 ⇒ x − y = y − z = z − x = 0 ⇒ x = y = z . In this case we are just looking for number of ways of selecting one element from set { − 1 0 , − 9 , − 8 . . . , 9 , 1 0 } which is 2 1 . But we have also counted the pair ( 0 , 0 , 0 ) which already has been counted in first case.
So, total number of ordered pairs ( x , y , z ) is 3 3 1 + 2 1 − 1 = 3 5 1 .
According to the key technique, we can see there are: x+y+z=0 or x^2+y^2+z^2-xy-yz-xz=0. If x+y+z=0, we can count that there are 331 ordered triplets. If ^2+y^2+z^2-xy-yz-xz=0, there are 1/2[(x+y)^2+(y+z)^2+(x+z)^2]=0, x=y=z, so there are 10-(-10)+1=21 ordered triplets. Besides, (0,0,0) appears in both of these two situations and is counted twice. So the answer equals to 331+21-1=351.
x^3 + y^3 +z^3 = 3xyz x^3 + y^3 +z^3 - 3xyz= 0 x^3 + (x^2)y +(x^2)z + y^3 +(y^2)x +(y^2)z + z^3 +(z^2)x +(z^2)y - (x^2)y - (y^2)x - xyz - (x^2)z - (z^2)x- xyz - (y^2)z- (z^2)y - xyz = 0 (x+y+z)(x^2 + y^2 + z^2 - xy- yz- xz) = 0 x+ y +z = 0 or x^2 + y^2 + z^2 - xy- yz- xz = 0 For the former solution, if x=0, y= -z and we have (y,z) = (10,-10), (9,-9) ....(-9,9), (-10,10) - a total of 21 solutions. For x modulo = 1 (i.e. x= +/- 1), we have 20 solutions for y and z. For x modulo 2, we have 19 solutions and so on. For x modulo 10, we have 11 solutions. Hence it can be seen that for x+y+z = 0, we have a total of 2(11+12+13+...+19+20)+ 21 = 331 solutions,
As for the latter solution as stated in the factorised form, we let x^2 = xy, y^2 = xz and z^2 = yz. The first equation tells us that x= y and the last one tells us that y=z, hence x=y=z. We therefore have (x,y,z) = (-10,-10,-10) all the way to (10,10,10) which gives us 21 solutions here. But do note that (0,0,0) is counted in this scenario as well as if x+y+z = 0.
Hence there are 331+21-1 = 351 solutions altogether.
Rearranging and factoring the given equation,
2 1 ( x + y + z ) [ ( x − y ) 2 + ( x − z ) 2 + ( y − z ) 2 ] = 0
If the latter is equal to 0,
x=y=z
We have 10-(-10)+1=21 solutions in total.
If the former is equal to 0,
Rearranging x + y + z = 0 ,
y + z = − x
Claim : there are 21-|x| solutions for each − 1 0 ≤ x ≤ 1 0 .
Proof: for − 1 0 ≤ x ≤ 0 we have
(y,z) = (-x-10,10), (-x-9,9),...,(10.-x-10)
With x+21 = 21-|x| solutions.
for 1 ≤ x ≤ 1 0 we have
(y,z) = (10-x,-10),(9-x,-9),...,(-10,10-x)
With 21-x = 21-|x| solutions.
Therefore, the number of solutions in this case is
11+12+13+14+...+20+21+20+...+14+13+12+11=331
Note that only the triplet (0,0,0) appeared in both cases.
Therefore, the total number of triplets that satisfy the conditions is 331+21-1=351.
If x 3 + y 3 + z 3 − 3 x y z = 0 , then we factor this into ( x + y + z ) ( x 2 + y 2 + z 2 − x y − x z − y z ) = 0 . Either x + y + z = 0 or ( x − y ) 2 + ( y − z ) 2 + ( z − x ) 2 = 0 , which is equivalent to x = y = z . If x = y = z , we obviously have 21 solutions. If x + y + z = 0 , we have a little casework to do. Since ( 0 , 0 , 0 ) is already covered, we can do casework on how many 0's there are. Case 1: One of x, y, and z are 0. Then WLOG x = 0 (multiply by 3). We then get y = − z , so there are 20 possibilities, which gives a total of 60 in this case. Case 2: None of x, y, and z are 0. In this case, we can WLOG x > 0, y < 0, and z < 0, so we need to multiply by 6 in the end. Then x = ( − y ) + ( − z ) where x, -y, and -z are positive. This is just the number of ways to have 2 positive integers add up to less than or equal to 10. This is 10 choose 2 or 45. Thus this case gives 45*6 = 270. In total, this gives 2 1 + 6 0 + 2 7 0 = 3 5 1 ways.
we can divide into 2 case
case 1 if x=y=z so x 3 + y 3 + z 3 = 3 x y z that satisfy too all integer but because − 1 0 ≤ x , y , z ≤ 1 0 so there is only 21 triplets
case 2 we know x 3 + y 3 + z 3 = ( x + y + z ) ( x 2 + y 2 + z 2 ) − ( x y + x z + y z ) ( x + y + z ) + 3 x y z if x + y + z = 0 we can conclude that x 3 + y 3 + z 3 = 3 x y z set: a = x + 1 0 b = y + 1 0 c = z + 1 0
we get 0 ≤ a , b , c ≤ 2 0 the problems change become a + b + c = 3 0
use inklusi eksklusi to solve so we get, ( 2 3 2 ) − 3 ( 2 1 1 ) = 3 3 1
in case 2 we get 331 triplets
but in case 1 and case 2 we found same solution there is ( 0 , 0 , 0 )
Total of them is 331+21-1=351
for x+y+z=0, number = 6* (1+……+10) +1
( x^3+y^3+z^3 = 3xyz )\ ( \Leftrightarrow (x+y+z)( (x-y)^2 +(y-z)^2 + (z-x)^2) = 0 )\ Either the right bracket is 0 (i.e. ( x=y=z )\ ), or ( x+y+z )\ After a bit a calculation (which is too messy to type), the first case has 20 triples and the second has 331, which makes a total of 351.
From the factorization, 0 = x 3 + y 3 + z 3 − 3 x y z = ( x + y + z ) ( x 2 + y 2 + z 2 − x y − y z − z x ) = 2 1 ( x + y + z ) ( ( x − y ) 2 + ( y − z ) 2 + ( z − x ) 2 ) , we have the following cases:
Case 1: x = y = z . There are 1 0 − ( − 1 0 ) + 1 = 2 1 such triplets.
Case 2: x + y + z = 0 . For z = k , we want x + y = − k subject to − 1 0 ≤ x , y ≤ 1 0 . Consider the square grid where − 1 0 ≤ x , y ≤ 1 0 , and the line x + y = − k intersecting. It follows that there are 2 1 − ∣ k ∣ possible pairs for ( x , y ) . Now, taking the sum from k = − 1 0 to k = 1 0 , there are a total of 1 1 + 1 2 + 1 3 + … + 2 1 + … + 1 1 = 3 3 1 such triplets.
The only triplet that is double counted (in both cases) is ( 0 , 0 , 0 ) . Thus there are 2 1 + 3 3 1 − 1 = 3 5 1 triplets that satisfy the equation.
Sir i cant understand what the square grid you are talking about
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Firstly, we can factorize x 3 + y 3 + z 3 − 3 x y z into 2 1 ( x + y + z ) ( ( x − y ) 2 + ( y − z ) 2 + ( z − x ) 2 ) . So if, x 3 + y 3 + z 3 = 3 x y z then 2 1 ( x + y + z ) ( ( x − y ) 2 + ( y − z ) 2 + ( z − x ) 2 ) = 0 ⇒ ( x + y + z ) = 0 or ( x − y ) 2 + ( y − z ) 2 + ( z − x ) 2 ) = 0 ⇒ ( x + y + z ) = 0 or x = y = z
Case 1: The last condition will be valid for 21 values that is: − 1 0 ≤ x = y = z ≤ 1 0
Case 2: The challenge is to count number of solutions of first term. Lets use the substitution X = 1 0 + x , Y = 1 0 + y , Z = 1 0 + z . and count number of solution of the equation X + Y + Z = 3 0 where 0 ≤ X , Y , Z ≤ 2 0 . If we did not have any such constraint then number of solution by the stars and stripes would be: ( 2 3 2 ) = 4 9 6 . If one of the variables was over 21 then number of solution would be: 3* ( 2 1 1 ) = 1 6 5 Note, we cannot have more than one variable over 21, since the sum is only 30. Hence our desired count applying Inclusion-Exclusion principle is: 4 9 6 − 1 6 5 = 3 3 1
Both term will be valid for only 1 value that is: x = y = z = 0 .
Hence again applying inclusion exclusion principle, the total number of solution is: 3 3 1 + 2 1 − 1 = 3 5 1
[Edits for clarity - Calvin]