It's Only Algebra

Algebra Level 5

{ a + b f g = 0 b + c g h = 0 d + e i j = 0 b c + d e g + h i + j = 0 \begin{cases}a+b-f-g=0\\ b+c-g-h=0\\ d+e-i-j=0\\ b-c+d-e-g+h-i+j=0\end{cases}

a , b , c , d , e , f , g , h , i , j a,b,c,d,e,f,g,h,i, j are all numbers, each of which can only be either 0 or 1, and they satisfy the system of equations above.

Find the total number of distinct, possible values of a + b + c + d + e + f + g + h + i + j a+b+c+d+e+f+g+h+i+j .

2 3 4 5 6 7 8 9

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1 solution

Michael Mendrin
Jan 28, 2017

This is a rip-off from Calvin Lin's problem, which he gives a full explanation of the math behind this

Summing Up Certain Roots of Unity

The only possible totals are ( 0 , 2 , 4 , 5 , 6 , 8 , 10 ) (0, 2, 4, 5, 6, 8, 10) , or 7 7 in all

More direct solutions are welcome, it'd be interesting to see the different ways of solving this

Edit: To simplify solving this problem, consider the transform a = 1 a a=1-a' , and do the same for all of the variables. The 1 -1 drops out in all four equations, and we divide everything by 1 -1 to end up where we started from. So that means that given any solution, there exist a complement solution where for every 1 1 , there is now a 0 0 , and vice versa. So, for example, if we proved that there exists no solution such that the total is 3 3 , then there exists no solution such that the total is 7 7 . It's very easy to show that there can't be any solutions such that the total is 1 1 , so that means there is no solution such that the total is 9 9 .

As an example of where the total can be 5 5 for a solution (which means that its complement is also a solution), we have, using the exact same four equations as displayed, but with variables replaced by integers, we have, letting ( a , c , d , g , j ) = ( 1 , 1 , 1 , 1 , 1 ) (a, c, d, g, j)=(1,1,1,1,1)

{ 1 + 0 0 1 = 0 0 + 1 1 0 = 0 1 + 0 0 1 = 0 0 1 + 1 0 1 + 0 0 + 1 = 0 \begin{cases}1+0-0-1=0\\ 0+1-1-0=0\\ 1+0-0-1=0\\ 0-1+1-0-1+0-0+1=0\end{cases}

It is not hard to show that the total can be any of the even numbers, ( 0 , 2 , 4 , 6 , 8 , 10 ) (0, 2, 4, 6, 8, 10) , after which we're left with proving that the total cannot be 3 3 , which implies it cannot be 7 7 either. To prove the case for even totals, the four equations can be rearranged as follows

{ ( a f ) + ( b g ) = 0 ( b g ) + ( c h ) = 0 ( d i ) + ( e j ) = 0 ( b g ) ( c h ) + ( d i ) ( e j ) = 0 \begin{cases}(a-f)+(b-g)=0\\ (b-g)+(c-h)=0\\ (d-i)+(e-j)=0\\ (b-g)-(c-h)+(d-i)-(e-j)=0\end{cases}

and from this we can show that

( a f ) = ( b g ) = ( c h ) = ( d i ) = ( e j ) (a-f)=-(b-g)=(c-h)=(d-i)=-(e-j)

which means if we set this value at 0 0 , we can flip any pair of variables arbitrarily to create another solution, thus either raising or lowering the total by 2 2 s. Hence, the total can be of any even number from 0 0 to 10 10 .

Finally, we'll look at the special case where the total is 3 3 . Consider the original set of equations, and start with assuming that all of the variables are 0 0 which is a solution.

{ a + b f g = 0 b + c g h = 0 d + e i j = 0 b c + d e g + h i + j = 0 \begin{cases}a+b-f-g=0\\ b+c-g-h=0\\ d+e-i-j=0\\ b-c+d-e-g+h-i+j=0\end{cases}

For the total to be 3 3 , we have to flip 3 3 of the variables from 0 0 to 1 1 without affecting the sum of of each equation which is 0 0 . Clearly, it is not possible that an odd number of variables can be flipped in any one equation---either none or two can be flipped. A little analysis of the equations given shows this to be impossible. For example, if a a and f f are flipped, since there are no instances of a a and f f elsewhere, then the 3rd flipped variable will change the sum of 0 0 of any equation it appears.

Hence, no solutions exist for which the total is 3 3 , and consequently 7 7 as well. And so we have the final list of possible totals ( 0 , 2 , 4 , 5 , 6 , 8 , 10 ) (0, 2, 4, 5, 6, 8, 10) , or 7 distinct totals.

@Michael Mendrin , i am not getting total sum as 5 :- (( Could u plz tell the value of all variables to get it.

sanyam goel - 4 years, 4 months ago

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( a , c , d , g , i ) = ( 1 , 1 , 1 , 1 , 1 ) (a, c, d, g, i)=(1, 1, 1, 1, 1)

I had made a slight mistake, accidentally transposing a couple of variables. Problem fixed. But this solution is not unique.


Michael Mendrin - 4 years, 4 months ago

See my edited response to you. Thanks for pointing out an error.

Michael Mendrin - 4 years, 4 months ago

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