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Fdambro294
Ah, it’s (c - z)^2 ?

Posted from my mobile device

Yes, it is. Edited. Thank you!
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IMO D

There are 3 pairs of numbers -: a,x ; b,y ; c,z .
For the sum to be 1 as in the equation given , possible values of sqr(a-x) ,sqr(b-y) and sqr(c-z) could be 1,0,0 or 0,1,0 or 0,0,1.

For the time being lets suppose sqr(a-x) =1 and sqr(b-y) and sqr(c-z) are 0.
Possible values of a are 4,3,2,1 .
When a=4 ; y=3 -> 1 possible combination
when a=3 ; y = 2 or 4 -> so 2 possible combinations ....(as it is square of the difference(a-x) , the value is 1)
when a=2 ; y= 1 or 3 -> 2 possible combinations
when a=1 ; y= 2 or 0 -> 2 possible combinations
when a=0 ; y=1 -> 1 possible combination .

So we have a total 8 possible combinations for (a,x)

Now as (b-y) and (c-z) need to have value 0 , y and z will take the same value as b and c respectively .
What values can 'b' take -: 4,3,2,1,0 -> 5 values ; Corresponding to each of the 5 values of 'b' y will have only 1 value ( b and y need to be same to give sqr(b-y) = 0) -> total possible combinations -> 5*1 =5

What values can 'c' take -: 4,3,2,1,0 -> 5 values ; Corresponding to each of the 5 values of 'c' z will have only 1 value ( c and z need to be same to give sqr(c-z) = 0) -> total possible combinations -> 5*1=5

Total possible combinations = 8*5*5 = 200 .

Now in this example we chose (a-x) to have value 1 and (b-y) and (c-z) to have value 0 . Similarly , 2 other scenarios are also possible where (b-y) and (c-z) have values 1 . So total 3 different scenarios are possible.

so in total the number of possible non negative values of (a,x,b,y,c,z) are 200 * 3 = 600
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How many non-negative integer solutions, a, b, c, x, y and z, with all six integers less than or equal to 4, does the equation: \((a-x)^{2}+(b-y)^{2}+(c-z)^{2}=1\) have?

(A) 300
(B) 400
(C) 500
(D) 600
(E) 700

Using the given integers 0, 1, 2, 3, 4:
Number of ways to yield an absolute difference of 1 = 8
Number of ways to yield a difference of 0 = 5

Let:
O = an absolute difference of 1
Z = a difference of 0

One way for the three terms to yield a sum of 1:
O+Z+Z

Number of options for O = 8
Number of options for the first Z = 5
Number of options for the second Z = 5
To combine these options, we multiply:
8*5*5 = 200

Since O can be the first term, the second term, or the third term -- a total of 3 options -- we multiply by 3:
200*3 = 600

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