Question: Wilma, Xavier, Yaska and Zakir are four young friends, who have a passion for integers. One day, each of them selects one integer and writes it on a wall. The writing on the wall shows that Xavier and Zakir picked positive integers, Yaska picked a negative one, while Wilma’s integer is either negative, zero or positive. If their integers are denoted by the first letters of their respective names, the following is true:
W4 + X3 + Y2 + Z ≤ 4
X3 + Z ≥ 2
W4 + Y2 ≤ 2
Y2 + Z ≥ 3
Given the above, which of these can W2 + X2 + Y2 + Z2 possibly evaluate to?
Explanation:
Given that X, Z are positive Y is negative and W can be either positive or zero or negative.
The given conditions are:
W4 + X3 + Y2 + Z ≤ 4
X3 + Z ≥ 2
W4 + Y2 ≤ 2
Y2 + Z ≥ 3
For W4 + Y2 ≤ 2. Since Y is negative but Y2 is always positive and must be less than 2 because W4 is a nonnegative value. Hence Y = -1 is the only possibility. For W this can take any value among -1, 0, 1.
Y2 + Z ≥ 3. Since Y = -1, Z must be at least equal to 2 so the value of Y2 + Z ≥ 3 is greater than 2.
X is a positive value and must at least be equal to 1.
The condition: W2 + X2 + Y2 + Z2 here has all the independent values:
X2 , Y2 , Z2 , W2 are non-negative.
W4 + X3 + Y2 + Z ≤ 4:
Since the value of Z is at least equal to 2 the value of Y2 is equal to 1.
Since X is a positive number in order to have the condition of W4 + X3 + Y2 + Z ≤ 4 satisfied. The value of Z must be the minimum possible so that X3 + Y2 + Z to have a value equal to 4 when X takes the minimum possible positive value equal to 1.
Hence X must be 1. W must be equal to 0 so that:
W4 + X3 + Y2 + Z ≤ 4. The sum = (0 + 1 + 1 + 2) = 4. The only possible case.
The value of W2 + X2 + Y2 + Z2 = (0 + 1 + 1 + 4) = 6.