Question

Wilma, Xavier, Yaska and Zakir are four young friends, who have a passion for integers. One day, each of them selects one inWilma, 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?

• A. 0
• B. 6
• C. 4
• D. 1
• E. 9

Answer 1

The Correct Option is B

To determine the possible value of \( W^2 + X^2 + Y^2 + Z^2 \), we first need to analyze the given inequalities involving the integers \( W, X, Y, \) and \( Z \).

1. We are given the following inequalities:
   • \( W^4 + X^3 + Y^2 + Z \leq 4 \)
   • \( X^3 + Z \geq 2 \)
   • \( W^4 + Y^2 \leq 2 \)
   • \( Y^2 + Z \geq 3 \)
2. Analyzing \( W, X, Y, \) and \( Z \):
   • \( X \) and \( Z \) must be positive integers.
   • \( Y \) is a negative integer.
   • \( W \) can be a negative integer, zero, or a positive integer.
3. First, consider the combination of these inequalities:
   • From \( X^3 + Z \geq 2 \), we can conclude that since both \( X \) and \( Z \) are positive, the smallest values can be \( X = 1, Z = 1 \).
   • From \( Y^2 + Z \geq 3 \) and substituting \( Z = 1 \), we get \( Y^2 + 1 \geq 3 \) which gives \( Y^2 \geq 2 \). Therefore, \( Y \) can be one of \(-2, -3, \text{or lower}\).
4. Finding a valid solution set:
   • Let's consider \( W = 0 \) because it keeps the power terms low.
   • Let the simplest positive values for \( X \) and \( Z \) be 1 each, i.e., \( X = 1, Z = 1 \).
   • Then for \( Y^2 \geq 2 \): if \( Y = -2\), then \( Y^2 = 4 \).
   • Substituting back, check:
     • \( W^4 + X^3 + Y^2 + Z = 0^4 + 1^3 + 4 + 1 = 6 \leq 4 \) does not satisfy.
     • Need to adjust \( W \) to positive to reduce \( Y \).
     • Trying \( W = 1\):
     • Value \( W^4 = 1^4 = 1\).
     • Recalculate \( W^4 + X^3 + Y^2 + Z = 1 + 1 + 4 + 1 = 7 \) incorrect.
     • Consider \( Z = 2 \to Y = -1\) then.
     • This now balances with \( Y^2 + Z \geq 3 \) and \( X^3 = 1\).
5. Thus, the valid set could be \( W = 0, X = 1, Y = -1, Z = 2 \)
6. For this set:
   • \( W^2 + X^2 + Y^2 + Z^2 = 0^2 + 1^2 + 1^2 + 2^2 = 0 + 1 + 1 + 4 = 6. \)

Therefore, the possible value that \( W^2 + X^2 + Y^2 + Z^2 \) evaluates to is 6.

Answer 2

To solve the given problem, let's analyze the constraints step by step:

1. \(W^4 + X^3 + Y^2 + Z \leq 4\)

2. \(X^3 + Z \geq 2\)

3. \(W^4 + Y^2 \leq 2\)

4. \(Y^2 + Z \geq 3\) 

Given that Xavier and Zakir picked positive integers and Yaska picked a negative integer, Wilma's integer can be negative, zero or positive. We must identify reasonable values for \(W\), \(X\), \(Y\), and \(Z\) that fulfill all constraints and then evaluate \(W^2 + X^2 + Y^2 + Z^2\).

Step 1: Assume \(Y = -1\). Then \(Y^2 = 1\).

Step 2: From constraint 4, \(1 + Z \geq 3 \Rightarrow Z \geq 2\), thus the lowest integer for \(Z\) satisfying this is \(Z = 2\).

Step 3: With \(Z = 2\), constraint 2 becomes \(X^3 + 2 \geq 2 \Rightarrow X^3 \geq 0\). Since \(X\) is a positive integer, \(X = 1\) is the smallest possibility.

Step 4: Substitute \(X = 1\), \(Y = -1\), and \(Z = 2\) into constraint 1: \(W^4 + 1 + 1 + 2 \leq 4 \Rightarrow W^4 \leq 0\). Therefore, \(W = 0\).

Given \(W = 0\), \(X = 1\), \(Y = -1\), \(Z = 2\), calculate \(W^2 + X^2 + Y^2 + Z^2 = 0^2 + 1^2 + (-1)^2 + 2^2 = 0 + 1 + 1 + 4 = 6\).

The possible value is thus: 6.