Given properly implemented equals() and hashCode(), the code prints "x = 1111". Which statement must always be true?\n\nx = 0;\nif (x1.hashCode() != x2.hashCode()) x = x + 1;\nif (x3.equals(x4)) x = x + 10;\nif (!x5.equals(x6)) x = x + 100;\nif (x7.hashCode() == x8.hashCode()) x = x + 1000;\nSystem.out.println("x = " + x);

Introduction / Context:
The problem requires deducing which relationship must hold given four conditional increments produced the total 1111. Each increment corresponds to one Boolean condition being true. With equals() and hashCode() implemented correctly, we use the contract to infer implications.

Given Data / Assumptions:

• Printed value is 1111 → every conditional branch executed (1 + 10 + 100 + 1000).
• Thus: x1.hashCode() != x2.hashCode(), x3.equals(x4), !x5.equals(x6), and x7.hashCode() == x8.hashCode() are all true.
• equals() and hashCode() are consistent with the Java contract.

Concept / Approach:
The equals–hashCode contract guarantees: if a.equals(b) is true, then a.hashCode() == b.hashCode() must be true. The converse is not required. If a.equals(b) is false, there is no guarantee about hash codes (they may be equal or different).

Step-by-Step Solution:

Verification / Alternative check:
Construct sample objects to satisfy all four conditions; only the relation between x3 and x4 is logically forced both in equals and hashCode dimensions.

Why Other Options Are Wrong:

• x2.equals(x1) or x1.equals(x2): would contradict the first condition because equals true implies matching hash codes.
• x5.hashCode() != x6.hashCode(): not implied; unequal objects may still collide.
• x8.equals(x7): not implied by equal hash codes alone.

Common Pitfalls:
Believing unequal objects must have different hash codes; collisions are allowed. Also, thinking equal hash codes force equals() to be true (they do not).

Final Answer:
x3.hashCode() == x4.hashCode()