Fill in the three relational operators (from left to right) between the letters in the chain K _ L _ M _ N so that the statement guarantees the relation N < K. Choose the only option that always makes N strictly less than K.

Introduction / Context:
This problem is a symbolic-inequality chain question. We must place three comparison operators between K, L, M, and N so that the overall chain necessarily implies N < K regardless of the actual equalities allowed by the symbols. The goal is to test careful reasoning about transitive relations (>, <, =) and certainty versus possibility.

Given Data / Assumptions:

• We have the pattern: K _ L _ M _ N (three blanks).
• Answer choices provide ordered triples of comparison symbols to insert left to right.
• We must ensure that, after inserting, the statement guarantees N < K (i.e., K > N) with no ambiguity.

Concept / Approach:
The key is to interpret each candidate chain and deduce whether K must be greater than N. Transitivity of inequalities applies: if K ≥ L and L = M and M > N, then K ≥ M and thus K > N. Conversely, if any choice allows N ≥ K in some assignment, it cannot “definitely” make N < K.

Step-by-Step Solution:

Verification / Alternative check:
Construct counterexamples for (b)–(d) by picking simple numbers (e.g., set M = L = 5, then adjust K and N according to each chain) to see N < K fails. For (a), any assignment obeying L = M and M > N makes K ≥ M > N, which forces K > N.

Why Other Options Are Wrong:

• (b) Allows N = M to exceed K (since K ≤ L < M); hence N > K can occur.
• (c) Allows equality K = L = M while N is larger; then N > K.
• (d) Only ensures K > M and N > M; it does not impose any order between K and N.

Common Pitfalls:

• Assuming “> M” on both sides implies K > N (it does not).
• Ignoring that “≥” permits equality, which must still lead to strict K > N to satisfy the requirement.
• Overlooking that “definitely holds true” means no counterexample may exist.

Final Answer:
≥ , = , >