Find the preimage under a constant +10 shift: AEFJ : KOPT :: ? : QUVZ

Difficulty: Easy

Correct Answer: GKLP

Explanation:


Introduction / Context:
Here the coding rule is a constant Caesar shift of +10. You must reverse the shift to find which input would encode to the given output, practicing inverse reasoning on uniform ciphers.


Given Data / Assumptions:

  • AEFJ → KOPT shows +10 for each letter.
  • We need ? → QUVZ using the same +10 rule.
  • Use modular wrap-around with A=1 … Z=26.


Concept / Approach:
If encoded = original + 10, then original = encoded − 10. Subtract 10 from each of Q, U, V, Z to obtain the preimage.


Step-by-Step Solution:

Q(17) − 10 = G(7).U(21) − 10 = K(11).V(22) − 10 = L(12).Z(26) − 10 = P(16).Thus the input word is GKLP.


Verification / Alternative check:
Apply +10 to GKLP → QUVZ, recovering the target perfectly. This confirms the uniform shift assumption and implementation.


Why Other Options Are Wrong:

  • GLKP/HKLP/HKQL: Each has at least one letter that fails the exact −10 back-shift from Q, U, V, Z.


Common Pitfalls:
Confusing addition and subtraction or miscounting wrap-around from A or Z. Carefully compute each letter independently.


Final Answer:
GKLP

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