If √(1 + 27/169) = 1 + N/13, determine the integer N that satisfies the identity.

Introduction / Context: This problem checks algebraic manipulation and recognition of perfect squares under radicals. Converting to a single fraction inside the square root simplifies the comparison.

Given Data / Assumptions:

• Equation: √(1 + 27/169) = 1 + N/13.
• N is an integer.

Concept / Approach: Evaluate the left-hand side exactly by combining fractions. Compare the simplified exact value with the right-hand side to find N.

Step-by-Step Solution: Compute inside radical: 1 + 27/169 = (169/169) + (27/169) = 196/169. √(196/169) = √196 / √169 = 14 / 13. Set 14/13 = 1 + N/13. Then N/13 = 14/13 − 1 = 1/13 ⇒ N = 1.

Verification / Alternative check: Substitute N = 1 into the RHS: 1 + 1/13 = 14/13, which matches the LHS exactly.

Why Other Options Are Wrong: 3, 5, and 7 give 1 + N/13 values greater than 14/13 and thus do not equal √(196/169).

Common Pitfalls: Forgetting that √(a/b) = √a / √b, or approximating decimals instead of using exact fractions can create rounding errors. Here, exact arithmetic is clean and quick.

Final Answer: 1