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CAT 2025Slot 2QAQues & Sol

AlgebraHard

Question

If m and n are integers such that $(m+2n)(2m+n)=27$, then the maximum possible value of $2m-3n$ is

Solution

1. Concept Used

  • Topic: Algebra – Integer Factorization & Linear Systems
  • Formula: $$ (m+2n)(2m+n) = 27, \quad \text{maximize} \quad 2m - 3n $$

2. Calculation

Since (m) and (n) are integers, both factors ((m+2n)) and ((2m+n)) must be integers whose product equals 27.

The complete list of integer factor pairs ((a, b)) such that (a \cdot b = 27) is: $$(1, 27),\ (3, 9),\ (9, 3),\ (27, 1),\ (-1, -27),\ (-3, -9),\ (-9, -3),\ (-27, -1)$$

Now observe a key structural constraint. If we set (a = m + 2n) and (b = 2m + n), then: $$a + b = (m + 2n) + (2m + n) = 3(m + n)$$

This means (a + b) must be divisible by 3. Checking each factor pair:

  • ((1, 27)): sum (= 28) → not divisible by 3 ❌
  • ((3, 9)): sum (= 12) → divisible by 3 ✅
  • ((9, 3)): sum (= 12) → divisible by 3 ✅
  • ((27, 1)): sum (= 28) → not divisible by 3 ❌
  • ((-1, -27)): sum (= -28) → not divisible by 3 ❌
  • ((-3, -9)): sum (= -12) → divisible by 3 ✅
  • ((-9, -3)): sum (= -12) → divisible by 3 ✅
  • ((-27, -1)): sum (= -28) → not divisible by 3 ❌

So the valid pairs are: ((3,9),\ (9,3),\ (-3,-9),\ (-9,-3)).

Case 1: ((a, b) = (3, 9)) $$m + 2n = 3 \quad \text{and} \quad 2m + n = 9$$ Subtracting the first from twice the second: (3m = 15 \Rightarrow m = 5), then (n = -1). $$2m - 3n = 2(5) - 3(-1) = 10 + 3 = 13$$

Case 2: ((a, b) = (9, 3)) $$m + 2n = 9 \quad \text{and} \quad 2m + n = 3$$ Solving: (m = -1,\ n = 5). $$2m - 3n = 2(-1) - 3(5) = -2 - 15 = -17$$

Case 3: ((a, b) = (-3, -9)) $$m + 2n = -3 \quad \text{and} \quad 2m + n = -9$$ Solving: (m = -5,\ n = 1). $$2m - 3n = 2(-5) - 3(1) = -10 - 3 = -13$$

Case 4: ((a, b) = (-9, -3)) $$m + 2n = -9 \quad \text{and} \quad 2m + n = -3$$ Solving: (m = 1,\ n = -5). $$2m - 3n = 2(1) - 3(-5) = 2 + 15 = 17$$

Comparing all values: (13,\ -17,\ -13,\ \mathbf{17}).

The maximum value is 17, achieved at ((m, n) = (1, -5)).

3. Solution

Answer = 17 ✅

The maximum possible value of (2m - 3n) is 17, obtained when (m = 1) and (n = -5).