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CAT 2021 Slot 3 — QA Question 22

ModulusMedium

The number of distinct pairs of integers (m, n) satisfying |1 + mn| < |m + n| < 5 is

Answer & solution

Correct answer: 12

Answer: 12

Solution

Medium

Both sides are non-negative, so squaring the first inequality is safe and turns it into a clean factored condition. That condition forces one of m,nm,n to be 00; the outer bound ∣m+n∣<5|m+n|<5 then limits the other.

1

Square ∣1+mn∣<∣m+n∣|1+mn|<|m+n|.

(1+mn)2<(m+n)2⇒ 1+m2n2+2mn<m2+n2+2mn⇒ m2+n2−m2n2−1>0\begin{aligned} &(1+mn)^2 < (m+n)^2\\ &\Rightarrow\ 1 + m^2n^2 + 2mn < m^2 + n^2 + 2mn\\ &\Rightarrow\ m^2 + n^2 - m^2n^2 - 1 > 0 \end{aligned}
2

Factor. Group to a product of two factors.

(m2−1)(1−n2)>0⇒ (m2−1)(n2−1)<0\begin{aligned} &(m^2-1)(1-n^2) > 0\\ &\Rightarrow\ (m^2-1)(n^2-1) < 0 \end{aligned}

So exactly one of m2,n2m^2,n^2 is below 11 and the other above 11. For integers, "below 11" forces that variable to be 00.

3

Case n=0n=0. Then m2>1m^2>1, and the bound ∣m+n∣<5|m+n|<5 becomes ∣m∣<5|m|<5.

1<∣m∣<5⇒ m=±2,±3,±4(6 pairs)\begin{aligned} &1 < |m| < 5\\ &\Rightarrow\ m = \pm2,\pm3,\pm4 \quad\text{(6 pairs)} \end{aligned}

(Each also satisfies ∣1+mn∣=1<∣m∣|1+mn|=1<|m|, so all qualify.)

4

Case m=0m=0. By symmetry, 1<∣n∣<51<|n|<5.

n=±2,±3,±4(6 pairs)⇒ Total=6+6=12\begin{aligned} &n = \pm2,\pm3,\pm4 \quad\text{(6 pairs)}\\ &\Rightarrow\ \text{Total} = 6 + 6 = 12 \end{aligned}
12 pairs12\ \text{pairs}

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CAT 2021 Slot 3 QA Q22 — Modulus solution | TheCATExam