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CAT 2019 Slot 2 — QA Question 8

FactorsEasy

What is the largest positive integer such that (n2+7n+12)(n2-n-12) is also positive integer?

Answer & solution

Correct answer: 12

  • A

    6

  • B

    8

  • C

    16

  • 12

Solution

Easy

Factor numerator and denominator; a common factor cancels, leaving n+4n−4\dfrac{n+4}{n-4}. Split off the integer part so the expression becomes 1+8n−41+\dfrac{8}{n-4} — an integer exactly when n−4n-4 divides 88. The largest divisor gives the largest nn.

1

Factorise and cancel. n2+7n+12=(n+3)(n+4)n^{2}+7n+12=(n+3)(n+4) and n2−n−12=(n−4)(n+3)n^{2}-n-12=(n-4)(n+3).

n2+7n+12n2−n−12=(n+3)(n+4)(n−4)(n+3)=n+4n−4(n≠−3)\begin{aligned} &\frac{n^{2}+7n+12}{n^{2}-n-12}=\frac{(n+3)(n+4)}{(n-4)(n+3)}=\frac{n+4}{n-4} \quad (n\neq -3) \end{aligned}
2

Separate the integer part. Write n+4=(n−4)+8n+4=(n-4)+8.

n+4n−4=(n−4)+8n−4=1+8n−4\begin{aligned} &\frac{n+4}{n-4}=\frac{(n-4)+8}{n-4}=1+\frac{8}{n-4} \end{aligned}
3

Make it an integer and maximise. From step 2 the value is an integer iff n−4n-4 divides 88. The largest such divisor is 88.

n−4=8⇒ n=12\begin{aligned} &n-4=8\\ &\Rightarrow\ n=12 \end{aligned}
Largest n=12\text{Largest }n=12

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CAT 2019 Slot 2 QA Q8: What is the largest positive integer such that ( n 2 + 7 n + 12 ) ( n 2 - n - 12 ) is also positive integer? — Solution | TheCATExam