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CAT 2023 Slot 2 — QA Question 20

Number TheoryEasy

If p2 + q2 - 29 = 2pq - 20 = 52 - 2pq, then the difference between the maximum and minimum possible value of (p3 - q3) is

Answer & solution

Correct answer: 378

  • A

    243

  • 378

  • C

    189

  • D

    486

Solution

Easy

A chained equation X=Y=ZX=Y=Z really gives two equations. Pairing them turns the messy expressions into the clean identities (p−q)2(p-q)^2 and (p+q)2(p+q)^2. From the two values of p−qp-q and p+qp+q you get all (p,q)(p,q) pairs, then read off the largest and smallest p3−q3p^3-q^3.

1

First pairing p2+q2−29=2pq−20p^2+q^2-29=2pq-20:

p2+q2−2pq=29−20⇒ (p−q)2=9⇒ p−q=±3\begin{aligned} &p^2+q^2-2pq=29-20\\ &\Rightarrow\ (p-q)^2=9\\ &\Rightarrow\ p-q=\pm 3 \end{aligned}
2

Second pairing p2+q2−29=52−2pqp^2+q^2-29=52-2pq:

p2+q2+2pq=52+29⇒ (p+q)2=81⇒ p+q=±9\begin{aligned} &p^2+q^2+2pq=52+29\\ &\Rightarrow\ (p+q)^2=81\\ &\Rightarrow\ p+q=\pm 9 \end{aligned}
3

All four sign combinations give four (p,q)(p,q) pairs:

p−q=3, p+q=9:(p,q)=(6,3)⇒p3−q3=216−27=189p−q=−3, p+q=9:(p,q)=(3,6)⇒p3−q3=27−216=−189p−q=3, p+q=−9:(p,q)=(−3,−6)⇒p3−q3=−27+216=189p−q=−3, p+q=−9:(p,q)=(−6,−3)⇒p3−q3=−216+27=−189\begin{aligned} &p-q=3,\ p+q=9:&&(p,q)=(6,3)&&\Rightarrow p^3-q^3=216-27=189\\ &p-q=-3,\ p+q=9:&&(p,q)=(3,6)&&\Rightarrow p^3-q^3=27-216=-189\\ &p-q=3,\ p+q=-9:&&(p,q)=(-3,-6)&&\Rightarrow p^3-q^3=-27+216=189\\ &p-q=-3,\ p+q=-9:&&(p,q)=(-6,-3)&&\Rightarrow p^3-q^3=-216+27=-189 \end{aligned}
4

Max minus min:

max⁡=189,min⁡=−189difference=189−(−189)=378\begin{aligned} &\max=189,\quad \min=-189\\ &\text{difference}=189-(-189)=378 \end{aligned}

Difference =378=\mathbf{378}

By symmetry the four values are just ±189\pm 189, so the answer is simply 2×189=3782\times 189=378 — you only ever need to compute one pair.

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CAT 2023 Slot 2 QA Q20 — Number Theory solution | TheCATExam