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GATE DA 2026 | Aptitude | Memory Based | Question 5

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2 2 votes

Distinct digits $P,Q,R,S,X,Y$ are chosen from ${0,1,2,\dots,9}$. Let $PQ$ and $RS$ be two digit numbers formed by the digits $(P,Q)$ and $(R,S)$, respectively, such that $PQ$ and $RS$ are consecutive integers. If
$$(PQ)^2 + (RS)^2 = X YP$$
where $XYP$ is a three digit number, then the value of $Y$ is:

  1. $4$
     
  2. $5$
     
  3. $6$
     
  4. $7$

2 Answers

3 3 votes
since both are sequential and note that all p,q,r,s are diffrent also so, only valid pq and rs pair are (19,20),(29,30),(39,40),(49,50),(59,60),(69,70),(79,80)

and further it is asked that sum of square of pq and rs is (3 Digit) so, only pair (19,20) sum of product will result in 3 digit i.e.

$19^2 + 20^2$ = 361 + 400 = 761 and like this  Y = 6 that is

OPTION C
• edited by
1 1 vote
Here 100<= (PQ)^2 + (RS)^2 <=999 and PQ and RS be Consecutive RS and PQ differ by 1 . In PQ = P*10 + Q means P in tens.

So If PQ^2 + RS^2 = X*100+Y*10+P so Unit Digit X*100+Y*10+P of equals to P .

By Trying Some Cases Like

(P = 2)

Like PQ = 21 and RS=22 , PQ^2 + RS^2 = 925 here unit digit is 5 so wrong

PQ = 20 and RS = 21 , PQ^2 + RS^2 = 841 unit digit is 1 but p is 2 so wrong .

P=1

PQ = 19 RS = 20 ,PQ^2 + RS^2 = 761 here Unit Digit is P = 1  So answer is C(6)
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