CSAT Solved Papers/ 2021/Q60

2021 CSAT — Q60

Quant Number theory 2.5 marks Hard

Consider the following multiplication problem:

(PQ) \times 3 = RQQ,

where $P$ , $Q$ and $R$ are different digits and $R \neq 0$ .

What is the value of $(P + R) \div Q$ ?

A 1
B 2 Answer
C 5
D Cannot be determined due to insufficient data

Worked rationale

Write the numbers by place value: $\overline{PQ} = 10P + Q$ and $\overline{RQQ} = 100R + 11Q$ . Then

3(10P + Q) = 100R + 11Q \;\Rightarrow\; 30P + 3Q = 100R + 11Q \;\Rightarrow\; 30P = 100R + 8Q \;\Rightarrow\; 15P = 50R + 4Q.

Units constraint: the units digit of $3Q$ must be $Q$ , so $3Q \equiv Q \pmod{10}$ , i.e. $2Q \equiv 0 \pmod{10}$ , giving $Q\in\{0, 5\}$ .

$Q = 0$ : $15P = 50R \Rightarrow 3P = 10R$ , impossible for nonzero digits ( $R\neq 0$ ).
$Q = 5$ : $15P = 50R + 20 \Rightarrow 3P = 10R + 4$ . For a digit $P$ , $R = 2$ gives $3P = 24 \Rightarrow P = 8$ .

So $P = 8, Q = 5, R = 2$ — distinct, $R\neq 0$ . Check: $85\times 3 = 255 = \overline{RQQ}$ ✓.

(P + R)\div Q = (8 + 2)\div 5 = 10\div 5 = 2.

Answer: (b) 2.

Why the other options miss

A

an arithmetic slip: solves the digits but mis-evaluates $(8+2)/5$ or picks a wrong $P$ .
C

answered a different combination: forms $(P+R)\times Q$ or another combination instead of the quotient asked.
D

gave up too early: concludes the puzzle is underdetermined, missing that the units constraint forces a unique $(P,Q,R)$ .

Specialist insight

The units digit does the heavy lifting: $3Q$ ending in $Q$ forces $Q\in\{0,5\}$ , and $Q=0$ dies immediately, leaving $Q=5$ and the clean $3P = 10R + 4$ . Only $R=2$ yields a digit $P=8$ , so the solution is unique — which is exactly why “cannot be determined” is the trap. In cryptarithms, always extract the units-digit congruence first; it usually collapses the search to one or two cases.

The trap, in one line

$3Q\equiv Q\pmod{10}\Rightarrow Q=5$ , then $3P=10R+4\Rightarrow(P,R)=(8,2)$ ; $(8+2)/5 = 2 \Rightarrow$ (b).

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