CSAT Solved Papers/ 2023/Q46

2023 CSAT — Q46

Quant Number theory 2.5 marks Hard

Let $pp$ , $qq$ and $rr$ be $2$ -digit numbers where $p < q < r$ . If $pp + qq + rr = tt0$ , where $tt0$ is a $3$ -digit number whose last digit is zero, then consider the following statements:

$p$ has $5$ possible values.
$q$ has $6$ possible values.

Which of the above statements is/are correct?

A 1 only
B 2 only
C Both 1 and 2 Answer
D Neither 1 nor 2

Worked rationale

The repdigit $pp = 11p$ , $qq = 11q$ , $rr = 11r$ , and $tt0 = 110t$ (digits $t, t, 0$ ). So

11(p + q + r) = 110\,t \;\Rightarrow\; p + q + r = 10\,t.

With $p < q < r$ digits in $1$ – $9$ , the sum ranges $6 \le p+q+r \le 24$ , so $10t \in \{10, 20\}$ — i.e. $p+q+r = 10$ or $20$ .

Sum $= 10$ : increasing triples $(1,2,7), (1,3,6), (1,4,5), (2,3,5)$ .

Sum $= 20$ : increasing triples $(3,8,9), (4,7,9), (5,6,9), (5,7,8)$ .

Eight triples in all. Read off the values:

$p$ -values $\{1,1,1,2,3,4,5,5\} = \{1,2,3,4,5\}$ — $5$ distinct. Statement 1 ✓
$q$ -values $\{2,3,4,3,8,7,6,7\} = \{2,3,4,6,7,8\}$ — $6$ distinct. Statement 2 ✓

Answer: (c) Both 1 and 2.

Why the other options miss

A

missed a case: misses a triple in the sum- $20$ family (e.g. drops $(5,7,8)$ ), losing a $q$ -value and rejecting Statement 2.
B

missed a case: misses a sum- $10$ triple (e.g. $(2,3,5)$ ), shrinking the $p$ -set below $5$ .
D

wrong formula: mis-reduces $tt0$ (e.g. treats it as $100t$ , not $110t$ ), getting a wrong sum constraint and discarding both.

Specialist insight

Two structural reads do all the work: $\overline{pp} = 11p$ and $\overline{tt0} = 110t$ , which collapse the equation to $p+q+r = 10t$ . The bound $6 \le p+q+r \le 24$ then forces exactly two target sums ( $10$ and $20$ ). The only real labour is exhaustively listing the increasing triples for each sum — a missed triple is the whole trap, since it silently drops a $p$ - or $q$ -value. Counting distinct values (not triples) is the final discipline.

The trap, in one line

$11(p+q+r)=110t \Rightarrow p+q+r\in\{10,20\}$ ; the $8$ increasing triples give $5$ distinct $p$ and $6$ distinct $q$ $\Rightarrow$ (c).

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