CSAT Solved Papers/ 2023/Q16

2023 CSAT — Q16

Quant Number theory 2.5 marks Medium

A $3$ -digit number $ABC$ , on multiplication with $D$ gives $37DD$ where $A, B, C$ and $D$ are different non-zero digits. What is the value of $A + B + C$ ?

A 18 Answer
B 16
C 15
D Cannot be determined due to insufficient data

Worked rationale

The product $37DD$ is a $4$ -digit number $3700 + 11D$ (digits $3, 7, D, D$ ). So

ABC \times D = 3700 + 11D, \qquad ABC = \frac{3700 + 11D}{D}.

$ABC$ must be a $3$ -digit integer, so sweep $D = 1,\dots,9$ and require divisibility and distinct non-zero digits:

$D = 4$ : $\frac{3700 + 44}{4} = \frac{3744}{4} = 936$ . Check $936 \times 4 = 3744 = 37\,44$ ✓. Digits $A,B,C,D = 9,3,6,4$ — all distinct, all non-zero. Valid.
$D = 5$ : $\frac{3755}{5} = 751$ , and $751\times 5 = 3755$ , but then $A,B,C,D = 7,5,1,5$ — $B = D = 5$ repeats. Rejected.
$D = 2,3,6,7,8,9$ : $3700 + 11D$ is not divisible by $D$ (e.g. $3766/6, 3777/7, 3788/8$ are non-integers).
$D = 1$ : $3711/1 = 3711$ is $4$ -digit, not $ABC$ . Rejected.

The unique solution is $ABC = 936,\ D = 4$ , so

A + B + C = 9 + 3 + 6 = 18.

Answer: (a) 18.

Why the other options miss

B

an arithmetic slip: a division error producing a wrong $ABC$ (e.g. mis-dividing $3744$ ) whose digits sum to $16$ .
C

missed a case: accepts the $D=5,\ ABC=751$ branch ( $7+5+1 = 13$ , or a near variant) without checking the distinct-digit constraint $B \neq D$ .
D

solved the wrong question: gives up at “two unknowns ( $ABC$ and $D$ )” without noticing $37DD = 3700+11D$ turns it into a one-variable divisibility sweep with a unique answer.

Specialist insight

The unlock is reading $37DD$ as the number $3700 + 11D$ , which converts a scary cryptarithm into ” $ABC = (3700+11D)/D$ , test nine values of $D$ .” Divisibility kills most $D$ ; the distinct-digit rule kills the $D=5$ near-miss. The deadliest distractor is (d) “cannot be determined” — it tempts anyone who doesn’t see that the puzzle is fully constrained. CSAT rewards the candidate who tries the structure before declaring under-determination.

The trap, in one line

Read $37DD = 3700+11D$ ; only $D=4$ gives an integer $ABC=936$ with all digits distinct $\Rightarrow A+B+C=18$ .

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