Vedadots
Q1516/80Q17
Q16·CSAT · Prelims 2023

Cryptarithm ABC x D = 37DD

ReasoningCryptarithmeticCoding-decoding● Hard

Question

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?

Options

a

12

b

18

Answer
c

36

d

72

Explanation

Translate the cryptarithm structure into a basic division form: 37DD \div D = ABC. Expand the 4-digit number: 37DD = 3700 + 11D. So we have: 3700 + 11D/D = 3700/D + 11 = ABC.

Since ⟨MATH⟩ABC⟨/MATH⟩ must be an integer, ⟨MATH⟩D⟨/MATH⟩ must be a perfect single-digit divisor of 3700. The single-digit divisors of 3700 are 1, 2, 4, and 5.

If D = 5: 3755 \div 5 = 751. Here A=7, B=5, C=1. However, this violates the constraint that A, B, C, D must be different digits (since D=5 and B=5 repeat).
If D = 4: 3744 \div 4 = 936. Here A=9, B=3, C=6, D=4. All four digits \{9, 3, 6, 4\} are non-zero and completely distinct. This satisfies all constraints perfectly.

Calculate the required sum: A + B + C = 9 + 3 + 6 = 18.

Convert multiplication cryptarithms with repeating trailing digits into algebraic fraction expansions to easily isolate the single-digit divisor candidate.

Answer: (b).

Question details

Year

2023

Paper

CSAT

Question

Q16

Section

Logical & Analytical Reasoning

Sub-topic

Cryptarithmetic

Type

Coding-decoding

Difficulty

Hard

Source hint

Cryptarithmetic puzzle

Same sub-topic — other years

Cryptarithmetic has appeared in multiple CSAT papers:

Q39

2025

Cryptarithm PQR - PS = PPT

ReasoningHard

See all questions on Cryptarithmetic

Browse every tagged question across all years

Explore →
Back to CSAT 2023