Cryptarithm ABC + DEF = 1111
Question
If ABC and DEF are both 3-digit numbers such that A, B, C, D, E and F are distinct non-zero digits such that ABC + DEF = 1111, then what is the value of A + B + C + D + E + F?
Options
28
29
30
31
Explanation
Align the column addition vertically to observe the carry-over dynamics :
Let c1 be the carry from units to tens, c2 the carry from tens to hundreds, and c3 the carry from hundreds to thousands. Then the column equations are: C + F = 1 + 10c1 B + E + c1 = 1 + 10c2 A + D + c2 = 1 + 10c3 Since ABC + DEF = 1111, the carry into the thousands place c3 = 1. Adding the three equations gives: (A + B + C + D + E + F) + (c1 + c2) = 13 + 10(c1 + c2) Hence S = A + B + C + D + E + F = 13 + 9(c1 + c2). Each of c1 and c2 is 0 or 1, but c1 cannot be 0 (units sum cannot be 1 because C and F are non-zero), and similarly c2 cannot be 0, so c1 + c2 = 2. Therefore S = 13 + 92 = 31.
Answer: (d).
Question details
Year
2023
Paper
CSAT
Question
Q9
Section
Logical & Analytical Reasoning
Sub-topic
Cryptarithmetic
Type
Coding-decoding
Difficulty
Hard
Source hint
Cryptarithmetic puzzle
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