A hard question on number puzzles for P4–P5, with a step-by-step solution, diagrams and the answer. Number puzzles and made-up operations appear in the Hua Xia Cup from the P1 and P2 first rounds to the P6 finals.
The question
In this sum, different letters stand for different digits and the same letter always stands for the same digit.
No number starts with 0.
ABCD
ABC
AB
+A
2342
What is the four-digit number ABCD?
Show the hints (3)
💡Hint 1: A appears in all four numbers: in the thousands, hundreds, tens and ones. What are the four As worth together? And B, C, D?
💡Hint 2: The total = 1111 × A + 111 × B + 11 × C + D = 2342.
💡Hint 3: 111 × B + 11 × C + D is at most 1107, less than 1111, so A is the whole-number part of 2342 ÷ 1111.
Worked solution
Have a go first, then read the solution step by step.
1.A appears once in each of the thousands, hundreds, tens and ones: 1111 × A. B gives 111 × B, C gives 11 × C, and D just D.
2.So 1111 × A + 111 × B + 11 × C + D = 2342.
3.111 × 9 + 11 × 9 + 9 = 1107, less than 1111. 2342 ÷ 1111 = 2 remainder 120, so A = 2.
4.11 × 9 + 9 = 108, less than 111. 120 ÷ 111 = 1 remainder 9, so B = 1.
5.D is less than 11. 9 ÷ 11 = 0 remainder 9, so C = 0 and D = 9.
6.ABCD = 2109. Check: 2109 + 210 + 21 + 2 = 2342.
⚠️Common mistake: taking the digits of 2342 as A, B, C, D. The columns carry, so they differ.
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