--- title: "Cryptarithmetic WRONG+WRONG=RIGHT" subject: "Artificial Intelligence" chapter: 2 key: "semester_6/artificial_intelligence:cryptarithmetic-wrong-wrong-right" source: https://www.bctengineer.com/q/semester_6/artificial_intelligence/cryptarithmetic-wrong-wrong-right site: BCT Engineering --- # Cryptarithmetic WRONG+WRONG=RIGHT Artificial Intelligence · Chapter 2: Problem Solving - Asked in: 2070 Bhadra, 5+3 marks - Video: [Artificial Intelligence | Tutorial #2 | Crypt Arithmetic Problem](https://www.youtube.com/watch?v=2E-oGMJWUjQ) — RANJI RAJ - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/artificial_intelligence/cryptarithmetic-wrong-wrong-right - Whole chapter: https://www.bctengineer.com/md/semester_6/artificial_intelligence/chapter-2 ## Question Solve the following crypto-arithmetic problem, where different letters denote different integers and identical letters denote the same integer: WRONG + WRONG = RIGHT. Explain the steps that you have followed. ## Short answer Solve as a CSP with column carries. A valid assignment is W=3, R=7, O=0, N=8, G=1, I=4, H=6, T=2 (37081+37081=74162). ## Answer Treat it as a **CSP**: variables W, R, O, N, G, I, H, T; domain 0-9; all different; W and R (leading letters) not zero. ```text W R O N G + W R O N G ----------- R I G H T ``` No carry goes out of the leftmost column, as both numbers have 5 digits. #### Column equations (c1...c4 carries) 1. G + G = T + 10·c1 2. N + N + c1 = H + 10·c2 3. O + O + c2 = G + 10·c3 4. R + R + c3 = I + 10·c4 5. W + W + c4 = R #### Steps followed 1. **Column 5:** R = 2W + c4, so R ≥ 2 and R is even when c4 = 0; also R ≤ 9 means W ≤ 4. 2. **Column 1:** T = 2G mod 10, so T is even. 3. Pick **W = 3, c4 = 1** ⇒ R = 2·3 + 1 = **7**. 4. **Column 4:** 7 + 7 + c3 = I + 10 (c4 = 1) ⇒ I = 4 + c3. Take c3 = 0 ⇒ **I = 4**. 5. **Column 3:** O + O + c2 = G (c3 = 0). Take c2 = 1: G is odd; choose **O = 0, G = 1**. 6. **Column 1:** G + G = 2 ⇒ **T = 2, c1 = 0**. 7. **Column 2:** N + N + 0 = H + 10·c2 with c2 = 1 ⇒ 2N = H + 10. Choose N = 8 ⇒ **H = 6**. 8. **Check distinct digits:** W = 3, R = 7, O = 0, N = 8, G = 1, I = 4, H = 6, T = 2. All different ✓. #### Verification ```text 3 7 0 8 1 + 3 7 0 8 1 ----------- 7 4 1 6 2 ``` 37081 × 2 = 74162 = RIGHT ✓ **Answer: W = 3, R = 7, O = 0, N = 8, G = 1, I = 4, H = 6, T = 2** (37081 + 37081 = 74162). *Note:* a computer search shows 21 valid assignments for this puzzle (for example 24153 + 24153 = 48306), so more than one answer is correct. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In WRONG + WRONG = RIGHT, what does column 5 give? A. R = W B. W = 2R C. R = c4 D. R = 2W + c4 Answer: D. W + W + c4 = R. (Key point: Column 5) 2. Why is T even? A. T = 2G mod 10, a doubled value B. T is always 0 C. G is even D. c1 is 1 Answer: A. Column 1 gives T = 2G mod 10, which is even. (Key point: Column 1 parity) 3. With W = 3 and c4 = 1, what is R? A. 6 B. 7 C. 3 D. 4 Answer: B. R = 2*3 + 1 = 7. (Key point: Choosing W and R) 4. What sum results? A. 37018 + 37018 = 74036 B. 37081 + 37081 = 74126 C. 37081 + 37081 = 74162 D. 36081 + 36081 = 72162 Answer: C. W=3, R=7, O=0, N=8, G=1, I=4, H=6, T=2 gives 74162. (Key point: Verification) ## Related in Artificial Intelligence - [Cryptarithmetic RIGHT+RIGHT=WRONG](https://www.bctengineer.com/question.md?key=semester_6%2Fartificial_intelligence%3Acryptarithmetic-right-right-wrong) — Artificial Intelligence, chapter 2, asked 3× - [Cryptarithmetic LOGIC+LOGIC=PROLOG](https://www.bctengineer.com/question.md?key=semester_6%2Fartificial_intelligence%3Acryptarithmetic-logic-prolog) — Artificial Intelligence, chapter 2, asked 3× - [CSP and SWIM+WEAR=RELAX](https://www.bctengineer.com/question.md?key=semester_6%2Fartificial_intelligence%3Acsp-swim-wear-relax) — Artificial Intelligence, chapter 2