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February 1, 2025 16:45
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flash-twentyfive
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<USER> | |
### 4. Race to 2025! | |
``` | |
G N A J | |
+ N A | |
----------- | |
2 0 2 5 | |
``` | |
In this coded addition, different letters represent different digits, and the same digit is always represented by the same letter. Additionally, none of the multi-digit numbers start with zero. | |
**What is the value of NA?** | |
<ASSISTANT> | |
Here's how we can solve this puzzle: | |
1. **Focus on the units column:** J + A = 5 (or ends in 5). Since no two letters can represent the same digit, and we are adding two single digit numbers the sum must be less than 20. | |
2. **Consider the tens column:** A + N = 2 (or ends in 2). Since there might be a carry-over from the units column, A + N could also equal 12. | |
3. **Analyze the hundreds column:** N + (nothing or 1 if there was a carry over from the tens) = 0 (or ends in zero). This means N must be 0, or if there was a carry-over N must be 9. | |
4. **Consider the thousands column:** G + (nothing or 1 if there was a carry over from the hundreds) = 2. | |
5. **Try N = 0:** If N = 0, then from the tens column, A + 0 = 2, so A = 2. From the units column, J + 2 = 5, so J = 3. From the hundreds column, 0 + (nothing) = 0 which is possible. From the thousands column, G + (nothing) = 2, so G = 2. However, A and G cannot both be 2 so N cannot be zero. | |
6. **Try N = 9:** If N = 9, from the tens column, A + 9 = 12, so A = 3. From the units column, J + 3 = 5, so J = 2. From the hundreds column, 9 + (1 from the carry-over) = 10, which ends in 0. From the thousands column, G + (1 from the carry-over) = 2, so G = 1. | |
Therefore, G = 1, N = 9, A = 3, and J = 2. | |
NA = 93. | |
Final Answer: The final answer is $\boxed{93}$ |
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