Letter N Digit 67420B

1. **State the problem:** We have the addition $AMC + AMC + AMC + AMC + AMC + AMC + AMC + AMC = FUN$, where each letter is a different non-zero digit. We want to find the digit represented by $N$. 2. **Rewrite the sum:** Since $AMC$ is added 8 times, this is equivalent to: $$8 \times AMC = FUN$$ 3. **Express $AMC$ and $FUN$ as numbers:** Let $A$, $M$, $C$, $F$, $U$, and $N$ be digits. Then: $$AMC = 100A + 10M + C$$ $$FUN = 100F + 10U + N$$ 4. **Equation:** $$8 \times (100A + 10M + C) = 100F + 10U + N$$ 5. **Analyze the multiplication:** Since $A$, $M$, $C$, $F$, $U$, $N$ are digits 1-9 and distinct, and $8 \times AMC = FUN$, $FUN$ must be a three-digit number. 6. **Consider the last digit $N$:** The last digit of $FUN$ is $N$, which equals the last digit of $8 \times AMC$. The last digit of $8 \times AMC$ depends on the last digit of $AMC$, which is $C$. So: $$N = (8 \times C) \mod 10$$ 7. **Try possible values for $C$ (1 to 9):** - $C=1 \Rightarrow N=8$ - $C=2 \Rightarrow N=6$ - $C=3 \Rightarrow N=4$ - $C=4 \Rightarrow N=2$ - $C=5 \Rightarrow N=0$ (not allowed, digits are non-zero) - $C=6 \Rightarrow N=8$ - $C=7 \Rightarrow N=6$ - $C=8 \Rightarrow N=4$ - $C=9 \Rightarrow N=2$ Possible $N$ values are $2,4,6,8$. 8. **Check the carry and digit constraints:** Since $8 \times AMC$ is a three-digit number $FUN$, $F$ must be 1 or less because $8 \times 999 = 7992$ (4 digits), so $F$ is likely 7 or less. Try $F=7$ (since $8 \times 8xx$ can be near 7xx): Try $A=1$ (smallest digit), then $AMC$ is about 100+. Calculate $8 \times 123 = 984$ (digits 9,8,4) which fits pattern $FUN$. Check if digits are distinct and non-zero: $A=1$, $M=2$, $C=3$, $F=9$, $U=8$, $N=4$. Digits are distinct and non-zero. Check last digit $N=4$ matches $8 \times C = 8 \times 3 = 24$, last digit 4. This matches. 9. **Answer:** The digit $N$ represents is $4$. **Final answer:** $\boxed{4}$