We are given that a positive integer $N$ is represented as $abc$ in base 5 and $cba$ in base 9. We want to find $N$ in base 10 and base 4.

1. Problem Description

We are given that a positive integer $N$ is represented as $abc$ in base 5 and $cba$ in base

9. We want to find $N$ in base 10 and base

4.

2. Solution Steps

The digits in base 5 are $0, 1, 2, 3, 4$. The digits in base 9 are $0, 1, 2, 3, 4, 5, 6, 7, 8$. Therefore, $0 \le a \le 4$, $0 \le b \le 4$, $0 \le c \le 4$ and $0 \le a \le 8$, $0 \le b \le 8$, $0 \le c \le 8$. Combining these, we have $1 \le a \le 4$ (since $a$ is the leading digit), $0 \le b \le 4$, and $0 \le c \le 4$. Since $c$ is the leading digit in base 9, we also have $1 \le c \le 8$.

We have:

$N = a \cdot 5^2 + b \cdot 5^1 + c \cdot 5^0 = 25a + 5b + c$

$N = c \cdot 9^2 + b \cdot 9^1 + a \cdot 9^0 = 81c + 9b + a$

Equating the two expressions for $N$:

$25a + 5b + c = 81c + 9b + a$

$24a - 4b - 80c = 0$

$6a - b - 20c = 0$

$b = 6a - 20c$

From the inequalities above, $1 \le a \le 4$, $0 \le b \le 4$, $1 \le c \le 4$.

Substituting $b = 6a - 20c$, we get $0 \le 6a - 20c \le 4$.

$6a \le 4 + 20c$

$6a \ge 20c$

If $c=1$, then $20 \le 6a \le 24$, so $20/6 \le a \le 4$. Thus, $a$ can be $4$.

If $a=4$, then $b = 6(4) - 20(1) = 24 - 20 = 4$. So, $a=4, b=4, c=1$.

Then, $N = 25a + 5b + c = 25(4) + 5(4) + 1 = 100 + 20 + 1 = 121$

$N = 81c + 9b + a = 81(1) + 9(4) + 4 = 81 + 36 + 4 = 121$

So $N=121$ in base

1

0.

To convert 121 to base 4:

$121 \div 4 = 30$ remainder 1

$30 \div 4 = 7$ remainder 2

$7 \div 4 = 1$ remainder 3

$1 \div 4 = 0$ remainder 1

So, $121 = 1321_4$

From the original problem,

$A=1, B=4, C=0, D=4, E=1, F=4$

$G=2, H=5, I=5, J=8, K=1, L=9$

$M=6, N=2, O=0$

$a=4, b=4, c=1$

$P=4, Q=4, R=1$

$STU = 121$

$VWXY = 1321$

3. Final Answer

$A=1, B=4, C=0, D=4, E=1, F=4$

$G=2, H=5, I=5, J=8, K=1, L=9$

$M=6, N=2, O=0$

$a=4, b=4, c=1$

$P=4, Q=4, R=1$

$STU = 121$

$VWXY = 1321$