In the six-digit number $1 A B C D E$, each letter represents a digit. Given that $1 A B C D E \times 3=A B C D E 1$, the value of $A+B+C+D+E$ is
(A) 29
(B) 26
(C) 22
(D) 30
(E) 28
The units digit of the product $1 A B C D E \times 3$ is 1 , and so the units digit of $E \times 3$ must equal 1 . Therefore, the only possible value of $E$ is 7 .

Substituting $E=7$, we get

| $1 A B C D 7$ |
| ---: |
| $\times \quad 3$ |
| $A B C D 71$ |

Since $7 \times 3=21,2$ is carried to the tens column.

Thus, the units digit of $D \times 3+2$ is 7 , and so the units digit of $D \times 3$ is 5 .

Therefore, the only possible value of $D$ is 5 .

Substituting $D=5$, we get

| $1 A B C 57$ |
| ---: |
| $\times \quad 3$ |
| $A B C 571$ |

Since $5 \times 3=15,1$ is carried to the hundreds column.

Thus, the units digit of $C \times 3+1$ is 5 , and so the units digit of $C \times 3$ is 4 .

Therefore, the only possible value of $C$ is 8 .

Substituting $C=8$, we get

| $1 A B 857$ |
| ---: |
| $\times \quad 3$ |
| $A B 8571$ |

Since $8 \times 3=24,2$ is carried to the thousands column.

Thus, the units digit of $B \times 3+2$ is 8 , and so the units digit of $B \times 3$ is 6 .

Therefore, the only possible value of $B$ is 2 .

Substituting $B=2$, we get

| $1 A 2857$ |
| ---: |
| $\times \quad 3$ |
| $A 28571$ |

Since $2 \times 3=6$, there is no carry to the ten thousands column.

Thus, the units digit of $A \times 3$ is 2 .

Therefore, the only possible value of $A$ is 4 .

Substituting $A=4$, we get

| 142857 |
| ---: |
| $\times \quad 3$ |
| 428571 |

Checking, we see that the product is correct and so $A+B+C+D+E=4+2+8+5+7=26$.