, 2017

The units digit of the product $1ABCDE\times3$ is 1, and so the units digit of $E\times3$ must equal 1.

Therefore, the only possible value of $E$ is 7.

Substituting $E=7$, we get

[[IMAGE0]]

Since $7\times3=21$, 2 is carried to the tens column.

Thus, the units digit of $D\times3+2$ is 7, and so the units digit of $D\times3$ is 5.

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

Substituting $D=5$, we get

[[IMAGE1]]

Since $5\times3=15$, 1 is carried to the hundreds column.

Thus, the units digit of $C\times3+1$ is 5, and so the units digit of $C\times3$ is 4.

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

Substituting $C=8$, we get

[[IMAGE2]]

Since $8\times3=24$, 2 is carried to the thousands column.

Thus, the units digit of $B\times3+2$ is 8, and so the units digit of $B\times3$ is 6.

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

Substituting $B=2$, we get

[[IMAGE3]]

Since $2\times3=6$, there is no carry to the ten thousands column.

Thus, the units digit of $A\times3$ is 2.

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

Substituting $A=4$, we get

[[IMAGE4]]

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