, 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 $A+B+C+D+E=4+2+8+5+7=26$.

Points $P,Q,R,S,T$ divide the bottom edge of the park into six segments of equal length, each of which has length $600\div6=100$ m.

If Betty and Ann had met for the first time at point $Q$, then Betty would have walked a total distance of $600+400+4\times100=1400$ m and Ann would have walked a total distance of $400+2\times100=600$ m.

When they meet, the time that Betty has been walking is equal to the time that Ann has been walking and so the ratio of Betty’s speed to Ann’s speed is equal to the ratio of the distance that Betty has walked to the distance that Ann has walked.

That is, if they had met for the first time at point $Q$, the ratio of their speed’s would be $1400:600$ or $14:6$ or $7:3$.

Similarly, if Betty and Ann had met for the first time at point $R$, then Betty would have walked a total distance of $600+400+3\times100=1300$ m and Ann would have walked a total distance of $400+3\times100=700$ m.

In this case, the ratio of their speed’s would be $1300:700$ or $13:7$.

When Betty and Ann actually meet for the first time, they are between $Q$ and $R$.

Thus Betty has walked less distance than she would have had they met at $Q$ and more distance than she would have had they met at $R$.

That is, the ratio of Betty’s speed to Ann’s speed must be less than $7:3$ and greater than $13:7$.

We must determine which of the five given answers is a ratio that is less than $7:3$ and greater than $13:7$.

One way to do this is to convert each ratio into a mixed fraction.

That is, we must determine which of the five answers is less than $7:3=\frac73=2\frac13$ and greater than $13:7=\frac{13}{7}=1\frac67$.

Converting the answers, we get $\frac53=1\frac23, \frac94=2\frac14, \frac{11}{6}=1\frac56, \frac{12}{5}=2\frac25,$ and $\frac{17}{7}=2\frac37$.

Of the five given answers, the only fraction that is less than $2\frac13$ and greater than $1\frac67$ is $2\frac14$.

If Betty and Ann meet for the first time between $Q$ and $R$, then the ratio of Betty’s speed to Ann’s speed could be $9:4$.