This is a discrete math problem (and the joke is that we are "discreet" about these problems).
0.615 = 615/1,000 = 123/200
Every time Earth goes around 123 times (123 years) Venus will have gone around 200 times.
So we know that after 123 years, they will be lined up again AND that they will be lined up on the same line!
Possibly they could have been lined up before, but in another direction.
This will occur if the number of orbits they have traveled has the exact same decimal fraction.
If X represents the number of years (with fraction), then we could write:
(200/123)*X - X = K
where K must be an integer (no fractional part).
(200/123)*X is the number of orbits that Venus will have done.
X is the number of orbits for Earth -- one per year.
If we can find two numbers (X and K) such that the problem is solved, then we will have found an answer.
Since we know K must be an integer (easier to guess at), we will solve for X relative to K:
(200/123)*X - X = K
Put X in evidence:
[(200/123) - 1]*X = K
we know that 1 = 123/123, so we can solve the square bracket:
(77/123)*X = K
Move the factor to the other side:
X = K * (123/77) = number of years to satisfy K
Now, we simply set K = 1 (the easiest integer)
(in discrete math, if you are forced to guess, always begin with the easiest guess)
for K = 1, we get X = 123/77 (= 1.5974026... year)
Let's try it:
After 1.5974026 years, Earth will have gone 1.5974026... orbits.
After 1 orbit it will have passed over the start point, so it is 0.5974026... of the way on its second orbit. (it you want to be technical, since one orbit represents 360 degrees, then Earth is now 215 degrees past its starting point).
After 1.5974026 Earth-years, Venus will have gone for
1.5974026.../ 0.615 = 2.5974026...
After 1 orbit it passes over its start point. Same after 2.
Therefore, Venus is now 0.5974026... of the way on its third orbit, meaning it is 215 degrees past its original starting point.
Earth and Venus are both in the same direction from the Sun (215 degrees relative to their starting point). Therefore, they must be in line.
Try it with K = 2, 3, or 7 or 92, etc. It will always work.
If K = 77, you get the special case:
X = K*(123/77) = 123 (which is the one we began with)
Earth will have gone exactly 123 orbits (back to its starting point, because it is 123.0000000...) and Venus will have gone 200.0000000... times around. Not only are they in line, they are back on the original line.
---
The way you formulate the question still leaves the possibility that they could be in line, but not on the same side of the sun (the fractional difference would be exactly 0.5).
K = 0.5
X = 0.5 * 123/77 = 123/154 = 0.7987... year
Earth will have gone 0.7987... (three-quarters of the way around)
while Venus will have gone:
0.7987... / 0.615 = 1.2987...
A difference of exactly 0.5 (meaning 180 degrees).
Earth, the Sun and Venus are lined up, but with Venus being on the other side of the Sun (called "superior conjunction")
solutions with K being an integer will "predict" inferior conjunctions (Venus between Earth and the Sun) while the solutions with K being a "half integer" (1.5, 2.5, 3.5...) will predict the superior conjunctions.
---
Unfortunately in real life, the orbits are ellipses, and they are not even perfect ellipses (because of interference from other planets' gravity).