A rope burns irregularly in 16 minutes and costs 32 rupees, while a second rope burns also irregularly in 7 minutes and costs 14 rupees. Both can be lit only at one end and can be turned off and lit again as many times we want, until they are completely burned. In what way can we count 1 minute by using such ropes and how much will it cost? (Multiple ropes can be used but we are looking for the combination that will cost less).

My attempt: We light the two ropes simultaneously so when the second is fully burnt, 7 minutes have passed, so the remaining from the first one will be burnt in 9 minutes. Then we light a second of 7 mins and this way we have one of 2 minutes. Following the same procedure twice, we can have another one of 2 minutes. Then we light another of 7 minutes, along with the two of 2 minutes one after the other, and we have one of 3 minutes left (7-2-2). Then with the one of 3 minutes and another one of 2 minutes we can count 1 minute.

...but this would be a disaster, cost-wise :)

Any other ideas are most welcome!!!

New answer - 116 rupees

So, apparently we can get even cheaper than 120 rupees, this time by buying only a single 16-minute rope, and six 7-minute ropes (cost is $1 times 32 + 6 times 14 = 116$). This is how you do it:

1. Create a 2-minute rope with the 16-minute rope and two 7-minute ropes (remaining ropes: $4 times 7, 1 times 2$)


2. Use the 2-minute rope to turn two 7-minute ropes into 5-minute ropes (remaining ropes: $2 times 7, 2 times 5$)


3. Use one of the 5-minute ropes to turn the remaining 7-minute ropes into 2-minute ropes (remaining ropes: $1 times 5, 2 times 2$)


4. Use the two 2-minute ropes to reduce the 5-minute rope to the 1-minute rope you really wanted all along

I no longer dare to say this might be optimal.

Old answer - 120 rupees

The cheapest I have been able to work out so far is 120 rupees: 2 16-minute ropes and 4 7-minute ropes (cost is $2 times 32 + 4 times 14 = 120$). You do this as follows:

1. Light the two 16-minute ropes and a 7-minute rope to create two 9-minute ropes (remaining ropes: $2 times 9$, $3 times 7$)


2. Use a 9-minute rope and a 7-minute rope to create a 2-minute rope (remaining ropes: $1 times 9$, $2 times 7$, $1 times 2$)


3. Light the two 7-minute ropes and the 2-minute rope to create two 5-minute ropes (remaining ropes: $1 times 9$, $2 times 5$)


4. Use the 9-minute rope and one of the 5-minute ropes to create a 4-minute rope (remaining ropes: $1 times 5$, $1 times 4$)


5. Use your 5-minute rope and 4-minute rope to create the desired 1-minute rope

I'm reasonably confident this is optimal, but I have no idea how to conclusively prove it.

Start by creating a 2 minute rope as you described. Then start burning three 7 minutes rope at the same time, for 2 min. You are left with three 5 min ropes. Start burning a 16 minute ropes for three times five minutes.

THIS IS NOT AN ANSWER.

Just some thoughts that might help - please let me know if I've misunderstood any questions/answers.

The costs for the 2 kinds of rope are just double the ropes' burn times, so really we're looking to minimise the number of rope-minutes we buy.

The asker's solution involves a [7 minute] rope and 3 [2 minute] ropes.
We can make 3 [2 minute] ropes from 3 [16 minute] ropes lit simultaneously with a [7 minute], and then another [7 minute] we light after the first [7 minute] burns.
This means we buy 3 [7 minute] ropes, and 3 [16 minute] ropes, for a total of 69 minutes purchased, 138 rupees.

Andrei's solution involves creating 3 [5 minute] ropes, and burning them alongside a [16 minute].
We can create 3 [5 minute] ropes from 5 [7 minute ropes] and a [16 minute] rope - first the [16] burns with the first [7], then with the second [7], then we light the last 3 [7]s and they end up as [5 minute] ropes.
This means we buy 2 [16 minute] ropes and 5 [7 minute] ropes, for a total of 67 minutes purchases, 134 rupees.

If we skipped any tricks with creating [x minute] ropes (x!=7,x!=16), we would be burning 7 [7 minute ropes] for 49 minutes, and 3 [16 minute] ropes for 48 minutes, giving us a [1 minute] rope made from the 7th [7 minute] rope. This costs 194 rupees.

I written up these answers in their optimal forms - to create Y [X minute] ropes I assume we can do the process that creates 1 [X minute] rope, but when we light up the rope that ends up being [X minutes] we buy Y of them instead of 1. This is how we can create 1 [5 minute] rope by purchasing 1 [16], 3[7]s, the end rope having started off as a [7], but we can create 101 [5 minute] ropes by purchasing 1 [16] and 103 [7]s, 2 of which are burnt in the process. This is obviously cheaper than buying 1[16], 3[7]s, 101 times.

We have to begin by burning at least 1 [16] and at least 1 [7] - otherwise we're wasting rope, since can't make any guesses or burn at both ends. We must burn at least 1 other [16], because otherwise we're stuck with just a ridiculous number of [7]s (9, by making 4 [5]s (uses up 6) and then burning those as a [20] alongside 3 more [7]s).

So we have to use at least 2 [16]s. Andrei's solution uses that many, and only 5 [7]s - to use only 4 [7]s I think is impossible, given that the 5 is already an optimisation.
If we used 3 [16]s we're on par with the asker's solution, so we'd have to use only 2 [7]s - this also seems pretty impossible, not least because we'd be stuck on even numbers.