## Crystal strength testing brainteaser

By David, 8 July 2020 Brainteaser

**Difficulty: Taxing**

Terri’s destructive testing lab is very popular. The most important piece of equipment is an electric-powered ram that smashes into samples to see how tough they are. It’s powered by capacitors, and by changing the number of capacitors that are plugged in, Terri can control how powerful the ram is.

Late in the day, a geologist knocked on Terri’s door. They had three identical crystals and wanted them to be tested by the ram.

Terri put the first crystal in place, connected 10 capacitors and fired the ram. The crystal shattered.

Terri now only has two crystals left to test with. The geologist wants to know the smallest number of capacitors that will smash the crystal. And there’s only time to run five more tests before the lab closes for the day.

Luckily, if a crystal survives a test, it will remain undamaged and can be reused.

Can you work out a plan for the remaining tests to get the work done in time?

For advanced mathematicians, can you make sure the testing takes fewer than five attempts?

**Scroll down or click for a hint, or the answer!**

## Brainteaser hint

If you could do 10 tests, which tests would you do? Remember, you have two crystals, so you can use one to get a ballpark, and the other to refine your guess.

## Brainteaser answer

One solution is to test a crystal with 5 capacitors:

• If it breaks, test the remaining crystal until it smashes, starting with 1 capacitor, then 2, then 3, then 4.

• If it doesn’t break, test the crystal again until it smashes, with 6 capacitors, then 7, then 8, then 9.

There’s actually a way to finish with only four tests! With the first crystal, try the following capacitor tests, stopping when the crystal breaks: 4, 7, 8, 9. Then, use the second crystal to refine your answer. For example, if it survives 4 and breaks at 7, then test 5 and 6.

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8 July, 2020 at 7:42 pm

I took a slightly different approach.

Try with [3]. If it smashes, then we know it’s [1, 2, 3] so try the second one with [1] and then [2].

If the second one smashes, then we know it’s [1] or [2]. If it survives, then it’s [3].

This needs 3 tests at worst case.

If [3] survives, it has to be [4] or higher. We try the first on [6] (second test).

If it smashes, then we try [4, 5]. This tells us if the target number is [4, 5, 6].

Total 4 tests, at worst case.

If [6] survives, it has to be [7] or higher. We now test [8] (third test)

If [8] smashes, we test [7]. This tells us if [7, 8] is the target number.

Total 4 tests, at worst case.

if [8] survives, it has to be [9] or [10].

Test with [9]. If it smashes, [9] is the target (forth test)

If [9] survives, the target is [10]

Total 4 tests.

This means that if the crystals are very fragile, then we only needed 3 tests at most, otherwise 4 tests.

10 July, 2020 at 1:04 pm

I arrived at the same answer as Martin for four tests. I think the proposed solution for four tests (starting at 4 capacitors) is actually wrong – if the first crystal breaks at 4 capacitors, you only have one crystal left to determine if the limit is 1, 2 or 3 capacitors, which isn’t possible to do with 100% certainty. Starting at 3 capacitors solves this problem.

10 July, 2020 at 1:44 pm

Hi Samuel,

I think it’s possible to complete in four tests if the first test is at four.

then test, 1, 2, 3 in order until you find the breaking point, you end up with a maximum of four tests to find if the breaking point is 1, 2, 3 or 4.

Well done on the solution!

12 July, 2020 at 10:20 pm

What about the first test at 2, if it breaks then test 1, if not try next test at 4, if it breaks then try 3, if not try the third test at 6 and back to 5. Fourth will be at 8 and the fifth will be at either 9 or 7. That has still only used the last 2 crystals and no more than 5 tests.