A combination lock with three dials, each numbered 1 through 8, is defective in that you only need to get two of the numbers right to open the lock. (For example, suppose the true combination is 4-2-7. Then 4-2-7 would open the lock, but so would 4-2-5, 4-2-2, 4-8-7 or 4-6-7. But not 2-4-7.)

Question

06-04-2021
Mathematics

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A combination lock with three dials, each numbered 1 through 8, is defective in that you only need to get two of the numbers right to open the lock. (For example, suppose the true combination is 4-2-7. Then 4-2-7 would open the lock, but so would 4-2-5, 4-2-2, 4-8-7 or 4-6-7. But not 2-4-7.)

Sagot :

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sandlee09 sandlee09 · Answer 1 · 2021-04-08T12:08:50-04:00

Answer:

64 combination attempts

Step-by-step explanation:

Since each dial in the lock is numbered from 1-8 then there are 8 possibilities for each dial. Out of the three dials, only 2 actually need to be correct in order for the lock to open therefore, we simply raise the number of possibilities for each dial to the power of 2 which should give us the total number of tries we need in order to guarantee that it opens. Assuming that you are making the combinations in numerical order 1-1-1, 1-1-2, 1-1-3, etc.

8^2 = 64 combination attempts

Sagot :

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