If there are 30 cans out of them one is poisoned if a person tastes very…

Key idea: use the fact that each mouse has two possible outcomes (alive or dead) in one 24-hour test. With m mice you can produce 2^m distinct outcome patterns, so you can distinguish up to 2^m different cans.

Compute the minimum m such that 2^m ≥ 30.

• 2^4 = 16, which is less than 30, so 4 mice are not enough.

• 2^5 = 32, which is at least 30, so 5 mice are enough.

Therefore the minimum number of mice required is 5.

Procedure to implement the test:

• Label the 5 mice M1 through M5.

• Assign each can a unique 5-bit binary code (for example, 00001, 00010, ..., up to 11110).

• For each can, give a sample to mouse Mi if and only if the i-th bit of that can's code is 1.

• After 24 hours, record which mice died. The pattern of dead/alive mice is the binary code for the poisoned can.

Brief example (smaller scale): with 3 mice you can distinguish up to 2^3 = 8 cans. Give samples according to 3-bit codes; the pattern of dead mice identifies which of the up-to-8 cans is poisoned.