men are sharing a big pile of bananas with a monkey. The first gives one banana to the monkey and takes some bananas, leaving th of the remaining pile. The second gives one banana to the monkey and takes some bananas, leaving th of the remaining pile. This pattern continues, with the th person leaving th of the pile after giving a single banana to the monkey each time. Finally, the th man gives one banana to the monkey, takes half of the pile of bananas, and leaves the rest for the monkey.
Let the smallest amount of total bananas needed such that each person receives a positive integer number of bananas be . What is ?
Note: The men do take some bananas for themselves.
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This problem is easy to just bash out. However, working with binary simplifies the problem immensely.
Note that subtracting 1 from a binary number N decreases the units digit by 1 (duh). However, dividing by 2 is the clever part: it moves the decimal place one place to the left. Dividing \by 2 n moves the decimal place n places to the left.
Thus, the number 1 1 0 1 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 2 works: we first subtract the 1 , cancelling out the first 1 . Then we move the decimal place 6 places to the left, arriving at the next 1 . We the subtract 1 again, cancelling it out. We move 5 places to the left, arriving at the next 1 . We can continue doing this, and the final number of bananas after the process is over is 1 , which is left for the monkey.
Thus, we just need to convert 1 1 0 1 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 2 to base 1 0 , which is 3 4 4 2 7 5 3 , and the answer is 7 5 3 .