There are three cups with the same amount of cubes in each one. A person removes two cubes at a time from the first cup and had one cube left over. Then the person removes three cubes at a time from the second cup and this time had two cubes left over. Lastly, the person removes 5 cubes at a time and now has 3 left over. What is the fewest number in these cups to make these outcomes possible?

Question

Xander Vance

Mathematics

15 September, 15:41

There are three cups with the same amount of cubes in each one. A person removes two cubes at a time from the first cup and had one cube left over. Then the person removes three cubes at a time from the second cup and this time had two cubes left over. Lastly, the person removes 5 cubes at a time and now has 3 left over. What is the fewest number in these cups to make these outcomes possible?

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Elsa · Answer 1

The fewest number in these cups to make these outcomes possible is 23 cubes all 3 cups because based on each person’s getting cubes at a time and the cubes that was left over were starting from one up to three that was left over in each of the cups.