Problem proposed by Bruno Holland and Samuel Vitosa
John has one piece of chain with n links, each weighing 1g. Each time a link of a chain is broken, we get 3 pieces. For example, if the piece has 9 links and we break the fourth, then we get pieces with the following number of links: 3, 1, 5. The weight of the broken link is still 1 g.
a) If n = 8, show an example of how you can separate only two links and get pieces that can be put together so that we can hold any weight from 1g to 8g.
b) If n = 16, show an example of how you can separate only two links and get pieces that can be put together so that we can hold any weight from 1g to 16g.
Send your solution to [email protected] Don’t forget to include your name, city and school, if possible. We will publish the solution next Sunday.
Solve the puzzle 170.
Note initially that X + X + Y is less than or equal to (2×9) +8 = 26, so it will be added in the tens place with a maximum of two units. Since Y is different from Z, by adding one or two units to Y, we should get the two-digit number ZZ, but since Y + 2 is less than or equal to 9 + 2 = 11, the only possibility is that we have ZZ = 11 , so Y = 9, and for the sum 2X + Y ending in 1 and X being different from 1, we should have X = 6.
In fact, these values make the sum: 6 + 6 + 99 = 111.
Bruno Holland Professor of Federal University of Goiás and Samuel Vitosa Professor at the Federal University of Bahia.
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