Problem number 29 states, Christina has 8 coins whose weights in grams are different positive integers. When Christina puts any 2 coins on one side of a balance and any 2 on the other side of the balance, the side containing the heaviest of the 4 coins is always the heavier side. What is the smallest possible weight of the heaviest coin? So we have the 8 coins which we will represent their weights in grams with the variables A through H. We know that their weights are all different positive integers, so we can say that A will be the lightest coin and that H will be the heaviest coin. We don't say that these coins have the same weights less than or equal to one another because the weights cannot be equal to one another. So our goal is to minimize Z which is the heaviest of the 4 coins in every case such that any coins that weigh less than the heaviest coin plus any other coin weigh less than the heaviest coin and any other coin. So for the lightest 4 coins, the worst case scenario would be if the second and third heaviest coins were paired up against the heaviest and least heavy coin. We don't have any values so let's choose the lowest numbers for A, B, and C and use our relation to compute D. So let's choose A equals 1, B equals 2, and C equals 3. So now the equation tells us that 2 plus 3 must be less than 1 plus D. Subtracting 1 from both sides, we get that D is greater than 4. The next biggest integer after 4 is 5, so let's say that D is equal to 5. Now let's do the same thing calculating E. So again in the worst case scenario the two heaviest coins, right now C and D would be paired up against E and the lightest coin, so A. So 3 plus 5 would be less than 1 plus E. So E would be greater than 7, so we would choose E equals 8. Let's do the same for F. So we would have 5 plus 8 would be less than 1 plus F. F must be greater than 12, so let's choose the next biggest one, F equals 13. Now let's try doing it for G. So 8 plus 13 would be less than 1 plus G, which would mean that G must be greater than 20, so the next biggest one would be G equals 21. Lastly let's do H. We would get 13 plus 21 is less than 1 plus H, so H must be greater than 33, so we would choose H equals 34. And that's the minimum value of the heaviest coin. You can notice that the weights of these coins follow the Fibonacci sequence, so if you're able to figure that out you could have just computed the ninth term in the Fibonacci sequence. That's a little shortcut. So the question asked us, what is the smallest possible weight of the heaviest coin? The answer is 34. Letter C.

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