Maths Olympiad Prep

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Algebra Difficulty 4.6 AIME Find the answer United States

Problem:

Ethan initially writes some numbers on a blackboard, each of which is either a 33 or a 55. He then repeatedly picks two numbers and replaces them with their sum, difference, product, or quotient (if the divisor is nonzero). Let f(n)f(n) denote the minimum number of numbers Ethan must initially write for him to be able to eventually write the number nn. For example, f(2025)6f(2025) \leq 6 because Ethan could start with 33, 33, 33, 33, 55, and 55 on the board, then repeatedly multiply two numbers at a time to eventually get 20252025.

Submit a comma-separated ordered 8-tuple of integers corresponding to the values of f(164)f(164), f(187)f(187), f(191)f(191), f(224)f(224), f(255)f(255), f(286)f(286), f(374)f(374), and f(479)f(479), in that order, or an X for any value you wish to leave blank. For instance, if you think f(164)=9f(164) = 9 and f(224)=8f(224) = 8, you should submit "9, X, X, 8, X, X, X, X". You will earn 0.6W(C+1)24\left|0.6^{W} \cdot \frac{(C + 1)^{2}}{4}\right| points, where CC is the number of correct answers you submit and WW is the number of incorrect (non-blank) answers.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution:

The following expressions represent optimal ways for Ethan to make each of the 8 given numbers.

164=3(5(5+5)+3)+5164 = 3(5(5 + 5) + 3) + 5

187=3(3+5)(3+5)5187 = 3(3 + 5)(3 + 5) - 5

191=55(3+5)33191 = 5 \cdot 5(3 + 5) - 3 \cdot 3

224=(3+5)(55+3)224 = (3 + 5)(5 \cdot 5 + 3)

255=55(5+5)+5255 = 5 \cdot 5(5 + 5) + 5

286=(3+5+5)(553)286 = (3 + 5 + 5)(5 \cdot 5 - 3)

374=35553/3=35555/5374 = 3 \cdot 5 \cdot 5 \cdot 5 - 3 / 3 = 3 \cdot 5 \cdot 5 \cdot 5 - 5 / 5

479=(553)(553)5=(355)(355)5479 = (5 \cdot 5 - 3)(5 \cdot 5 - 3) - 5 = (3 - 5 \cdot 5)(3 - 5 \cdot 5) - 5

It can be checked by code that these are optimal.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.