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Progetto Olimpiadi della Matematica - GARA di FEBBRAIO

Algebra Difficulty 4.1 AIME Find the answer Italy

How much is $\sqrt[4]{2^{20}+2^{27}+2^{31}+2^{32}+2^{37}+2^{40}}$ ?

Your answer

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

Solution

The answer is 1056. Once a factor $2^{20}$ has been collected under the fourth root, one is led to compute $\sqrt[4]{1+2^{7}+2^{11}+2^{12}+2^{17}+2^{20}}$ . At this point it is possible to recognize in the expression under the root the square of the trinomial $1+2^{6}+2^{10}=(2^{5}+1)^{2}$ or to directly verify that it is the fourth power of $2^{5}+1$ . The expression as a whole is therefore equal to $2^{5}(2^{5}+1)$ , that is 1056.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from the original; metadata (topic, difficulty) added by this project.