Problem from IYMC easy

Number Theory Level 2

sin ⁡ ( π + log ⁡ 2 ( 2 π ⋅ 2 π ) 2 5 − 2 4 − 2 3 ) 2 4 − 1 2 3 / 2 3 + log ⁡ 2 ( log ⁡ 3 9 15 + π 1 + ( − 1 ) 17 + 1 ) + ( − 1 ) 5 + ( − 1 ) 27 ( − 1 ) 766 = ? \small \sqrt{\sin \Bigl (\frac{π+\log_2(\sqrt{2^π\cdot 2^π})}{2^5-2^4-2^3}\Bigr)\sqrt[3]{\frac{2^{4-1}}{2^{3/2}}}+\log_2\Bigl(\log_39^{15}+π^{1+(-1)^{17}}+ 1\Bigr)+\frac{(-1)^5+(-1)^{27}}{(-1)^{766}} }= \ ?


The answer is 2.

Arifin Ikram by arifin ikram , 17, Bangladesh

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2 solutions

The given expression is equal to sin ⁡ π 4 × 2 3 2 3 + log ⁡ 2 ( 32 ) − 2 \sqrt {\sin \fracπ4\times \sqrt[3] {2^{\frac32}}+\log_2 (32)-2}

= 1 2 × 2 + 5 − 2 = 4 = 2 =\sqrt {\frac{1}{\sqrt 2 }\times \sqrt 2 +5-2}=\sqrt 4=\boxed 2 .

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Elijah L
Oct 24, 2020

The answer does not evaluate to 2 2 . As the question currently stands, the answer is 1 + log ⁡ 2 29 \sqrt{1+\log_2 29} .

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