Two Too Many Toos

Algebra Level 2

Evaluate 2 ( 22 ) ( 2 + 2 ) ( 222 22 ) ( 2222 + 222 + 22 + 2 ) ( ( 222 ) ( 2 ( 22 ) + 2 ( 22 ) ) ( 22 ( 2 ( 22 ) + 2 ( 22 ) ) ) ( 2 ( 2 ) ( 2 ) ( 2 ) + 2 + 2 ( 2 ) ( 2 ) + 2 + 2 2 + 2 ( 2 ) ( 2 ) + 2 + 2 2 + 2 + 2 ) 2 \sqrt[2]{\dfrac {2(22)(2+2)(222-22)(2222+222+22+2)} {((222)(2(22)+2(22))-(22(2(22)+2(22)))(2^{(2)(2)(2)+2}+2^{(2)(2)+2+\frac{2}{2}}+2^{(2)(2)+2}+2^{2+2}+2)}}


The answer is 2.

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1 solution

Timothy Cao
Apr 11, 2018

Evaluating 2 ( 22 ) ( 2 + 2 ) ( 222 22 ) ( 2222 + 222 + 22 + 2 ) ( ( 222 ) ( 2 ( 22 ) + 2 ( 22 ) ) ( 22 ( 2 ( 22 ) + 2 ( 22 ) ) ) ( 2 ( 2 ) ( 2 ) ( 2 ) + 2 + 2 ( 2 ) ( 2 ) + 2 + 2 2 + 2 ( 2 ) ( 2 ) + 2 + 2 2 + 2 + 2 ) 2 \sqrt[2]{\frac {2(22)(2+2)(222-22)(2222+222+22+2)} {((222)(2(22)+2(22))-(22(2(22)+2(22)))(2^{(2)(2)(2)+2}+2^{(2)(2)+2+\frac{2}{2}}+2^{(2)(2)+2}+2^{2+2}+2)}}

( 22 ) ( 2 + 2 ) ( 222 22 ) (22)(2+2)(222-22) is equivilient to ( 222 ) ( 2 ( 22 ) + 2 ( 22 ) ) 22 ( 2 ( 22 ) + 2 ( 22 ) ) (222)(2(22)+2(22))-22(2(22)+2(22)) after factoring

Canceling them out, we have

2 ( 2222 + 222 + 22 + 2 ) ( 2 ( 2 ) ( 2 ) ( 2 ) + 2 + 2 ( 2 ) ( 2 ) + 2 + 2 2 + 2 ( 2 ) ( 2 ) + 2 + 2 2 + 2 + 2 ) 2 \sqrt[2]{\frac {2(2222+222+22+2)} {(2^{(2)(2)(2)+2}+2^{(2)(2)+2+\frac{2}{2}}+2^{(2)(2)+2}+2^{2+2}+2)}}

= 2 ( 2468 ) ( 2 10 + 2 7 + 2 6 + 2 4 + 2 ) 2 =\sqrt[2]{\frac {2(2468)} {(2^{10}+2^{7}+2^{6}+2^{4}+2)}}

= 2 ( 2468 ) 1234 2 =\sqrt[2]{\frac {2(2468)} {1234}}

= 4 ( 1234 ) 1234 2 =\sqrt[2]{\frac {4(1234)} {1234}}

= 4 2 =\sqrt[2]{4}

= 2 =2 just as you guessed.

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