Ancestor of "Bodmas"

Algebra Level 3

The following information shows how the expression [ 1 2 + 4 × 8 ÷ 16 1-2+4\times8\div16 ] is growing every second:


[ 1 2 + 4 × 8 ÷ 16 1-2+4\times8\div16 ] at t=1 second

[ 1 2 + ( 32 64 + 128 × 256 ÷ 512 ) + 4 × 8 ÷ 16 1-2+(32-64+128\times256\div512)+4\times8\div16 ] at t=2 second

[ 1 2 + ( 32 64 + ( 1024 2048 + 4096 × 8192 ÷ 16384 ) + 128 × 256 ÷ 512 ) + 4 × 8 ÷ 16 \large1-2+(32-64+(1024-2048+4096\times8192\div16384)+128\times256\div512)+4\times8\div16 ] at t=3 second.

Then what will be the final result after simplification of the expression that we will get at t=60 seconds by applying BODMAS and correcting it to three decimal places

If you think this does not exist then enter 8888 as your answer.

If you are getting answer like w . x y z × 1 0 a w.xyz\times10^{a} after correction to three decimal places, then enter it as w . x y z E + a w.xyzE+a .

Clarification: E E represents E notation


This problem is a part of the set All-Zebra


The answer is 6.571E+88.

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1 solution

Abhay Tiwari
Apr 26, 2016

value of first expression= 2 0 2^{0}

value of second expression= 2 0 + 2 5 2^{0}+2^{5}

value of third expression= 2 0 + 2 5 + 2 10 2^{0}+2^{5}+2^{10}

Value of the 60th expression(after one minute)= 2 0 + 2 5 + 2 10 + + 2 285 + 2 290 + 2 295 2^{0}+2^{5}+2^{10}+\dots+2^{285}+2^{290}+2^{295}

Final result of the expression= 2 0 × 1 ( 2 5 ) 60 1 2 5 2^{0}\times\frac{\large1-(2^{5})^{60}}{1-2^{5}} = 6.571 E + 88 6.571E+88

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