Percentages calculation challenge

Algebra Level 4

1 000...000 ⏞ 2016 x digits × 1 %%%...%%% ⏞ 888 x terms of % 1 0 2016 = 1 000...000 ⏟ 202 x digits % \frac{1\overbrace{\text{000...000}}^{2016x\ \text{digits}} \times 1\overbrace{\text{\%\%\%...\%\%\%}}^{888x\ \text{terms of \%}}}{10^{2016}}=1\underbrace{\text{000...000}}_{202x\ \text{digits}}\text{\%}

Find x x which satisfies the above equation.


The answer is 53.

Tommy Li by Tommy Li , 22, Hong Kong

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1 solution

Tommy Li
Jul 3, 2016

1 000...000 ⏟ 2016 x digits × 1 %%%...%%% ⏟ 888 x terms of % 1 0 2016 = 1 000...000 ⏟ 202 x digits % \dfrac{1\underbrace{\text{000...000}}_{2016x\ \text{digits}} \times 1\underbrace{\text{\%\%\%...\%\%\%}}_{888x\ \text{terms of \%}}}{10^{2016}}=1\underbrace{\text{000...000}}_{202x\ \text{digits}}\text{\%}

1 0 2016 x × 1 0 888 x × ( − 2 ) 1 0 2016 = 1 0 202 x − 2 \dfrac{10^{2016x} \times 10^{888x \times (-2)}}{10^{2016}}=10^{202x-2}

1 0 2016 x + 888 x × ( − 2 ) − 2016 = 1 0 202 x − 2 10^{2016x+888x\times (-2)-2016} =10^{202x-2}

2016 x + 888 x × ( − 2 ) − 2016 = 202 x − 2 2016x+888x \times (-2)-2016=202x-2

38 x = 2014 38x=2014

x = 53 x=53

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Exactly way of solving - Great !

Maybe you should include a part where you substitute your exponents or set both sides in the log of 10 to make it more obvious to others - just a suggestion ;)

Lukas Henke - 4 years, 11 months ago

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