A small sum of digits ? Really?

Level 2

If $x = 9ab$ where $a$ is an integer consists of a sequence of $2014$ eights and the integer $b$ consists of a sequence of $2014$ fives. Then the sum of digits of $x$ is ?

The answer is 18126.

1 solution

Anirudh Sreekumar
Dec 10, 2017

$\begin{aligned}x&=9ab\\ &=9\dfrac{(10^{2014}-1)\times 8}{9}\times\dfrac{(10^{2014}-1)\times 5}{9}\\ &=\dfrac{(10^{2014}-1)^2\times 40}{9}\\ &=\dfrac{(10^{4029}-2\times10^{2015}+10)\times4}{9}\\ &=\dfrac{4}{9}\left(\underbrace{99999\cdots}_{\substack{2013\text{ 9s}}}8\underbrace{\cdots00000}_{\substack{2013\text{ 0s}}}10\right)\\ &=4\left(\underbrace{11111\cdots}_{\substack{2013\text{ 1s}}}0\underbrace{\cdots888888}_{\substack{2013\text{ 8s}}}90\right)\\ &=\left(\underbrace{44444\cdots}_{\substack{2013\text{ 4s}}}3\underbrace{\cdots555555}_{\substack{2013\text{ 5s}}}60\right)\\\\ \implies S(n)&=2013(5+4)+6+3\\ &=2014\times 9\\ &=\color{#EC7300}\boxed{\color{#333333}18126} \end{aligned}$

Can you please explain that how did you write $a$ as $\dfrac{10^{2014} - 1}{9} \times 8$

Vinayak Bansal - 3 years, 6 months ago

$\begin{aligned}10^{2014}-1&=\underbrace{99999\cdots}_{\substack{2014\text{ 9s}}}\\ \dfrac{\left(\underbrace{99999\cdots}_{\substack{2014\text{ 9s}}}\right)}{9}&=\underbrace{11111\cdots}_{\substack{2014\text{ 1s}}}\\ 8\times \underbrace{11111\cdots}_{\substack{2014\text{ 1s}}}&=\underbrace{888888\cdots}_{\substack{2014 \text{ 8s}}}=a\end{aligned}$

Anirudh Sreekumar - 3 years, 6 months ago

Thanks for your explanation

Vinayak Bansal - 3 years, 5 months ago

A small sum of digits ? Really?

The answer is 18126.

1 solution

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