Decimal-Integers!

Algebra Level 2

N = x ⏟ ’x’ have n-no of digits 999 … 999 ⏟ n − d i g i t s N = \large \dfrac{\underbrace{x}_{\text{'x' have n-no of digits}}}{\underbrace{999\ldots999}_{n-digits}} Is it possible to identify the decimal expansion of numbers, as stated above, without actual division?

No Yes

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Viki Zeta by Viki Zeta , 19, India

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1 solution

Viki Zeta
Sep 17, 2016

Look at the following examples: 1 9 = 0.1111 … 10 99 = 0.10101010 … 581 999 = 0.581581581 … 1945 9999 = 0.194519451945 … Now, let’s see why this happens N = a 10 + b 100 + c 1000 + a 10000 + … (or) 0. a b c a b c … 1000 N = 100 a + 10 b + c + a 10 + b 100 + c 1000 + … 1000 N = a + b + c + N 999 N = 100 a + 10 b + c 999 N = a b c ‾ since general form of 3 digit number is 100a + 10b + c _____________________________________________________________________________________ This similarly follows for any n-digit number N = a 10 + b 100 + c 1000 + … ⏟ n-digits + a 1 0 n + 1 + b 1 0 n + 2 + … ⏟ n-digits 1 0 n N = 1 0 n − 1 a + 1 0 n − 2 b + 1 0 n − 3 c + … ⏟ n-digits + a 10 + b 100 + … ⏟ n-digits 1 0 n N = 1 0 n − 1 a + 1 0 n − 2 b + 1 0 n − 3 c + … ⏟ n-digits + N 1 0 n N − N = 1 0 n − 1 a + 1 0 n − 2 b + 1 0 n − 3 c + … ⏟ n-digits ( 999 … 9 ⏟ n − d i g i t s ) N = 1 0 n − 1 a + 1 0 n − 2 b + 1 0 n − 3 c + … ⏟ n-digits ( 999 … 9 ⏟ n − d i g i t s ) N = a b c d … ‾ ⏟ n − d i g i t s using general form of n-digit number N = a b c d … ‾ ⏟ n − d i g i t s 999 … 9 ⏟ n − d i g i t s \text{Look at the following examples:} \\ \dfrac{1}{9} = 0.1111\ldots \\ \dfrac{10}{99} = 0.10101010\ldots \\ \dfrac{581}{999} = 0.581581581\ldots \\ \dfrac{1945}{9999} = 0.194519451945\ldots \\ \text{Now, let's see why this happens} \\ N = \dfrac{\color{#D61F06}{a}}{\color{#3D99F6}{10}} + \dfrac{\color{#D61F06}{b}}{\color{#3D99F6}{100}} + \dfrac{\color{#D61F06}{c}}{\color{#3D99F6}{1000}} + \dfrac{\color{#D61F06}{a}}{\color{#3D99F6}{10000}} + \ldots \text{ (or) }\color{#D61F06}{0}.\color{#3D99F6}{abcabc}\ldots \\ \color{#3D99F6}{1000}N = \color{#D61F06}{100}a + \color{#D61F06}{10}b + c + \dfrac{a}{\color{#D61F06}{10}} + \dfrac{b}{\color{#D61F06}{100}} + \dfrac{c}{\color{#D61F06}{1000}} + \ldots \\ 1000N = a + b + c + \color{#3D99F6}{N} \\ 999N = 100a + 10b + c \\ \color{#3D99F6}{999}N = \color{#D61F06}{\overline{abc}} \text{ since general form of 3 digit number is 100a + 10b + c} \\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\\ \text{This similarly follows for any n-digit number} \\ N = \underbrace{\dfrac{\color{#D61F06}{a}}{\color{#3D99F6}{10}} + \dfrac{\color{#D61F06}{b}}{\color{#3D99F6}{100}} + \dfrac{\color{#D61F06}{c}}{\color{#3D99F6}{1000}} + \ldots}_{\text{n-digits}} + \underbrace{\dfrac{\color{#D61F06}{a}}{\color{#3D99F6}{10^{n+1}}} + \dfrac{\color{#D61F06}{b}}{\color{#3D99F6}{10^{n+2}}} + \ldots}_{\text{n-digits}} \\ \color{#3D99F6}{10^{n}}N = \underbrace{10^{\color{#3D99F6}{n-1}}\color{#D61F06}{a} + 10^{\color{#3D99F6}{n-2}}\color{#D61F06}{b} + 10^{\color{#3D99F6}{n-3}}\color{#D61F06}{c} + \ldots}_{\text{n-digits}} + \underbrace{\dfrac{\color{#D61F06}{a}}{\color{#3D99F6}{10}} + \dfrac{\color{#D61F06}{b}}{\color{#3D99F6}{100}} + \ldots}_{\text{n-digits}} \\ \color{#3D99F6}{10^{n}}N = \underbrace{10^{\color{#3D99F6}{n-1}}\color{#D61F06}{a} + 10^{\color{#3D99F6}{n-2}}\color{#D61F06}{b} + 10^{\color{#3D99F6}{n-3}}\color{#D61F06}{c} + \ldots}_{\text{n-digits}} + N \\ \color{#3D99F6}{10^{n}}N - N = \underbrace{10^{\color{#3D99F6}{n-1}}\color{#D61F06}{a} + 10^{\color{#3D99F6}{n-2}}\color{#D61F06}{b} + 10^{\color{#3D99F6}{n-3}}\color{#D61F06}{c} + \ldots}_{\text{n-digits}} \\ (\underbrace{999\ldots9}_{n-digits})N = \underbrace{10^{\color{#3D99F6}{n-1}}\color{#D61F06}{a} + 10^{\color{#3D99F6}{n-2}}\color{#D61F06}{b} + 10^{\color{#3D99F6}{n-3}}\color{#D61F06}{c} + \ldots}_{\text{n-digits}} \\ (\underbrace{999\ldots9}_{n-digits})N = \underbrace{\overline{abcd\ldots}}_{n-digits} \text{ using general form of n-digit number} \\ N = \dfrac{\underbrace{\overline{abcd\ldots}}_{n-digits}}{\underbrace{999\ldots9}_{n-digits}}

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