Name: Modular Arithmetic
Uploaded: 2019-07-19
Duration: 4 min 48 s
Description: Modular arithmetic is a fundamental component of cryptography. This video explains the basics of modular arithmetic with the help of a few simple examples.

Question

Solve the equation

[x] \{x\} + x = 2 \{x\} + 10

, where

[\cdot]

denotes the greatest integer function and

\{\cdot\}

denotes the fractional part function.

Accepted Answer

By using $x = [x] + \{x\},$ the given equation $[x] \{x\} + x = 2 \{x\} + 10$ becomes

$[x] \{x\} + [x] + \{x\} = 2 \{x\} + 10$

$\Rightarrow [x] \{x\} + [x] - \{x\} = 10,$

$\Rightarrow [x] \{x\} - \{x\} + [x] - 1 = 10 - 1,$

$\Rightarrow ([x] - 1) \{x\} + ([x] - 1) = 9,$

$\Rightarrow ([x] - 1) (\{x\} + 1) = 9 .$

Since, $[x] - 1$ is an integer, so $\{x\} + 1$ is a rational number.

Let $\{x\} = \frac{m}{n},$ where $m$ and $\{\cdot\}$ are integers and we know that $0 \leq \{x\} < 1$

$\Rightarrow 0 \leq \frac{m}{n} < 1,$ $\Rightarrow 0 \leq m < n,$

Then, we get $([x] - 1) (\frac{m}{n} + 1) = 9$

$\Rightarrow ([x] - 1) (m + n) = 9 n . . . (i)$

When, $m = 0,$ then $\{x\} = 0,$ so $x = 10 .$

When, $m > 0,$ let $H . C . F . (m, n) = 1,$ then $H . C . F . (m + n, n) = 1,$

So, from equation $(i)$ $(m + n)$ be a divisor of $9$ i.e., $m + n$ can take values $1, 3$ or $9 .$

But $m + n = 1$ is not possible because of $m > 0$ and $m < n,$ then the possible values for $m = 0, n = 1$ and $\{x\} = 0,$ which we had already taken.

$(i)$ When, $m + n = 3,$ then $m = 1, n = 2,$

So $\{x\} = \frac{m}{n} = \frac{1}{2},$

Put this value in the equation $(i)$ , to get

$([x] - 1) (3) = 9 \times 2$ $\Rightarrow [x] - 1 = 6$ $\Rightarrow [x] = 7,$

$\Rightarrow x = [x] + \{x\} = 7 + \frac{1}{2} = 7 \frac{1}{2} .$

$(i)$ When $m + n = 9,$ then there are three possible cases

$m = 1, n = 8$ $\Rightarrow \{x\} = \frac{1}{8},$

Put this value in the equation $(i)$ , to get

$([x] - 1) (9) = 9 \times 8$ $\Rightarrow [x] - 1 = 8$ $\Rightarrow [x] = 9,$

$\Rightarrow x = 9 + \frac{1}{8} = 9 \frac{1}{8} .$

And, $m = 2, n = 7$ $\Rightarrow \{x\} = \frac{2}{7},$

$\Rightarrow ([x] - 1) (9) = 9 \times 7$ $\Rightarrow [x] - 1 = 7$ $\Rightarrow [x] = 8,$

$\Rightarrow x = 8 + \frac{2}{7} = 8 \frac{2}{7}$

And, $m = 4, n = 5$ $\Rightarrow \{x\} = \frac{4}{5},$

$\Rightarrow ([x] - 1) (9) = 9 \times 5$ $\Rightarrow [x] - 1 = 5$ $\Rightarrow [x] = 6$

$\Rightarrow x = 6 + \frac{4}{5} = 6 \frac{4}{5} .$

Thus, there are five solutions for $x : 10, 7 \frac{1}{2}, 9 \frac{1}{8}, 8 \frac{2}{7}, 6 \frac{4}{5} .$

Solve the equation $[x] \{x\} + x = 2 \{x\} + 10$ , where $[\cdot]$ denotes the greatest integer function and $\{\cdot\}$ denotes the fractional part function.

Important Questions on Number System

Learn and Prepare for any exam you want !

Solve the equation xx+x=2x+10, where · denotes the greatest integer function and · denotes the fractional part function.

Important Questions on Number System

Learn and Prepare for any exam you want !

Solve the equation $[x] \{x\} + x = 2 \{x\} + 10$ , where $[\cdot]$ denotes the greatest integer function and $\{\cdot\}$ denotes the fractional part function.