
When a positive integer $m$ is divided by 7, the remainder is 4. What is the remainder when $2m$ is divided by 7?
A) 9
B) 1
C) 4
D) 6
E) 8
Answer
135.6k+ views
Hint: If dividend is double and divisor is same so remainder is also double but it should be less than divisor if it is greater than divisor then, that remainder should be again divided by divisor to get the original reminder.
Complete step by step solution:
Given $m$ is integer that is dividend,
Suppose $m=7x+r$
Where $r$ is remainder i.e. 7 and $x$is any whole number.
So,$m=7x+4$
Now multiply 2 both sides to get the value of $2m$
$\Rightarrow 2m=2\left( 7x+4 \right)$
$\Rightarrow 2m=14x+8$
So, now we divide $2m$by 7
$\Rightarrow \dfrac{2m}{7}=\dfrac{7y+{{r}_{2}}}{7}$
Where $y$is any whole number and ${{r}_{2}}$ be the remainder
$\Rightarrow \dfrac{14x+8}{7}=\dfrac{7y+{{r}_{2}}}{7}$
$\Rightarrow 7\left( \dfrac{2x}{7} \right)+\dfrac{8}{7}=\dfrac{7y+{{r}_{2}}}{7}$
$\Rightarrow 7\left( \dfrac{2x}{7} \right)+\dfrac{7+1}{7}=\dfrac{7y+{{r}_{2}}}{7}$
$\Rightarrow 7\left( \dfrac{2x+1}{7} \right)+\dfrac{1}{7}=\dfrac{7y+{{r}_{2}}}{7}$
$\Rightarrow 7\left( \dfrac{2x+1}{7} \right)+\dfrac{1}{7}=\dfrac{7y}{7}+\dfrac{{{r}_{2}}}{7}$
Now compare the equation both side and we get ${{r}_{2}}=1$
Then the remainder will be 1
Hence option (B) is correct.
Additional information:
In division we will see the relationship between the dividend, divisor, quotient and remainder. The number which we divide is called the dividend. The number by which we divide is called the divisor. The result obtained is called the quotient. The number left over is called the remainder.
Dividend = divisor × quotient + remainder
Note: Don’t confuse over dividend and divisor, doubled remainder of doubled dividend is not always a correct, it should be less than divisor.
Complete step by step solution:
Given $m$ is integer that is dividend,
Suppose $m=7x+r$
Where $r$ is remainder i.e. 7 and $x$is any whole number.
So,$m=7x+4$
Now multiply 2 both sides to get the value of $2m$
$\Rightarrow 2m=2\left( 7x+4 \right)$
$\Rightarrow 2m=14x+8$
So, now we divide $2m$by 7
$\Rightarrow \dfrac{2m}{7}=\dfrac{7y+{{r}_{2}}}{7}$
Where $y$is any whole number and ${{r}_{2}}$ be the remainder
$\Rightarrow \dfrac{14x+8}{7}=\dfrac{7y+{{r}_{2}}}{7}$
$\Rightarrow 7\left( \dfrac{2x}{7} \right)+\dfrac{8}{7}=\dfrac{7y+{{r}_{2}}}{7}$
$\Rightarrow 7\left( \dfrac{2x}{7} \right)+\dfrac{7+1}{7}=\dfrac{7y+{{r}_{2}}}{7}$
$\Rightarrow 7\left( \dfrac{2x+1}{7} \right)+\dfrac{1}{7}=\dfrac{7y+{{r}_{2}}}{7}$
$\Rightarrow 7\left( \dfrac{2x+1}{7} \right)+\dfrac{1}{7}=\dfrac{7y}{7}+\dfrac{{{r}_{2}}}{7}$
Now compare the equation both side and we get ${{r}_{2}}=1$
Then the remainder will be 1
Hence option (B) is correct.
Additional information:
In division we will see the relationship between the dividend, divisor, quotient and remainder. The number which we divide is called the dividend. The number by which we divide is called the divisor. The result obtained is called the quotient. The number left over is called the remainder.
Dividend = divisor × quotient + remainder
Note: Don’t confuse over dividend and divisor, doubled remainder of doubled dividend is not always a correct, it should be less than divisor.
Recently Updated Pages
JEE Main 2021 July 25 Shift 2 Question Paper with Answer Key

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 20 Shift 2 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

How to find Oxidation Number - Important Concepts for JEE

Half-Life of Order Reactions - Important Concepts and Tips for JEE

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

An aeroplane left 50 minutes later than its schedu-class-11-maths-JEE_Main

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Elastic Collisions in One Dimension - JEE Important Topic

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Maths Chapter 8 Sequences and Series

NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections

NCERT Solutions for Class 11 Maths Chapter 9 Straight Lines

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths Chapter 13 Statistics
