Tag: fundemental theorem of arithmetic

Quizzes (0) Questions (21)

Questions Related to fundemental theorem of arithmetic

Multiple choice maths real number fundemental theorem of arithmetic real numbers on number line fundamental theorem of arithmetic

 The square of any positive odd integer for some integer $ m$ is of the form 

  1. 7m+1

  2. 8m+1

  3. 8m+3

  4. 7m+2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know that any positive odd integer a is of the form 4q + 1 or 4q + 3 where q is some integer.
Case-1: $a=4q+1$
$\Rightarrow a^2=16q^2+8q+1=8(2q^2+q)+1$
$=8m+1$,
where $m=2q^2+q=integer$.
Case-2: $a=4q+3$
$\Rightarrow a^2=16q^2+24q+9$
$=8(2q^2+3q+1)+1=8m+1$,
where $m=2q^2+3q+1=integer$.
Hence square of any positive odd integer is of the form $8m+1$ for some integer m.

Multiple choice maths real number fundemental theorem of arithmetic real numbers on number line fundamental theorem of arithmetic

We know that any odd positive integer is of the form $4q + 1 $ or $4q + 3$ for some integer $q.$
Thus, we have the following two cases.

  1. $n^2-1$ is divisible by 8
  2. $n^2+1$ is divisible by 8
  3. $n-1$ is divisible by 8
  4. $n+1$ is divisible by 8
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When $n=4q+1$
In this case, we have
$n^2-1=(4q+1)^2-1=16q^2+8q+1-1$
$=8q(2q+1)=8r$ where $r=q(2q+1)$ is an integer
$\Rightarrow n^2-1$ is divisible by 8.
Case-II: When $n=4q+3$
In this case, we have
$n^2-1=(4q+3)^2-1=16q^2+24q+9-1=16q2+24q+8$
$=8(2q^2+3q+1)=8(2q+1)(q+1)$
$=8r$ where $r=(2q+1)(q+1)$ is an integer.
$\Rightarrow n^2-1$ is divisible by 8
Hence $n^2-1$ is divisible by 8.

Multiple choice maths real number fundemental theorem of arithmetic real numbers on number line fundamental theorem of arithmetic

If any positive' even integer is of the form 4q or 4q + 2, then q belongs to:

  1. whole number

  2. rational number

  3. real number

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let a be any positive even integer and b=$4$.Then by division algorithm,
$a=4q+r$ for some integer $q \ge 0$ and $r=0,1,2,3$
So,
$a=4q$ or, 
$4q+1$,
$4q+2$
$4q+3$
Because $0 \ge r \ge 4$
Now,
$4q$i.e $2(2q)$ is an even number
$\therefore$ $4q+1$ is an odd number
$4q+2$ i.e. $2(2q+1)$ is an even number
$\therefore (4q+2)+1=4q+3$ is an odd number
Thus, We can say that any even integer can be written as in the form of $4q, 4q+2$ where $q$ is the whole number

Multiple choice maths real number fundemental theorem of arithmetic real numbers on number line fundamental theorem of arithmetic

A number when divided by  $156$  gives  $29$  as remainder. If the same number is divided by  $13$ , what will be the remainder?

  1. $4$
  2. $3$
  3. $5$
  4. $6$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given number = 156x + 29
=156x + 26 + 3
= 13 $\displaystyle \times $ 12x + 13 $\displaystyle \times $ 2 + 3
= 13(12x + 2) + 3
$\displaystyle \therefore $  When the number is divided by 13 the remainder will be 3

Multiple choice maths real number fundemental theorem of arithmetic real numbers on number line fundamental theorem of arithmetic

In a question on division the divisor is  $7$  times the quotient and  $3 $ times the remainder. If the remainder is  $28$  then what is the dividend?

  1. $1008$
  2. $1516$
  3. $1036$
  4. $2135$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Divisor $ = 3$ $\displaystyle \times $ Remainder = 3$\displaystyle \times $ $28=84$
Quotient = $\displaystyle \frac{1}{7}\times Divisor=\frac{1}{7}\times 84=12$
$\displaystyle \therefore $ Dividend = Divisor $\displaystyle \times $ Quotient + Remainder
$= 84$ $\displaystyle \times $  $12+ 28 = 1008 + 28 = 1036$

Multiple choice maths real number fundemental theorem of arithmetic real numbers on number line fundamental theorem of arithmetic

 One and only one out of  $n, n + 4, n + 8, n + 12\  and \ n + 16 $ is ......(where n is any positive integer)

  1. Divisible by 5

  2. Divisible by 4

  3. Divisible by 10

  4. Divisible by 12
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know that any positive integer is of the form 5q, 5q + 1 or 5q + 2, 5q + 3 or 5q + 4 for some integer q and one and onlyone of these possibilities can occur. So, we have the following cases:
Case-I: When $n=5q$
In this case, we have
$n=5q$, which is divisible by 5
Now, $n=5q$
$\Rightarrow n+4=5q+4$
$\Rightarrow n+4$ leaves remainder 4 when divided by 5
$\Rightarrow n+4$ is not divisible by 5.
Now $n+8=5q+8=5(q+1)+3=5m+3$, m is an integer.
Clearly, n+8 is not divisible by 5.
Again, $n+12=5q+12=5(q+2)+2=5m+2$, m in an integer.
Clearly n+12 is not divisible by 5.
Now $n+16=5q+16=5(q+13)+1=5m+1$, m is an integer
$\Rightarrow n+16$ is not divisible by 5
Thus, if n = 5q only one out of n, n + 4, n + 8, n +
12 and n + 16 is divlsible by 5,
Similarly, this result can be proved for the rest of .
the cases.

Multiple choice maths real number fundemental theorem of arithmetic real numbers on number line fundamental theorem of arithmetic

Sum of digits of the smallest number by which $1440$ should be multiplied so that it becomes a perfect cube is

  1. $4$
  2. $6$
  3. $7$
  4. $8$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\because  1440 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5$
The pairs of $2$ and $3$ and $5$ are incomplete to make it perfect cube.
$\therefore$ Smallest  number  to  be  multiplied  $=  2 \times 3 \times 5 \times 5 = 150$

$\therefore$  The  sum  of  its  digits  $= 1+5+0 = 6.$