Tag: fundemental theorem of arithmetic

The given statement is false.

This statement is false.

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.

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.

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

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

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 $  $12+ 28 = 1008 + 28 = 1036$

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.

Let $n = 4q + 3$. Then $2n = 8q + 6$   $= 4(2q + 1 ) + 2$.

Thus, when 2n is divided by 4, the remainder is 2.

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