Real Numbers Test - 20

Let us assume for some integer k we have 6^k = 10x + 6

∴ 6^k+1= 6.6^k = 6 (10x+6)

⇒ 6^k+1=60x+36

⇒ 6^k+1=60x+30+6

⇒ 6^k+1=10 (6x+3)+6

∴ If 6^k ends with 6,

Then 6^k+1 ends with 6 for all natural numbers.

Hence, the correct option is (B).

At any value of $n,5^n$ is not divisible by $3$, as we can check.

As we know $5$ is a prime number.

At $n=3$ we get

$\frac{5\times 5\times 5}{3}$, not divisible by $3$

So, at any value of $n,5^n$ is not divisible by $3$ as we now $5$ is not divisible by $3$.

Hence, the correct option is (B).

Factor of m is x²y⁵ = y³(x²y²)

Factor of n is x³y² = x(x²y²)

Therefore HCF (m, n)  is x²y².

Hence, the correct option is (B).

For the relation $x=qy+r, 0\le r\le y$ [By remainder theorem]

Given, $112=qx6+r$

So, here $r$ lies between $0\le r\le 6$

Hence, the possible values of $r$ is $0,1,2,3,4,5$.

Hence, the correct option is (B).

Let the three consecutive positive integers be $n,n+1$ and $n+2$.

Whenever a number is divided by $3$ , the remainder obtained is either $0,1$ or $2$ .

Therefore, $n=3p$ or $3p+1$ or $3p+2$, where $p$ is some integer.

If $n=3p,$ then $n$ is divisible by $3$ .

If $n=3p+1$, then $n+2=3p+1+2=3p+3=3(p+1)$ is divisible by $3$ .

If $n=3p+2,$ then $n+1=3p+2+1=3p+3=3(p+1)$ is divisible by $3$.

So, we can say that one of the numbers among $n,n+1$ and $n+2$ is always divisible by $3$ that is:

$n(n+1)(n+2)$ is divisible by $3$.

Similarly, whenever a number is divided by $2$ , the remainder obtained is either $0$ or $1$.

Therefore, $n=2q$ or $2q+1,$ where $q$ is some integer.

If $n=2q,$ then $n$ and $n+2=2q+2=2(q+1)$ is divisible by $2$.

If $n=2q+1$, then $n+1=2q+1+1=2q+2=2(q+1)$ is divisible by $2$.

So, we can say that one of the numbers among $n,n+1$ and $n+2$ is always divisible by $2$.

Since, $n(n+1)(n+2)$ is divisible by $2$ and $3$.

Thus, $n(n+1)(n+2)$ is divisible by $6$.

Hence, the correct option is (C).

As per Euclid's Division Lemma

If $a$ and $b$ are $2$ positive integers, then

$a=bq+r$ where \(0 \leq r

Let positive integer be $a$ and $b=2$

Hence $a=2q+r$

where $(0\le r<2)$

$r$ is an integer greater than or equal to $0$ and less than $2$.

Hence $r$ can be either $0$ or $1$.

If $r=0$

Our equation becomes

$a=2q+r$

$a=2q+0$

$a=2q$

This will always be an even integer.

If $r=1$

Our equation becomes

$a=2q+r$

$a=2q+1$

This will always be an odd integer.

Thus, $q$ is an integer.

Hence, correct option is (C).

Case i: Let $n$ be an even positive integer. When $\mathrm{n}=2\mathrm{q}$

In this case, we have ${\mathrm{n}}^2-\mathrm{n}=(2\mathrm{q}{)}^2-2\mathrm{q}=4{\mathrm{q}}^2-2\mathrm{q}=2\mathrm{q}(2\mathrm{q}-1)$

${\mathrm{n}}^2-\mathrm{n}=2\mathrm{r},$ where $\mathrm{r}=\mathrm{q}(2\mathrm{q}-1)$

${\mathrm{n}}^2-\mathrm{n}$ is divisible by $2$.

Case ii: Let $n$ be an odd positive integer. When $\mathrm{n}=2q+1$

In this case

${\mathrm{n}}^2-\mathrm{n}=(2\mathrm{q}+1{)}^2-(2\mathrm{q}+1)=(2\mathrm{q}+1)(2\mathrm{q}+1-1)=2\mathrm{q}(2\mathrm{q}+1)$

${\mathrm{n}}^2-\mathrm{n}=2\mathrm{r},$ where $\mathrm{r}=\mathrm{q}(2\mathrm{q}+1)$

${\mathrm{n}}^2-\mathrm{n}$ is divisible by $2$.

$\therefore {\mathrm{n}}^2-\mathrm{n}$ is divisible by $2$ for every integer $n$.

Hence, the correct option is (D).

Any odd positive integer is in the form of $4p+1$ or $4p+3$ for some integer $p$.

Let $n=4p+1$

$(n^2-1)=(4p+1{)}^2-1=16p^2+8p+1=16p^2+8p=8p(2p+1)$

$\Rightarrow (n^2-1)$ is divisible by $8$.

Now, $n=4p+3$

$(n^2-1)=(4p+3{)}^2-1=16p^2+24p+9-1=16p^2+24p+8=8(2p^2+3p+1)$

$\Rightarrow n^2-1$ is divisible by $8$.

Therefore, $n^2-1$ is divisible by $8$ if $n$ is an odd positive integer. 

Hence, the correct option is (A).

According to Euclid's Division algorithm, any two numbers x and y can be expressed as:

$x=qy+r$

$x>y$ ----(given)

To find the value of $x=27,y=5$, put the value in division algorithm,

$x=qy+r$

$\Rightarrow 27=q\times 5+r$

$\Rightarrow 27=5\times 5+2$

$\Rightarrow q=5, r=2$

Hence, the correct option is (B).

Any positive integer is of the form $4q+1$ or $4q+3$

As per Euclid's Division lemma. If $a$ and $b$ are two positive integers, then. $a=bq+r$

Where $0\le r<b$.

Let positive integers be a and $b=4$

Hence, $a=bq+r$

Where, $(0\le r<4)$

$R$ is an integer greater than or equal to $0$ and less than $4$. Hence, $r$ can be either $0,1,2$ and $3$.

Now, If $r=1$

Then, our be equation is becomes 

$a=bq+r$

$a=4q+1$

This will always be odd integer. 

Now, If $r=3$

Then, our be equation is becomes 

$a=bq+r$

$a=4q+3$

This will always be odd integer.

Thus, any positive odd integer is of the form $4q+1,$ or $4q+3$.

Hence, the correct option is (B).

Given : a = 3q + r

By Euclid’s division algorithm if a and b are positive integers then  there exists a unique pair of integers q and r satisfying a = bq +r  and 0 ≤ r < b.

By comparing the given equation with euclid division algorithm

We will have a = 3q + r  and b = 3  

The value of r can take 0 ≤ r < 3

Hence, the correct option is (C).