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, \({n}=3 {p}\) or \(3{p}+1\) or \(3 {p}+2\), where \({p}\) is some integer.

If \({n}=3 {p},\) then \({n}\) is divisible by \(3\) .

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

If \(n=3 p+2,\) then \(n+1=3 p+2+1=3 p+3=3(p+1)\) 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, \({n}=2 {q}\) or \(2 {q}+1,\) where \({q}\) is some integer.

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

If \({n}=2{q}+1\), then \({n}+1=2 {q}+1+1=2{q}+2=2({q}+1)\) 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

\(a=b q+r\) where \(0 \leq r

Let positive integer be \(a\) and \(b=2\)

Hence \(a=2 q+r\)

where \((0 \leq r<2)\)

\({r}\) 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

\(a=2 q+r\)

\(a=2 q+0\)

\(a=2 q\)

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 \(\mathrm n=2 q+1\)

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 \(4 p+1\) or \(4 p+3\) for some integer \(p\).

Let \({n}=4{p}+1\)

\(\left(n^{2}-1\right)=(4 p+1)^{2}-1=16 p^{2}+8 p+1=16 p^{2}+8 p=8 p(2 p+1)\)

\(\Rightarrow\left({n}^{2}-1\right)\) is divisible by \(8\).

Now, \(n=4p+3\)

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

\(\Rightarrow {n}^{2}-1\) 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 \(4 {q}+1\) or \(4 {q}+3\)

As per Euclid's Division lemma. If \(a\) and \(b\) are two positive integers, then. \(a=b q+r\)

Where \(0 \leq {r}<{b}\).

Let positive integers be a and \({b}=4\)

Hence, \(a={bq}+{r}\)

Where, \((0 \leq {r}<4)\)

\({R}\) 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 

\({a}={bq}+{r}\)

\(a=4 q+1\)

This will always be odd integer. 

Now, If \({r}=3\)

Then, our be equation is becomes 

\(a=b q+r\)

\(a=4 q+3\)

This will always be odd integer.

Thus, any positive odd integer is of the form \(4 {q}+1,\) or \(4 {q}+3\).

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).