Q: For any Positive integer \$n\$, prove that \${n}^{3}-n\$ divisible by \$6\$.

The condition for any number to be divisible by \$6\$ is that the number must be individually divisible by \$3\$ and \$2.\$ Check whether \$ n^{3}-n \$ is divisible by \$3.\$\$ n^{3}-n=n\left ( n+1 \right )\left ( n-1 \right ) \$ When a number is divided by \$3\$ then by the remainder theorem, the remainder obtained is either \$0\$ or \$1\$ or \$2.\$\$ n=3p \$ or \$ n=3p+1 \$ or \$ n=3p+2 \$, where \$ p \$ is some integer.If \$ n=3p \$, then the number is divisible by \$3.\$If \$ n=3p+1 \$, then \$ n-1=3p+1-1=3p \$. The number is divisible by \$3.\$If \$ n=3p+2 \$, then \$ n+1=3p+2+1=3\left ( p+1 \right ) \$. The number is divisible by \$3.\$ So, any number in the form of \$ n^{3}-n=n\left ( n+1 \right )\left ( n-1 \right ) \$ is divisible by \$3.\$ Check whether \$ n^{3}-n \$ is divisible by \$2.\$When a number is divided by \$2\$, the remainder obtained is either \$0\$ or \$1\$ by the remainder theorem.\$ n=2p \$ or \$ n=2p+1 \$, where \$ p \$ is some integer.If \$ n=2p \$, then the number is divisible by \$2.\$If \$ n=2p+1 \$ then \$ n-1=2p+1-1=2p \$. The number is divisible by \$2.\$ So, any number in the form of \$ n^{3}-n=n\left ( n+1 \right )\left ( n-1 \right ) \$ is divisible by \$2.\$ Since, the given number \$ n^{3}-n=n\left ( n+1 \right )\left ( n-1 \right ) \$ is divisible by both \$3\$ and \$2.\$ Therefore, according to the divisibility rule of 6, the given number is divisible by \$6\$. Hence, \$ n^{3}-n=n\left ( n+1 \right )\left ( n-1 \right ) \$ is divisible by \$6.\$

Q: Round off the given numbers to the nearest tens, hundreds and thousands.

\$75847\$ - Nearest Ten: \$75850\$\$75847\$ - Nearest Hundred: \$75800\$\$75847\$ - Nearest Thousand: \$76000\$\$75847\$ - Nearest Ten Thousand: \$80000\$.

Q: Show that only one of the number \$n, n+2\$ and \$n+4\$ is divisible by \$3\$.

We know that any positive integer of the form \$3q\$ or, \$3q+1\$ or \$3q+2\$ for some integer \$q\$ and one and only one of these possibilities can occur.So, we have following cases:Case-I: When \$n = 3q\$In this case, we have\$n= 3q\$, which is divisible by \$3\$Now, \$n = 3q\$\$n+2 = 3q+2\$\$n+2\$ leaves remainder \$2\$ when divided by \$3\$Again, \$n = 3q\$\$n+4 = 3q+4=3(q+1)+1\$\$n+4\$ leaves remainder \$1\$ when divided by \$3\$\$n+4\$ is not divisible by \$3\$.Thus, \$n\$ is divisible by \$3\$ but \$n+2\$ and \$n+4\$ are not divisible by \$3\$. Case-II: when \$n = 3q+1\$In this case, we have\$n= 3q+1,\$\$n\$ leaves remainder \$1\$ when divided by \$3\$.\$n\$ is divisible by \$3\$Now, \$n = 3q+1\$\$n+2 = (3q+1)+2=3(q+1)\$\$n+2\$ is divisible by \$3\$.Again, \$n = 3q+1\$\$n+4 = 3q+1+4=3q+5=3(q+1)+2\$\$n+4\$ leaves remainder \$2\$ when divided by \$3\$\$n+4\$ is not divisible by \$3\$.Thus, \$n+2\$ is divisible by \$3\$ but \$n\$ and \$n+4\$ are not divisible by \$3\$. Case-III: When \$n= 3q+2\$In this case, we have\$n= 3q+2\$\$n\$ leaves remainder \$2\$ when divided by \$3\$.\$n\$ is not divisible by \$3\$.Now, \$n = 3q+2\$\$n+2 = 3q+2+2=3(q+1)+1\$\$n+2\$ leaves remainder \$1\$ when divided by \$3\$\$n+2\$ is not divisible by \$3\$.Again, \$n = 3q+2\$\$n+4 = 3q+2+4=3(q+2)\$\$n+4\$ is divisible by \$3\$.Hence, \$n+4\$ is divisible by \$3\$ but \$n\$ and \$n+2\$ are not divisible by \$3\$.

Q: Find the values of the letters in the following and give reasons for the steps involved.\$\ \ \ \ \ \ 4\ \ A\\\underline{\ +\ 9\ \ 8\ \ }\\\underline{\ \ C \ \ B\ \ 3\ \ }\$

\$A + 8 = 3\$, then A has to be 5 because unit digit of the sum has to be 3. There is a carry-over of \$1\$.So, \$4 + 9 + 1 = 14\$\$A = 5, B = 4\$ and \$C = 1\$

Q: If \$21y5\$ is multiple of \$9\$, where \$y\$ is a digit, what is the value of \$y\$?

Since the number \$21y5\$ is a multiple of \$9\$.So, the sum of its digits \$2 + 1 + y + 5 = 8 + y\$ is a multiple of \$9\$.\$\therefore (8 + y)\$ is either \$0\$ or \$9\$ or \$18\$ or ...Since \$y\$ is a digit, so \$(8 + y)\$ must be equal to \$9\$.i.e., \$8 + y = 9\$\$ \Rightarrow y = 9 -8 = 1\$

Q: Round these numbers to the nearest tens.\$28, 32, 52, 41, 39, 48, 64, 59, 99, 215, 1453, 2936\$.

\$28=30\$\$32=30\$\$52=50\$\$41=40\$\$39=40\$\$48=50\$\$64=60\$\$59=60\$\$99=100\$\$215=220\$\$1453=1450\$\$2936=2940\$.

Q: The value of A is :

As the right hand digit of the sum of \$A + B + C = C,\$it means that \$A+ B = 10\$ so \$A\$ or \$B,\$ none can be \$0.\$Also, the sum of \$A + B + C\$ can not exceed \$19.\$ So, \$1\$ is carried to the second column. Now, \$1 +A + B + C\$ gives \$A\$ as the right digit of the sum. So, \$A= C +1.\$ Further, again \$1\$ is carried over to the third column as Acannot be 0 and the sum of \$1 + A + B + C\$ can be at most 19.On adding \$A + B + C + 1\$ we get BA, thus here \$B = 1.\$Using \$A+ B = 10,A= 10 -1= 9\$Also, \$A= C + 1\$\$\Rightarrow C =A-1= 9-1=8.\$So, \$A= 9, B = 1\$ and \$C = 8\$ satisfy all the conditions.So, option A is correct.

Q: In the natural numbers from \$10\$ to \$250\$, how many are divisible by \$4\$?

\$\textbf{Step-1: Forming the Arithmetic Progression}\$ \$\text{Now, here the smallest number divisible by}\$ \$4\$ \$\text{is}\$ \$12.\$ \$a=12,d=4\$\$\textbf{Step-2: Finding the value of n}\$ \$\text{The}\$ \$n^{th}\$ \$\text{term in an A.P. is given by}\$ \$t_n=a+(n-1)d\$ \$248=12+(n-1)4\$ \$248=12+4n-4\$ \$4n=248-8\$ \$4n=240\$ \$n=60\$\$\textbf{Hence, between 10 to 250, 60 numbers are divisible by 4.}\$

Question 1

For any Positive integer $n$, prove that ${n}^{3}-n$ divisible by $6$.

Accepted Answer

The condition for any number to be divisible by $6$ is&#10;that the number must be individually divisible by $3$ and $2.$ Check whether $ n^{3}-n $ is divisible by $3.$$ n^{3}-n=n\left ( n+1 \right )\left ( n-1 \right )&#10;$ When a number is divided by $3$ then by the remainder&#10;theorem, the remainder obtained is either $0$ or $1$ or $2.$$ n=3p $ or $ n=3p+1 $ or $ n=3p+2 $, where $&#10;p $ is some&#10;integer.If $ n=3p $, then the number is divisible by $3.$If $ n=3p+1 $, then $ n-1=3p+1-1=3p $. The&#10;number is divisible by $3.$If $ n=3p+2 $, then $ n+1=3p+2+1=3\left ( p+1&#10;\right ) $. The number is divisible by $3.$ So, any number in the form of $ n^{3}-n=n\left (&#10;n+1 \right )\left ( n-1 \right ) $ is divisible by $3.$ Check whether $ n^{3}-n $ is divisible by $2.$When a number is divided by $2$, the remainder&#10;obtained is either $0$ or $1$ by the remainder theorem.$ n=2p $ or $ n=2p+1 $, where $&#10;p $  is some&#10;integer.If $ n=2p $, then the number is divisible by $2.$If $ n=2p+1 $ then $ n-1=2p+1-1=2p $. The number&#10;is divisible by $2.$ So, any number in the form of $ n^{3}-n=n\left (&#10;n+1 \right )\left ( n-1 \right ) $ is divisible by $2.$ Since, the given number $ n^{3}-n=n\left ( n+1&#10;\right )\left ( n-1 \right ) $ is divisible by both $3$ and $2.$ Therefore, according&#10;to the divisibility rule of 6, the given number is divisible by $6$. &#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;Hence, $ n^{3}-n=n\left ( n+1 \right )\left ( n-1&#10;\right ) $ is divisible by $6.$

Question 2

Round off the given numbers to the nearest tens, hundreds and thousands.

Accepted Answer

$75847$ - Nearest Ten: $75850$$75847$ - Nearest Hundred: $75800$$75847$ - Nearest Thousand: $76000$$75847$ - Nearest Ten Thousand: $80000$.

Question 3

Show that only one of the number $n, n+2$ and $n+4$ is divisible by $3$.

Accepted Answer

We know that any positive integer of the form $3q$ or, $3q+1$ or $3q+2$ for some integer $q$ and one and only one of these possibilities can occur.So, we have following cases:Case-I: When $n = 3q$In this case, we have$n= 3q$, which is divisible by $3$Now, $n = 3q$$n+2 = 3q+2$$n+2$ leaves remainder $2$ when divided by $3$Again, $n = 3q$$n+4 = 3q+4=3(q+1)+1$$n+4$ leaves remainder $1$ when divided by $3$$n+4$ is not divisible by $3$.Thus, $n$ is divisible by $3$ but $n+2$ and $n+4$ are not divisible by $3$. Case-II: when $n = 3q+1$In this case, we have$n= 3q+1,$$n$ leaves remainder $1$ when divided by $3$.$n$ is divisible by $3$Now, $n = 3q+1$$n+2 = (3q+1)+2=3(q+1)$$n+2$ is divisible by $3$.Again, $n = 3q+1$$n+4 = 3q+1+4=3q+5=3(q+1)+2$$n+4$ leaves remainder $2$ when divided by $3$$n+4$ is not divisible by $3$.Thus, $n+2$ is divisible by $3$ but $n$ and $n+4$ are not divisible by $3$. Case-III: When $n= 3q+2$In this case, we have$n= 3q+2$$n$ leaves remainder $2$ when divided by $3$.$n$ is not divisible by $3$.Now, $n = 3q+2$$n+2 = 3q+2+2=3(q+1)+1$$n+2$ leaves remainder $1$ when divided by $3$$n+2$ is not divisible by $3$.Again, $n = 3q+2$$n+4 = 3q+2+4=3(q+2)$$n+4$ is divisible by $3$.Hence, $n+4$ is divisible by $3$ but $n$ and $n+2$ are not divisible by $3$.

Question 4

$n^2 - 1$ is divisible by $8$, if $n$ is ___ number.

Accepted Answer

Any odd positive integer is in the form of 4p + 1 or 4p+ 3 for some integer p. Let $n = 4p+ 1$, $(n^2 &#8211; 1) = (4p + 1)^2 &#8211; 1 = 16p^2 + 8p + 1 = 16p^2 + 8p = 8p (2p + 1)$$\Rightarrow  (n^2 &#8211; 1)$ is divisible by 8. $(n^2 &#8211; 1) = (4p + 3)^2 &#8211; 1 = 16p^2 + 24p + 9 &#8211; 1 = 16p^2 + 24p + 8 = 8(2p^2 + 3p + 1)$$\Rightarrow  n^2&#8211; 1$ is divisible by 8. Therefore, $n^2&#8211; 1$ is divisible by 8 if n is an odd positive integer.

Question 5

Find the values of the letters in the following and give reasons for the steps involved.$\ \ \ \ \ \ 4\ \ A\\underline{\ +\ 9\ \ 8\ \ }\\underline{\ \ C \ \ B\ \ 3\ \ }$

Accepted Answer

$A + 8 = 3$, then A has to be 5 because unit digit of the sum has to be 3. There is a carry-over of $1$.So, $4 + 9 + 1 = 14$$A = 5, B = 4$ and $C = 1$

Question 6

If $21y5$ is multiple of $9$, where $y$ is a digit, what is the value of $y$?

Accepted Answer

Since the number $21y5$ is a multiple of $9$.So, the sum of its digits $2 + 1 + y + 5 = 8 + y$ is a multiple of $9$.$\therefore (8 + y)$ is either $0$ or $9$ or $18$ or ...Since $y$ is a digit, so $(8 + y)$ must be equal to $9$.i.e., $8 + y = 9$$ \Rightarrow y = 9 -8 = 1$

Question 7

Round these numbers to the nearest tens.$28, 32, 52, 41, 39, 48, 64, 59, 99, 215, 1453, 2936$.

Accepted Answer

$28=30$$32=30$$52=50$$41=40$$39=40$$48=50$$64=60$$59=60$$99=100$$215=220$$1453=1450$$2936=2940$.

Question 8

The value of A is :

Accepted Answer

As the right hand digit of the sum of $A + B + C = C,$it means that $A+ B = 10$ so $A$ or $B,$ none can be $0.$Also, the sum of $A + B + C$ can not exceed $19.$ So, $1$ is carried to the second column. Now, $1 +A + B + C$ gives $A$ as the right digit of the sum. So, $A= C +1.$ Further, again $1$ is carried over to the third column as Acannot be 0 and the sum of $1 + A + B + C$ can be at most 19.On adding $A + B + C + 1$ we get BA, thus here $B = 1.$Using $A+ B = 10,A= 10 -1= 9$Also, $A= C + 1$$\Rightarrow C =A-1= 9-1=8.$So, $A= 9, B = 1$ and $C = 8$ satisfy all the conditions.So, option A is correct.

Question 9

In the natural numbers from $10$ to $250$, how many are divisible by $4$?

Accepted Answer

$\textbf{Step-1: Forming the Arithmetic Progression}$                $\text{Now, here the smallest number divisible by}$ $4$ $\text{is}$ $12.$                $a=12,d=4$$\textbf{Step-2: Finding the value of n}$                $\text{The}$ $n^{th}$ $\text{term in an A.P. is given by}$                $t_n=a+(n-1)d$                $248=12+(n-1)4$                $248=12+4n-4$                $4n=248-8$                $4n=240$                $n=60$$\textbf{Hence, between 10 to 250, 60 numbers are divisible by 4.}$

Question 10

State whether true or false:If a number is divisible by 9, it must be divisible by 3.

Accepted Answer

Since 9 is divisible by 3, every number divisible by 9 will be also be divisible by 3. This happens because in the divisibilty test formula of 9, the sum of the digits should be divisible by 9. If this sum is divisible by 9, it automatically is divisible by 3 since 3 is a factor of 9.