1. A number $N$ is divided by $D$ resulted in quotient $q$ and remainder $r$. Then $N$ can be written as
a. $Dq - r$
b. $\dfrac{D}{q} - r$
c. $Dq + r$
d. $Dr + q$

Answer: D

Explanation:
$\quad\require{enclose}
\begin{array}{rll}
q &\\[0pt]
D \enclose{longdiv}{\;N}\kern-.2ex \\[0pt]
\underline{Dq}{\phantom{}} \\[0pt]
r\phantom{} \\[0pt]
\end{array}$
$\therefore$ $N-Dq = r$
$\therefore$ $N = Dq + r$ Note: The above result is very important and you should learn how to write a division in this format)

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2. One to ten numbers are written side by side. If this number is divided by 8 then what is the remainder?
a. 5
b. 6
c. 7
d. 0

Answer: B

Explanation:
If one to ten numbers are written side by side, last three digits are $910$.
$\quad\require{enclose}
\begin{array}{rll}
113 &\\[0pt]
8 \enclose{longdiv}{910}\kern-.2ex \\[0pt]
\underline{8}{\phantom{00}} \\[0pt]
11\phantom{0} \\[0pt]
\underline {8}{\phantom{0}} \\[0pt]
30\phantom{} \\[0pt]
\underline{24}\phantom{} \\[0pt]
\boxed6\phantom{}
\end{array}$

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3. What is the value of $k$ if $8975k$ is divisible by 7
a. 11
b. 16
c. 7
d. 9

Answer: B

Explanation:
$\quad\require{enclose}
\begin{array}{rll}
128\,\, &\\[0pt]
7 \enclose{longdiv}{8975k}\kern-.2ex \\[0pt]
\underline{7}{\phantom{0000}} \\[0pt]
19\phantom{000} \\[0pt]
\underline {14}{\phantom{000}} \\[0pt]
57\phantom{00} \\[0pt]
\underline{56}\phantom{00} \\[0pt]
1k\phantom{0} \\[0pt]
\underline{1k}\phantom{0} \\[0pt]
0\phantom{0}
\end{array}$
$\therefore$ $1k$ has to be a multiple of $7$.
The only value possible is $14$.
So $k = 4$

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4. A number when divided by 69 leaves a remainder of 50. Find the remainder when it is divided by 23.
a. 10
b. 20
c. 6
d. 4

Answer: D

Explanation:
The number $N$ can be written as $N = 69q + 50$.
If the above expression is divided by $23$, $69q$ gives a remainder $0$ as it is a multiple of $23$.
When $50$ is divided by $23$, the remainder is $4$.

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5. A number when divided by a 23 leaves a remainder of 5. Find the remainder when the original number is multiplied by 20 and then divided by the same divisor.
a. 8
b. 10
c. 19
d. 20

Answer: A

Explanation:
The number $N$ can be written as $N = 23q + 5$.
If the above expression is multiplied by $20$, Then $20N = 460q + 100$.
When $460q$ is divided by 23, it gives a remainder $0$ as it is a multiple of $23$.
When $100$ is divided by $23$, the remainder is $8$.

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6. A number when divided by a 23 leaves a remainder of 5. Find the remainder when square of the number is divided by the same divisor.
a. 1
b. 2
c. 4
d. 8

Answer: B

Explanation:
The number $N$ can be written as $N = 23q + 5$.
If the above expression is squared, then $N^2 = {\left( {23q + 5} \right)^2}$ = ${23^2} \times {q^2} + 2 \times 23q \times 5 + {25}$
${23^2} \times {q^2}$, $2 \times 23q \times 5$ are multiples of 23, thus give remainder $0$.
When $25$ is divided by $23$, the remainder is $2$.

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7. If $358k867$ is divisible by $11$ exactly, then $k=$
a. 1
b. 2
c. 3
d. 4

Answer: D

Explanation:
If $358k867$ has to be divisible by $11$, then the difference between the sum of the digits in the even places and odd places must be either $0$ or multiple of $11$.
$\therefore$ $(3 + 8 + 8 + 7) - (5 + k + 6)$ $= 26 - 11 - k$ $= 15 - k$
For $k=4$ the above expression gives remainder zero.

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8. The value of the digit in the number $69k3164$ such that it is divisible by $22$ is
a. 5
b. 6
c. 7
d. 0

Answer: C

Explanation:
If the given number has to be divisible by 22, it has to be divisible by 2 as well as 11.
The given number is an even number so it is divisible by 2.
If $69k3164$ has to be divisible by $11$, then the difference between the sum of the digits in the even places and odd places must be either $0$ or multiple of $11$.
$\Rightarrow$ $(6 + k + 1 + 4) - (9 + 3 + 6)$ $= (11+k) - 18$ $=k - 7$
$\therefore$ $k = 7$

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9. ${9^{16}} - {5^{16}}$ is divisible by
a. 7
b. 8
c. 56
d. All of the above

Answer: D

Explanation:
$a^n - b^n$ is always divisible by $a-b$ for all values of $n$
${9^{16}} - {5^{16}}$ $ = {\left( {{9^2}} \right)^8} - {\left( {{5^2}} \right)^8}$ $ = {\left( {81} \right)^8} - {\left( {25} \right)^8}$
$\therefore {\left( {81} \right)^8} - {\left( {25} \right)^8}$ is divisible by $81-25 = 56$
As the given expression is divisible by $56$, it is also divisible by the divisors of $56$.
$\therefore$ it is also divisible by $7$ and $8$ too.

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10. ${7^{10}} + {4^{10}}$ is divisible by
a. 11
b. 13
c. 65
d. B and C only

Answer: D

Explanation:
$a^n + b^n$ is always divisible by $a+b$ when $n$ is $odd$.
$\therefore $ We have to rewrite the above expression so that the index is $odd$.
${7^{10}} + {4^{10}} = {\left( {{7^2}} \right)^5} + {\left( {{4^2}} \right)^5}$ $ = {\left( {49} \right)^5} + {\left( {16} \right)^5}$
$\therefore {\left( {49} \right)^5} + {\left( {16} \right)^5}$ is divisible by $49+16 = 65$
As the given expression is divisible by $65$, it is also divisible by the divisors of $56$.
$\therefore$ it is also divisible by $5$ and $13$ too.
Note that, the given expression is not divisible by $(7 + 4)$ i.e.,$11$ as index is $even$

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11. ${11^{20}} - {4^{20}}$ is divisible by
a. 5
b. 15
c. 137
d. all of the above

Answer: D

Explanation:
$a^n - b^n$ is divisible by $a+b$ when $n$ is $even$.
${11^{20}} - {4^{20}}$ is divisible by $11+4=15$.
$\therefore $ ${11^{20}} - {4^{20}}$ is divisible by divisors of $15$ too. i.e.,$5$.
${11^{20}} - {4^{20}}$ $= {\left( {{{11}^2}} \right)^{10}} - {\left( {{4^2}} \right)^{10}}$ $ = {\left( {121} \right)^{10}} - {\left( {16} \right)^{10}}$
$\therefore {11^{20}} - {4^{20}}$ is divisible by $121+16 = 137$

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12. $K=23^x + 7^x$ where $x$ is an odd natural number. Which of the following need not be a factor of $K$.
a. 10
b. 15
c. 3
d. 8

Answer: D

Explanation:
If $x$ is odd, then $23^x + 7^x$ is exactly divisible by $23+7$. So $K$ is divisible by 30.
As $k$ is divisible by 30, it is also divisible by its factors, 10, 15, and 3 as well.
So $K$ need not be divisible by 8.