Answer: Option A. -> 7
:
A
Unit digit of $6^{28}$ is 6 and unit digit of $3^{22}$ is 9 (since $22=4k+2$). So, unit digit of given expression is $(16-9)=7$.

Answer: Option D. -> none of these
:
D
Find the least value and the highest value of $\frac{B}{A}$
Least $\frac{B}{A}=\frac{50}{(-1)}=-50$
Highest $\frac{B}{A}=\frac{50}{0}=\infty$
$(-50,\infty )$

Answer: Option D. -> none of these
:
D
option d
$\frac{{10}^1}{13}{\mid }_R=10$
$\frac{{10}^2}{13}{\mid }_R=9$
$\frac{{10}^3}{13}{\mid }_R=(-1)$
Thus,
$\frac{{10}^6}{13}{\mid }_R=1$
Since, $\frac{{10}^6}{13}{\mid }_R=1=\frac{{10}^3\times {10}^3}{13}{\mid }_R=(-1)\times (-1)=1$
Hence we get ${10}^{90}{\mid }_R=1$, we are left with ${10}^1$ which will give a remainder of 10, hence the remainder will be 10 or option d.

Answer: Option A. -> 1
:
A
629 divided by 21 remainder is -1. Thus the question changes to $\frac{-1^{24}}{21}$ Remainder =+1

Answer: Option B. -> even
:
B
OPTION (B)
Observe the pattern
112358
Every $3^{rd}$ term is even. Hence a term which is a multiple of 3 will be even. 66 is a multiple of 3. Hence, the ${66}^{th}$ term is even

Answer: Option D. -> none of these
:
D
Take an example of 11111111. This can be any of the 8 digits.
It would be divisible by 13 and 7 as if we could divide it into equal even groups of 3.
For 19, we need to divide it into even groups of 8. This is not true
This number is divisible by 11, where the rule of divisibility is sum of digits at odd places- sum of digits at even places.
Hence, Answer is none of these.

Answer: Option D. -> none of these
:
D
Answer=option d
Prime factorize $24000=2^6\times 3\times 5^3$
Take out $2^6$ and then find the number of factors. This will give all the odd factors
In this case, number of factors $=(3+1)\times (1+1)=4\times 2=8$

Answer: Option D. -> none of these
:
D
$p+q+r+s+t+u=2005$ of the 6 integers, the number of integers that can be odd is 5,3 or 1. In the expression $(-1{)}^p+(-1{)}^q+(-1{)}^r+(-1{)}^s+(-1{)}^t+(-1{)}^u$
1)When one integer is even and the others are odd, then only one among is +1 and others are -1 each. Hence the sum of the terms is -5 + 1 = -4.
2)When three of the integers are even, then three of the terms are +1 each and the remaining three are -1 each. Hence the sum is 0.
3)When five of the integers are even, then five terms will be +1 each and the other term is -1. Hence the sum is 4. $\therefore$ The minimum value of the sum is -4.

Answer: Option C. -> 7
:
C
Ans. (c)
${74}^{288}=(37\times 2{)}^{288}$ (refer demo tutorial for last 2 digits technique)
${37}^{288}\times 2^{288}$
$({37}^4{)}^{72}\times (2{)}^{288}$
$({37}^2\times {37}^2{)}^{72}\times (\_\_56)$
$(69\times 69{)}^{72}\times (\_\_56)$
$(\_\_61{)}^{72}\times (\_\_56)$
$\left(\_\_21\right)\times \left(\_\_56\right)=\_\_76.$ answer is 7

Answer: Option C. -> 610
:
C
option (c)
$T_{n+1}=T_{n-1}+T_n.{12}^{th}$ term$=144$
${14}^{th}$ term $={12}^{th}$ term $+{13}^{th}$ term
$144+{13}^{th}$ term$=377$
${13}^{th}=233$
${15}^{th}$ term $={13}^{th}$ term $+{14}^{th}$ term $=233+377=610$