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 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 $=(3+1)\times (1+1)=4\times 2=8$

option (c)

Any power of 10 when divided by 3 gives a remainder of 1. Also,7 divided by 3 remainder is 1. Therefore individual remainders are 1 and 1.

Thus, 1-1=0. Answer is 0

$p+q+r+s+t+u=2005$ 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.

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

Find the least value and the highest value of $\frac{B}{A}$

Least $\frac{B}{A}=\frac{50}{(-1)}=-50$

Highest $\frac{B}{A}=\frac{50}{0}=\infty$

$(-50,\infty )$

According to fermat's theorem " If P is a prime number and N is prime to P, then $N^p-N$ is divisible by P."

Open the brackets, $p^2[p^{27}-\frac{1}{p}]=p^{29}-p$ will always be divisible by 29, as it is prime.

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

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 $=233+377=610$

Any power of 9 when divided by 6, gives a remainder 3. hence answer $=\frac{11\times 3}{6}$((As there are $11$ terms)= $\frac{33}{6}$ = Remainder $=3$