To solve this question, we need to immediately recognise the fact that, $32768=8^5$

Substituting this in the above given equation, 

$\left(\dfrac{1}{8}\right)^k\times\ \left(\dfrac{1}{8}\right)^{5\times\ \dfrac{1}{3}}$ = $\left(\dfrac{1}{8}\right)\times\ \left(\dfrac{1}{8}\right)^{5\times\ \dfrac{1}{k}}$

Since the bases are equal, we can equate the powers on either side of the equation, 

$k+\frac{5}{3}=1+\frac{5}{k}$

$\frac{\left(3k+5\right)}{3}=\frac{\left(k+5\right)}{k}$

$3k^2+5k=3k+15$

$3k^2+2k-15=0$

Here in the given quadratic equation, the Discriminant is greater than 0,

 $2^2-\left(4\right)\left(3\right)\left(-15\right)$> 0

That means both the roots are real, hence we can simply take the sum of the roots of the quadratic equation in k, 

Which in a standard quadratic equation of the form $ax^2+bx+c$ is $-\frac{b}{a}$

Here, the sum of the real values of k is $-\frac{2}{3}$

When gives n distinct numbers, and asked to find the sum of the possible n distinct numbers that can be formed, 

We use the formula, $\left(10^{n-1}+10^{n-2}+10^{n-3}+...\right)\left(\left(n-1\right)!\right)\left(Sum\ of\ numbers\right)$

Inserting n=4, $\left(1000+100+10+1\right)\left(3!\right)\left(a+b+c+d\right)$

$\left(6666\right)\left(a+b+c+d\right)$

We are told that this value equals, $153310+n$

Since we are told that n is a single digit natural number, the total value cannot be that much greater and 6666 should perfectly divide $153310+n$

Upon dividing 153310 by 6666 we get the quotient as 22.99

Nearest value being 23, We take 6666x23 giving us the value 153318

Hence the value of n=8 and the value of a+b+c+d=23

Value of a+b+c+d+n=31. 

We must bring the right-hand side in the form so that everything has the same power. 

25 has factors 1, 5 and 25

The only common factor 40 and 25 have is 5 (other than 1 of course, which does not work)

So the right-hand side can be rewritten as $\left(2^5\right)^5\times\ \left(3^8\right)^5$

$\left(32\times\ 81\times\ 81\right)^5$

$\left(209952\right)^5$

Giving the value of m - n as 209952 - 5 = 209947

Therefore, Option D is the correct answer. 

We need to consider single-digit, two-digit and three-digit numbers

Single digit: There are only nine numbers for a single-digit number (not including zero since we are looking for positive integers only)

Double digits: The ten's digit can be chosen in 9 ways (not including 0), and the unit digit can be chosen in 9 ways (including 0 but excluding the digit used in the ten's place). Hence, a total of 81 numbers. 

Three-digit: The hundred's digit can be 1, 2, 3 or 4; hence, there are four options for the hundred's digit; for the ten's digit, there will be 9 options (numbers from 0 to 9 except the one chosen in the hundred's place); for unit's digit there would be 8 options. Hence, a total of 288 numbers

Giving the total numbers as 288+81+9 = 378

Therefore, 378 is the correct answer. 

Prime factorising 1134, we get  $1134 = 2 \times 3^4 \times 7$  and  $168 = 2^3 \times 3 \times 7$

$1134^n$ is a factor of 168  $\Rightarrow$ the factor of 2 should be at least 3, for 168 to be a factor  $\Rightarrow n = 3$

Now,  $1134^n = 1134^3 = 2^3 \times 3^{12} \times 7^3$  is a factor of  $168^m = \left(2^3 \times 3 \times 7\right)^m$  $\Rightarrow m = 12$ as power of 3 should be at least 12.

$\Rightarrow$ So, $m + n = 15$

It is given that $ a^m \cdot b^n = 144^{145} $, where $ a > 1 $ and $ b > 1 $.

144 can be written as $ 144 = 2^4 \times 3^2 $

Hence, $ a^m \cdot b^n = 144^{145} $ can be written as $ a^m \cdot b^n = \left(2^4 \times 3^2\right)^{145} = 2^{580} \times 3^{290} $

We know that $ 3^{290} $ is a natural number, which implies it can be written as $ a^1 $, where $ a > 1 $

Hence, the least possible value of m is 1. Similarly, the largest value of n is 580.

Hence, the largest value of (n-m) is (580-1) = 579

The correct option is C

It is given that k divides m+2n and 3m+4n.

Since k divides (m+2n), it implies k will also divide 3(m+2n). Therefore, k divides 3m+6n.

Similarly, we know that k divides 3m+4n.

We know that if two numbers a, and b both are divisible by c, then their difference (a-b) is also divisible by c.

By the same logic, we can say that {(3m+6n)-(3m+4n)} is divisible by k. Hence, 2n is also divisible by k.

Now, (m+2n) is divisible by k, it implies 2(m+2n) =2m+4n is also divisible by k.

Hence, {(3m+4n)-(2m+4n)} = m is also divisible by k.

Therefore, m, and 2n are also divisible by k.

The correct option is B

Since there are two distinct factors other than 1, and itself, which implies the total number of factors of N is 4.

It can be done in two ways. 

First case: N = $p^3$ (where p is a prime number)

Second case: N = $p_1$ * $p_2 $ (Where  $p_1$, $p_2 $ are the prime numbers)

From case 1, we can see that the numbers which is a cube of prime and less than 50 are 8, and 27 (2 numbers).

From case 2, we will get the numbers in the form (2*3), (2*5), (2*7), (2*11), (2*13), (2*17), (2*19), (2*23), (3*5), (3*7), (3*11), (3*13), (5*7) {(13 numbers)}

Hence, the total number of numbers having two distinct factors is (13+2) = 15.

We know that the number of factors of these two numbers is 15. We know that the factors of 15 are 1, 3, 5, and 15.

The number of factors of N is (p+1)⋅(q+1)(Where, N = $a^p ⋅ b^q$, and a, b are prime numbers).

Hence, the value of N will be least when (p+1) and (q+1) are as close as possible and a, and b are the least distinct prime numbers.

Hence, p + 1 = 3 => p = 2, and q + 1 = 5 => q = 4, and the prime numbers a, and b are 2, and 3, respectively.

Hence, the lowest value of N is N = $2^4$ × $3^2$ = $144$, and the second lowest value of N is N = $2^2$× $3^4$ = 324

Hence, the sum is (144 + 324) = 468

Given that the number of coins collected per week by two coin-collectors, A and B, are in the ratio 3: 4

Let us assume A collects 3c coins per week and B collects 4c coins per week.

Total number of coins collected by A in 5 weeks = 5*3c = 15c, which should be multiple of 7 => c should be multiple of 7.

Total number of coins collected by B in 3 weeks = 3*4c = 12c, which should be a multiple of 24 => c should be a multiple of 2.

So, the least possible value of c is lcm(2,7) = 14.

Coins sold by A in a week = 3c = 3*14 = 42.

The divisors of the given number will have the form $2^a*3^b*5^c*7^d$ with $0\le a\le6,\ 0\le b\le5,\ 0\le c\le3,\ 0\le d\le2$

Because the divisors should be in the form 3r+1, it cannot be divisible by 3, so (b=0). 

Reduce modulo $2 \bmod 3 = 2,\; 5 \bmod 3 = 2,\; 7 \bmod 3 = 1$

Hence,  $2^a5^c7^d\equiv 2^{a+c}\cdot1^d \equiv 2^{a+c}\pmod3$

2^k will be in the form 3r+1 only when K is even. So, we need a+c to be even

$a\in{0,\dots,6}$ has 4 even, 3 odd values 

$c\in{0,\dots,3}$ has 2 even, 2 odd

Number of (a,c) with (a+c) even is $4\cdot2 + 3\cdot2 = 8+6=14$

For each such pair, there are 3 choices for d = 0,1,2. Thus, total divisors in the form (3r+1) equals $14\times3=42$.