If all x, y and z are >1 then the sum of their factorials will be even and it cannot be a power of 3.
So one of them is 1 and it has to be z because it is the least among x, y and z.
So z =1.

If y>2 then 3 divides x!+y! and x!+y!+1 will not be divisible by 3.
So y = 2.

Now the equation is reduced to $$x!+3 = 3^w$$

If x=6 or more x! will be a factor of 9 and x!+3 can be written as $$3^2 k + 3 = 3(3k+1)$$ which cannot be a power of 3.
Also as x>y>z, the least value x can take is 3.

Now we know that x can only take the values of 3,4 and 5.
Substituting x=3, we get w=2
Substituting x=4, we get w=3.
Substituting x=5 will not give any solution such that 'w' is a natural number.

Thus the possible quadruplets of (x,y,z,w) are (3,2,1,2), (4,2,1,3).
Hence only 2 ordered quadruplets of (x,y,z,w) are possible.

As there are four integers, there will be atleast two integers whose moudlo 3 will be the same. Hence their difference will be divisible by 3. Hence the above product is always divisible by atleast 3
Also consider the following cases for a,b,c,d
Case 1: All 4 are even or All 4 are odd
Hence all the differences will be even and the product is divisible by atleast $$2^6$$
Case 2: 3 even and 1 odd
The pair wise differences of the even numbers will be even. Hence the product is divisible by atleast $$2^3$$
Case 3: 2 even 2 odd
The difference of even numbers will be even and difference of odd numbers will be even. Hence the product is divisible by atleast $$2^2$$
Case 4: 1 even 3 odd
The pair wise differences of the odd numbers will be even. Hence the product is divisible by atleast $$2^3$$

Combining all the cases, we will get that the product is always divisible by $$2^2$$ = 4
Hence the product is always divisible by 4 * 3 = 12

Given that the number is divisible by 18. Therefore, the number has to be divisible by 2 and 9.
Sum of the digits of the number should be divisible by 9, and the unit’s digit of the number should be either 0 or 6.
Smallest such number is 300096.
Therefore, the digit at the ten’s place is 9.
Hence, option (2) is the correct choice.

LCM of X and Y is a multiple of 7 and not necessarily a multiple of 49.

$$(245)_{x}+(162)_{x}-(427)_{x}=0$$
$$=>(2x^2+4x+5)+(x^2+6x+2)-(4x^2+2x+7)=0$$
$$=>x^2-8x=0$$
$$=>x=8$$

a = 3b + (1-b)/7.
So, every b of the form 7k+1 has a corresponding value for a.
For example, { (3,1), (23,8), (43,15)} etc.
a can take value from 3 to 1983 with a common difference of 20.
Let 1983 be the nth term of the AP with common difference 20 and first term 3.

Hence, 1983 = 3 + (n-1)20

n-1 = 1980/20 = 99

n = 100

Hence, the last possible term is the 100th term. Hence, there are 100 such pairs.

f(n) > n. So, n<120 For all n<100, The sum of the digits is less than 20 (as it is a two digit number) So, n+sum of its digits<120 or f(n)<120

So, the only numbers we need to look at are between 100 and 120. Of which, you can calculate that only 114 solves the equation.

X on division by 7 leaves a remainder of 2. So X is of the form 7k+2
Same number on division by 19 leaves a remainder of 1, So X is also of the form 19m + 1
Thus
7k+2 = 19m + 1
If we put m = 3, we get a number which is of the form 7k+2. Thus the smallest number which fits the given criteria is 19*3+1 =58
When 58 is divided by 23, the remainder would be 12.

If the length of the first story is odd, and the length of all the remaining stories are even, other than the first story, every story starts with an odd number

In this case, $$16^3 + 19^3$$ is divisible by $$16+19 = 35$$. Note that, $$16^3+19^3$$ is odd.
Similarly, $$17^3+18^3$$ is divisible by $$17+18=35$$. Also, note that $$17^3+18^3$$ is odd.

Hence, the sum of the four terms will be divisible by 35 and as the sum of two odd numbers is even, we know that the sum is divisible by 70.

Therefore, the remainder when $$16^3+17^3+18^3+19^3$$ is divided by 70 is 0

323 = 17 * 19

By Wilson's theorem we can say that.

16! + 1 is divisible by 17.
==> (16! +1)+85 is also divisible by 17.(Since 85 is divisible by 17)

18!+1 is divisible by 19.

18! + 1 = 0(mod 19)
18*17*16! + 1 = 0 (mod 19)
==> (-2)*(-1)*16! +1 = 0(mod 19) (Since 18= -1(mod19) and 17 = -2(mod 19))

==> 2*16! +1 = 0(mod 19)

This can only possible when 16! = 9(mod 19)
==> 16! + 10 = 0(mod 19)

==>16! + 86 = 0(mod 19)

Therefore the remainder when 16!+86 is divided by 323 is 0.

The number of divisors of all square numbers is odd.
So, the number of divisors of X = 60 - an odd number is also an odd number.
Hence, even X is a square number. Assuming X = $$P_{1}^{2a}$$*$$P_{2}^{2b}$$ . . .

So, (2a+1)*(2b+1)*. . (2k+1) + (4a+1)*(4b+1)*..(4k+1) = 60

The only solution to this is when a=1 and b=2 and the solutions for X<350 are $$2^{2}$$ * $$3^{4}$$ and $$2^{4}$$ * $$3^{2}$$ i.e X is 144 or 324.
Hence, number of solutions is 2

21 divides the number n.
So, the smallest divisor of n is either 2 or 3 (as 3 divides 21)
So, the largest divisor of n is either 21*2 = 42 or 21*3 = 63
Hence, n = 84 (21*2 = 42) and n = 189 (21*3=63)
So, number of values of n is 2.

135000 = $$2^3*3^3*5^4$$
10800 = $$2^4*3^3*5^2$$
Number of even factors of 135000 = 3*4*5 = 60
Number of even factors of 10800 = 4*4*3 = 48
But the numbers that have $$2^4$$ as a factor do not divide 135000. There are 4*3 = 12 such numbers.
=> 48 - 12 = 36 even factors divide both the numbers => 24 numbers do not divide 10800 but divide 135000.

As a summary, we are looking for even numbers which divide 135000 and not divide 10800.
We noted that there are 60 even factors of 135000 and there are 36 common even factors for 135000 and 10800.
Hence, the number of even factors which divide 135000 and not divide 10800 is 60-36 = 24

In Vijay’s novel, the page numbers are in base 8 where the 8 digits which are used are 0, 1,2,3,4,6,8, and 9.
The page number for a particular scene in Ajay's book is 189. Since Vijay's book consists of only 8 digits, the same scene in his book will be at the base 8 equivalent of 189.
This is because for only 8 digits are available for use in base 8.
For example, In base 8, the few numbers will be 0,1, 2, 3, 4, 5, 6, 7, 10, 11. etc.
Base 8 equivalent of 189 = $$2*8^2 + 7*8^1 + 5*8^0$$
Hence 189 will be written as 275 in base 8.

Every 100th page in the faulty book is equivalent to 100 in base 8, however,

Since, 275 is closer to 300, we can work backwards
let LHS be faulty book number and RHS be base 8 number
300-300, 299-277, 298-276, 296-275 => Hence, the number that will be printed is 296

OR

Every 100th page in the faulty book is equivalent to corresponding 100 in base 8, however,
200-200, 201-201, 202-202, 203-203, 204-204, 206-205, 208-206, 209-207

The entire sequence of numbers in 250s and 270s are absent in the faulty book
i.e. 248-246, 249-247, 260-250, 261-251.....

Therefore, 275 will be equivalent to 295, however since 5 cannot be printed
The number will be 296

$$7^{21}+49^{21}+343^{21}+2401^{21}$$ can be written as $$7^{21}+7^{42}+7^{63}+7^{84}$$.

Now consider the first term:
$$7^{21}\mod [7^{20}+1]$$
= $$[7^{21}+7-7]\mod[7^{20}+1]$$
= 7*$$[7^{20}+1]\mod [7^{20}+1]$$-7 $$\mod [7^{20}+1]$$
= 0-7 = -7

Now second term is $$7^{42}\mod [7^{20}+1]$$
= $$7^{21}\mod [7^{20}+1]$$ *$$7^{21}\mod [7^{20}+1]$$ = $$-7\times-7 = 49$$

Now consider the third time $$7^{63}\mod [7^{20}+1]$$
= $$7^{21}\mod [7^{20}+1] \times7^{21}\mod [7^{20}+1]* 7^{21}\mod [7^{20}+1]$$
= -7*-7*-7 = -343

The fourth term is $$7^{84}\mod [7^{20}+1]$$
= $$7^{21}\mod [7^{20}+1] \times7^{21}\mod [7^{20}+1]\times7^{21}\mod [7^{20}+1]\times7^{21}\mod [7^{20}+1]$$
= -7*-7*-7*-7 = 2401
$$[7^{21}+49^{21}+343^{21}+2401^{21}]\mod [7^{20}+1]$$
= -7+49-343+2401 = 2100

Let number a = 10m+n
So b = 10n+m
It is given that
$$a^{2} - b^{2} = k^{2}$$
So we have, $$(10m+n)^{2} - (10n+m)^{2} = k^{2}$$
=> $$99m^{2} - 99n^{2} = k^{2}$$
=> $$m^{2} - n^{2} = k^{2}/99$$
Since LHS is an integer, RHS must be an integer.
So $$k^{2}$$ must be a multiple of 99
99 = 11*9 => 9 is already a perfect square, so $$k^{2}$$ can be of the form $$99*11*a^{2}$$
If a > 1, then then m and n are not going to be 1 digit numbers.
Hence, the value of $$k^{2}$$ for which m and n are 2 digit numbers is 121*9 = 1089
=> k = 33
$$m^{2} - n^{2} = k^{2}/99$$ => $$m^{2} - n^{2} = 11$$
(m+n)(m-n) = 11
=> m = 6 and n = 5
So required numbers are 65 and 56
=> Hence a+b+k = 154

Let the two numbers be a and b respectively.
There are 2 possibilities => ab - a - b = 72 and a + b - ab = 72
Case 1: ab - a - b = 72
=> ab - a - b + 1 = 73
=> (a - 1)(b - 1) = 1 * 73 = (-1) * (-73)
a = 2, b = 74 and a = 0, b = - 72 are the solutions.

Case 2: a + b - ab = 72
=> ab - a - b + 1 = -71
=> (a - 1)(b - 1) = 1 * (-71) = (-1) * 71
=> a = 2, b = -70 and a = 0 and b = 72 are the solutions.
Hence, there are 4 solutions.

$$\frac{3}{a} + \frac{2}{b} = \frac{1}{18}$$

The given equation can be written as 3b + 2a =(ab/18)

54b+36a = ab

ab -36a - 54b = 0

In order to factorize , we add 36*54 = 1944 to both sides.

ab-36a-54b-1944 = 1944

(a-54)(b-36) = 1944.

This equation can be written as (a-54)(b-36) = 1944 = $$2^3 * 3^5$$
$$17 < a < 67$$

$$-37 < (a-54) < 13$$

For all values in the above range, which are factors of the number 1944, corresponding integral b value exists.

For $$17 < a < 67$$ the following ordered pairs satisfy the given equation.

(18, -18), (27, -36) (30, -45), (36, -72), (42, -126), (45,-180), (46, -207), (48, -288), (50,-450), (51,-612), (52, -936), (53, -1908), (55, 1980), (56, 1008), (57, 684), (58, 522), (60, 360), (62, 279), (63, 252), (66, 198).
i.e., total 20 pairs.

Both the series are essentially the same apart from the extra term $$k^{k!}$$ in the second series. For both of them to have the same units digit, it is necessary that $$k^{k!}$$ in the second series has a unit digit of 0. This can only happen when k is a multiple of 10.
Multiples of 10 less than or equal to 105 are 10,20,30,40,…100, i.e. 10.