XAT Number Systems Questions

We know that if the greatest common divisor of two numbers is 1, they are co-prime to each other.

Given that the greatest common divisor of the selected integer $$x$$ and any other integer between 2 and 40, both inclusive, is 1.

So, the selected number has to be a prime number, and that too greater than $$\dfrac{40}{2}$$ or $$20$$ because prime numbers less than 20, like 17, will have the greatest common divisor = 17 with their multiple, like 34 which is also in the range of selected integers.

Hence, the only possible values of $$x$$ are 23, 29, 31 and 37 i.e. a total of 4 values.

The given expression is,  $$\sqrt[3]{(3)^a(21)^{(3a - b)}(49)^{(2b + c)}}$$ and it can also be written as,

$$\sqrt[3]{(3)^a\ \left(3\right)^{3a\ -\ b}\ \ \left(7\right)^{3a\ -\ b}\ (7)^{4b+2c}}$$ $$=\ \sqrt[3]{\ \left(3\right)^{4a\ -\ b}\ \ (7)^{3a\ +\ 3b+2c}}$$

For the above expression to be a positive integer, the power of the expression must be an integer after applying the cube root for the expression inside. This means that the power of the expression inside must be a multiple of 3.

The expression is,

$$\sqrt[3]{\ \left(3\right)^{4a\ -\ b}\ \ (7)^{3a\ +\ 3b+2c}}\ =\ 3^{\frac{\left(4a\ -\ b\right)}{3}}\ \times\ 7^{\frac{\left(3a\ +\ 3b\ +\ 2c\right)}{3}}$$

The value   $$\dfrac{4a\ -\ b}{3}=\ \dfrac{3a\ +\ a\ -\ b}{3}\ =\ a\ +\dfrac{a\ -\ b}{3}$$   must be an integer which means that a - b must be a multiple of 3.

Similarly, the value   $$\dfrac{3a\ +\ 3b\ +\ 2c}{3}=\ a\ +\ b\ +\ \dfrac{2c}{3}$$   must also be an integer which gives us the condition that c must be a multiple of 3.

So, the necessary and sufficient conditions are a - b and c must be a multiple of 3.

Hence, the correct answer is option E.

abbb can be written as (1000*a+100*b+10*b+1*b) = 1000a+111b

Now, we need to check to the necessary and sufficient condition for which (1000a+111b) is divisible by a.

We know that 1000a is always divisible by 'a', hence, we need to check for which condition, 111b is always divisible by a.

111b can be written as (3*37*b) => (3b*37) must be divisible by a.

a can't be a factor of 37, which implies 'a' is a factor of 3b => 3b is divisible by 'a'

The correct option is E

Let's say the unknown number is x—the LCM of x, 990= 6930.

LCM is the greatest power of a prime number in those two numbers.

Now let's prime factorize 990 and 6930.

990=2*5*11*$$3^2$$

6930=2*$$3^2$$*5*7*11

The number 990 has all the powers of prime numbers, which are also present in 6930, except 7. So, x definitely has to have 7.

Then, it is given that the GCD of x and 550 is 110. So, x has to be a factor of 110. 110 is not a multiple of 7. So, we have to multiply the 110 by 7. That is 770, which is the minimum value of x.

The sum of digits of x is 7+7+0=14

Let the 4-digit locker key be $$abcd$$, where $$9\ge\ a,b,c,d>0$$, and they are all unique.

It is given that $$a\le\ 3\ \&\ \frac{b}{c}=a$$

It is also given that 'd' is the smallest number.

Case (i): $$a=1$$

Not possible, as 'd' is the smallest number, 'a' cannot be 1.

Case (ii): $$a=2$$

Then 'd' can take only one value, i.e. $$d=1$$ 

$$b=2c$$

 $$if\ c=3,\ then\ b=6$$

$$if\ c=4,\ then\ b=8$$

for $$c\ge\ 5$$, 'b' will not be a single-digit number.

Hence, two cases are possible, i.e. (2631, 2841).

Case (iii): $$a=3$$

'd' can take two values, i.e. $$d=1\ or\ d=2\ $$

& $$b=3c$$

If d = 1, then 'c' can take only one value. i.e. $$\ c=2,\ then\ b=6$$

If d = 2, then 'c' cannot take any value.

Hence one case is possible, i.e. (3621)

Amit has to try 3 different combinations. Option (E) is correct.

The first number Haruka entered is 1.

From Rule-1, the series will be 2, 4, 8, 16

From Rule 2, 16 will be followed by 7

Like-wise, the series will be 1, 2, 4, 8, 16, 7, 14, 5, 10, 1, 2, 4,......

The series is repeating itself for every 9 iterations.

Initially, input is 1 and after first iteration, result is 2. 

This implies,

9n = 1

9n + 1 = 2

9n + 2 = 4

.

.

.

.

9n + 7 = 5

9n + 8 = 10

In the question, it is given that key is pressed 68 times. 

68 = 9(7) + 5

68 is in the form of 9n + 5

This implies, 7 is the output.

The answer is option A.

Let 'm' be the Greatest common divisor and A, B, C, D, E, F & G be the 7 unique numbers.

When 'm' divides A, B, ......& G, we get seven unique quotients. Let the quotients be a, b, c, d, e, f, and g.

It is given,

A + B + C + D + E + F + G = 1740

m(a + b + c + d + e + f + g) = 1740

$$\ \ 1740=2^2\times3\times5\times29$$

m(a + b + c + d + e + f + g) = $$\ \ 1740=2^2\times3\times5\times29$$

29 is a prime number and cannot be factorised further.

Therefore, largest possible value m can take is 60.

Least possible sum of 7 unique numbers is 1 + 2 + 3 +...+ 7 = 28

Replacing 7 with 8, we will get 29.

This implies 60(1 + 2 + 3 .....+ 6 + 8) = 1740

Largest possible G.C.D is 60.

The answer is option B.

Let the positive integers Raju & Sarita chose be 'r' & 's'.

After doing the given operations, the final numbers they get are $$\frac{2r-20}{5}\ \&\frac{2s-20}{5}$$

Adding results we get, $$\frac{2r-20}{5}\ +\frac{2s-20}{5}\ =\ 16$$

$$2r-20\ +2s-20\ =\ 80$$

$$2r+2s\ =\ 80+40+40=120$$

$$r+s\ =60$$

For maximum value of |r-s|, one of 'r' & 's' has to be maximum and other has to be minimum.

As r & s are positive integers minimum value they can take is "1".

If $$r=1,\ then\ s=59$$

$$\left|r-s\right|=\left|1-59\right|=58$$.

If $$s=1,\ then\ r=59$$

$$\left|r-s\right|=\left|59-1\right|=58$$

Hence maximum value of $$\left|r-s\right|$$ is 58. Option B is correct.

From the given information we get the following table,

The possible maximum marks are shown below.

Therefore, sixth highest possible score is 94.

Option B is correct.

The supplier receives his orders from the five buyers once every 2 weeks, once every 6 weeks, once every 8 weeks, once every 4 weeks, and once every 3 weeks.

The number of occasions where all the five buyers place the order on the same day is :

The LCM of the 5-time frames during which the 5 buyers place their orders :

Hence the LCM is :

(2, 6, 8, 4, 3).

= 24 weeks.

Once every 24 weeks, all five of them place the order simultaneously.

A year has 53 weeks in total :

Hence all five of them place the orders after 24 weeks, 48 weeks.

Given that X, Z are positive Y is negative and W can be either positive or zero or negative.

The given conditions are :

$$W^{4}+X^{3}+Y^{2}+Z\leq4$$
$$X^{3}+Z\geq2$$
$$W^{4}+Y^{2}\leq2$$
$$Y^{2}+Z\geq3$$

For $$W^4+\ Y^2\ \le\ 2$$. Since Y is negative $$but\ Y^2$$ is always positive and must be less than 2 because $$W^4$$ is a nonnegative value. Hence Y = -1 is the only possibility. For W this can take any value among -1, 0, 1.

$$Y^2+Z\ \ge\ 3$$. Since Y = -1, Z must be at least equal to 2 so the value of $$Y^2+Z\ \ge\ 3$$ is greater than 2.

X is a positive value and must at least be equal to 1.

The condition: $$W^{2}+X^{2}+Y^{2}+Z^{2}$$  here has all the independent values: $$X^2,\ Y^2,\ Z^2,\ W^2$$are nonnegative.

$$W^{4}+X^{3}+Y^{2}+Z\leq4$$ : 

Since the value of Z is at least equal to 2 the value of $$Y^2$$ is equal to 1.

Since X is a positive number in order to have the condition of $$W^{4}+X^{3}+Y^{2}+Z\leq4$$ satisfied. The value of Z must be the minimum possible so that $$X^3+Y^2+Z$$ to have a value equal to 4 when X takes the minimum possible positive value equal to 1.

Hence X must be 1. W must be equal to 0 so that :

$$W^{4}+X^{3}+Y^{2}+Z\leq4$$. = The sum = (0+1+1+2) = 4. The only possible case.

The value of $$W^{2}+X^{2}+Y^{2}+Z^{2}$$ = (0+1+1+4) = 6.

From the given conditions :

Considering the three-digit number to be a b c.

With the given conditions :

a = 2b, c = 3b.

Hence the number is of the form : 2b b 3b.

Since all three of the values must be less than 10 and non-negative:

This takes values : b = 1, b = 2, b = 3.

Hence the possible numbers are : (213, 426, 639) :

The interchanged number must be greater than the original by 198.

Hence the different rearrangements for the three numbers are :

213 : (312, 321, 132, 123, 231).

426 : (462, 624, 642, 246, 264)

639 : ( 693, 963, 936, 396, 369)

The only possible value which is higher than the original by 198 is :

(426, 624).

The middle digit is 2.

The number of oranges, apples, guavas, and pears = 126, 162, 198, and 306.

Each box must contain an equal number of fruits with only one type of fruit. The additional condition provided is that there should be a minimum number of boxes in total.

The distribution is possible in multiple ways in such a way that distribution in each box is placed in such that each box contains a certain number of fruits n which is a factor for all the four given number of fruits :

Arrangement of 1 fruit of one kind in a basket.

2 is a factor of 126, 162, 198, and 306. So we can place 2 fruits of a particular kind in a basket.

Since we were asked for the minimum number of boxes this is possible when a maximum number of fruits of a kind are placed in a box.

Hence each box must contain the Highest common factor for the four numbers :

The prime factorization for the four numbers :

126 : $$2\cdot7\cdot9$$, 162 : $$2\cdot9\cdot9$$, 198 : $$2\cdot9\cdot11$$, $$306\ =\ 2\cdot9\cdot17$$

The HCF is 18.

The number of boxes required for each :

$$\frac{126}{18},\ \frac{162}{18},\ \frac{198}{18},\ \frac{306}{18}$$

7+9+11+17 = 44.

Flight become znmorl

Let's assume $$k>3$$

So flight will become thgilf -> znmrol. Hence the value of k will be 6

Let the 6 digit number be  _ _ _ 42_

It is divisible by 2,3,5,7,11,13

Since the number is divisible by both 2 and 5 the last digit of the number must be 0.

The number is also divisible by 3, 7, 11, and 13.

Hence the number must also be divisible by 7*11*13.

=7*11*13 = 1001.

A number which is a multiple of 1001 is of the form abcabc.

This is because abc*(1001) = abc*(1000+1) = abc000 + abc = abcabc.

Hence the number is 420420.

The first digit is 4.

Given that :

(A O B) = (A + B) * B.

For (5 O 2) = ( 5+2 )*2 = 14.

((14) O 5 )= ( 14 + 5 )*5 = 19*5 = 95.

Option A: $$\left(\frac{\pi}{2}\right) \left[\left(\frac{1}{\pi}\right) + 1\right] - \frac{\pi}{2}$$ =1/2

Option B: $$\sin^21^{\circ}+\sin^22^{\circ}+....+\sin^289^{\circ}=44+1/2\ $$ because $$\sin^2\left(89\right)=\cos^2\left(1\right)\ \&\ \sin^2\left(1\right)+\cos^2\left(1\right)=1$$

Option C: $$\sqrt{2}\left(3\sqrt{2} - \frac{4}{\sqrt{2}}\right) + \sqrt{3}$$ = $$6-4+\sqrt{3}$$=$$2+\sqrt{3}$$ which is non-terminating and non repeating.

Option D: $$\frac{\left(\sqrt[3]{729}\right)}{3} + \frac{22}{7}$$ = 3+22/7 = 6.142857142857....

Option E: $$\left(\frac{\pi}{4}\right) + \left(\frac{\pi}{4}\right)^2 + \left(\frac{\pi}{4}\right)^3 + ...$$ (infinite terms)= $$\frac{1}{1-\frac{\pi\ }{4}}=\frac{4}{4\ -\pi\ }\ $$ => $$(4 - \pi)[1 + \left(\frac{\pi}{4}\right) + \left(\frac{\pi}{4}\right)^2 + \left(\frac{\pi}{4}\right)^3 + ...$$ (infinite terms)] =4

Using Fermat's theorem :

If p is a prime number and a, p are co primes $$\left(a^{p-1}\right)\ mod\ p=1$$

Remainder when $$19^{20}$$ is divided by 7 = $$19^2$$ mod 7 =4. ( Here $$19^{20\ }=\ \left(\left(19\right)^6\right)^3\cdot\left(19\right)^2$$

Since the remainder for $$19^6$$ is 1 the remainder for $$\ 19^{20}$$ is  equivalent to the $$\frac{19^2}{7}$$ = 4.

Remainder when $$20^{19}$$ is divided by 7 = $$20^1$$ mod 7 =6.( Here $$\frac{20^6}{7}$$ the remainder is 1 and since 

$$20^{19}=\ \left(20^6\right)^3\cdot\left(20\right)^1\ =\ \frac{\left(1\cdot20\right)}{7}$$. The remainder is 6.

Remainder when $$19^{20} - 20^{19}$$ is divided by 7=4-6=-2=> 5. 

In the given statement, the expression becomes a whole number only when the powers of all the prime numbers are also whole numbers. 

Let us first simplify the expression a bit by expressing all terms in terms of prime numbers. 

 $$\sqrt[3]{7^a\times 35^{b+1} \times 20^{c+2}}$$

$$\Rightarrow \sqrt[3]{7^a\times 5^{b+1} \times 7^{b+1} \times 2^{2(c+2)} \times 5^{c+2}}$$

$$\Rightarrow \sqrt[3]{2^{2c+4} \times 5^{b+c+3} \times 7^{a+b+1}}$$

$$\Rightarrow 2^{\frac{2c+4}{3}} 5^{\frac{b+c+3}{3}} 7^{\frac{a+b+1}{3}} $$

Now, from the given options, we can put in values of the variables and check the exponents of all the numbers. 

Option A : a = 2, b = 1, c = 1 : 

In this case, we can see that exponent of 5 ie $$\frac{b+c+3}{3} = \frac{5}{3} $$ is not a whole number. 

Option B : a = 1, b = 2, c = 2

In this case, we can see that exponent of 2 ie $$\frac{2c+4}{3} = \frac{8}{3} $$ is not a whole number. 

Option C : a = 2, b = 1, c = 2

In this case, we can see that exponent of 2 ie $$\frac{2c+4}{3} = \frac{8}{3} $$ is not a whole number. 

Option D : a = 3, b = 1, c = 1

In this case, we can see that exponent of 5 ie $$\frac{b+c+3}{3} = \frac{5}{3} $$ is not a whole number. 

Option E : a = 3, b = 2, c = 1

In this case, we can see that all exponents are whole numbers. 

Thus, option E is the correct option.

$$10 + 10^3 + 10^6 + 10^9 = 10 + 1000 + 1000000 + 1000000000$$
=$$1001001010$$
Therefore, option D is the right answer.

Let us find out the difference between the times given to figure out the pattern. 
The times given are 1:55 pm, 2:03 pm, 2:11 pm, 2:24 pm, 2:45 pm, 3:19 pm and 4:14 pm.
The difference between 2 consecutive times given are  8 minutes, 8 minutes, 13 minutes, 21 minutes, 34 minutes, and 55 minutes.
We can observe that the difference between the times are in the Fibonacci series. 
8 + 13 = 21
21 + 13 = 34
34 + 21 = 55

The Fibonacci series is as follows:
1,1,2,3,5,8,13,21,34,55.
But the first difference in the times given is 8. 
Therefore, the missing time must be such that it divides the interval of 8 minutes into 3 minutes and 5 minutes. 
The missing time should be 1:58 pm and hence, option B is the right answer. 

All numbers from 1 to 9 repeat their last digits over a cycle of 4. 
63 can be written as 4k+3.
85 can be written as 4k+1. 
Some number's third power should yield the first digit as some number's first power.
$$2^3$$ will yield $$8$$ as the last digit (2 and 8 is a possible solution).X+Y = 10
$$3^3$$ will yield $$7$$ as the last digit (3 and 7 is a possible solution).X + Y = 10
$$4^3$$ will yield $$4$$ as the last digit and hence, can be eliminated. 
$$5$$ and $$6$$ yield $$5$$ and $$6$$ respectively as the last digit for any power and hence, can be eliminated.
$$7^3$$ will yield $$3$$ as the last digit (7 and 3 is a possible solution). X+Y=10.
$$8^3$$ will yield $$2$$ as the last digit. 8+2 =10.
As we can see, X+Y = 10 in all the cases. Therefore, option B is the right answer.

The sum of 4 consecutive numbers should be a power of 2. 
The powers of 2 are 1, 2, 4, 8, 16, 32, 64, 128. The maximum possible value that 4 consecutive days can take is 28 + 29 + 30 + 31 = 118.
We can eliminate 1, 2, 4, and 8 since sum of 4 consecutive integers is always greater than 9. 
Let the first day be x. 
x+x+1+x+2+x+3 = 16
=> 4x = 10
x = 2.5
We can eliminate 16.
Let us check for 32.
x+x+1+x+2+x+3 = 32
4x = 26.
x is not an integer. 
Let us check the case in which x is the last day of the month. 
Even in the month of February, the least value that x can take is 28.
28+1+2+3 = 34 > 32.
We can eliminate 32 as well. 
Let us check for 64.
4x+6 = 64
4x = 58
The sum of no 4 consecutive days in the same month can be expressed as 4k+6. 
Let us check the cases in which 2 months are involved.
29+30+1+2 = 62 < 64.
30 + 31 + 1 + 2 = 64. This is a possible combination. 
There are 7 months with 31 days in a year. We have to eliminate December since 1 will spill over to the 1st of January of the next year. 
Therefore, in a year, there will be 6 such instances. Therefore, option B is the right answer. 

It has been given that $$N = (11^{p + 7})(7^{q - 2})(5^{r + 1})(3^{s})$$ is a perfect cube. All the factors given are prime. Therefore, the power of each number should be a multiple of 3 or 0. 

$$p,q,r$$ and $$s$$ are positive integers. Therefore, only the power of the expressions in which some number is subtracted from these variables or these variables are subtracted from some number can be made $$0$$.

$$11^{p + 7}$$:

This expression must be made a perfect cube. The nearest perfect cube is $$11^9$$. Therefore, the least value that $$p$$ can take is $$9-7=2$$.

$$7^{q - 2}$$

The least value that $$q$$ can take is 2. If $$q=2$$, then the value of the expression $$7^{q-2}$$ will become $$7^0=1$$, without preventing the product from becoming a perfect cube. 

$$5^{r+1}$$:

The least value that $$r$$ can take is $$2$$.

$$3^{s})$$:

The least value that $$s$$ can take is $$3$$. 

Therefore, the least value of the expression $$p + q + r + s$$ is $$2+2+2+3=9$$.
Therefore, option E is the right answer.

Converting numbers in decimal system

=> $$2061_B = 2B^3 + 6B + 1$$

and $$601_B = 6B^2 + 1$$

Acc. to ques, => $$2B^3 + 6B + 1 + 6B^2 + 1 = 432$$

Solving above equation, we get : $$B = 5$$

$$\therefore 1010_5$$ in decimal system = $$1(5)^3 + 1(5) = 130$$

Note :-  This question is technically wrong as we cannot have digit 6 in Base 5. 

If $$(a + b)^{(a + b)}$$ is divisible by 500, 

$$500 = 2^2 \times 5^3$$

=> Least value of $$a + b = 2 \times 5 = 10$$

For least $$a$$ and $$b$$, let $$a = 1$$

=> $$b = 10 - 1 = 9$$

$$\therefore$$ Min $$(a \times b) = 1 \times 9 = 9$$

Since a,b,c are consecutive integers

=> $$a = b-1$$ and $$c = b+1$$

Expression : $$\frac{a^3+b^3+c^3+3abc}{(a+b+c)^2}$$

= $$\frac{(b - 1)^3 + b^3 + (b + 1)^3 + 3 (b - 1) b (b + 1)}{(b - 1 + b + b + 1)^2}$$

= $$\frac{b^3 + 3b + b^3 + b^3 + 3b + 3b^3 - 3b}{9 b^2}$$

= $$\frac{6 b^3 + 3 b}{9 b^2} = \frac{2 b^2 + 1}{3 b}$$

Putting different values of b from - 10 to 10, we can verify that only - 1 and 1 satisfies to get integer values for the expression.

Ans - (C)

Expression : $$\frac{a + b + c + d}{a + b + c - d}$$

To maximize the above expression, we have to minimize the denominator

Minimum value of the denominator = 1

So we can make $$a + b + c = 26$$ and $$d = 25$$   (as maximizing d will give denominator the least value).

So required maximum value = $$\frac{a + b + c + d}{a + b + c - d}$$

= $$\frac{26 + 25}{26 - 25} = 51$$

L.C.M. of 3,4,5,6 = 60

Number is of the form = $$60 k_1 + 2$$ -------------(i)

When divided by 11, it leaves 0 remainder so number will also be of the form = $$11 k_2$$ ---------(ii)

Hence equating (i) and (ii), we get,

$$60k_1 + 2 = 11k_2$$

$$60k_1 - 11k_2 = -2$$ or $$11k_2 - 60k_1 = 2$$ -----------(iii)

It means $$60k_1$$ will leave remainder 9 when divide by 11.

Lets consider values for 60k1, if k1=1, 60k1=60, reminder is 60mod11=5

120mod11 will be 5+5=10, 180mod11 will be 5+5+5=15, since 15>11, reminder will be 15-11=4

240mod11 reminder will be 4+5=9

$$\therefore$$ By remainder root $$\frac{4 k_1}{11}$$ should leave remainder as 9 or -2

=> Possible values of $$K_1 = 4, 15, 26, 37, 48, 59$$ (As 11 and 60 are co-prime)

$$\therefore$$ Required value = $$60 \times 59 + 2 = 3540 + 2 = 3542$$

Alternatively,
L.C.M. of 3,4,5,6 = 60
As the number 60k+2 is divisible by 11, 60k leaves a reminder of 9
60mod11=5, 120mod11=10, 180mod11=4, 240mod11=9
Hence the first number where both conditions are satisfied as 242.
As 60 and 11 are co-prime, the next number where this is true is 242+60*11
Hence, the numbers are in the form 242+660k
For 6th number, k=5   =>  3300+242=3542

It is evident that, 1 wrong 2 marks question would result in 2.33 deduction from the total (As negative in 2 marks question is 1/3 of a mark)

1 wrong of 1 mark question lead to deduction of 1.25 marks

1 unattempted of 1 mark question lead to deduction of 1.5 marks

1 unattempted of 2 marks question lead to deduction of 3 marks

$$\therefore$$ Rank of student who scores 27.5 = 5

Since the three digit number is divisible by 10, then the unit's digit is 0

Let the three digit number = $$a b 0$$

After the digits are interchanged, the new number is also divisible by 10, thus only a and b are interchanged.

=> New number = $$b a 0$$

Difference between number is divisible by 40

=> $$(100a + 10b) - (100b + 10a) = 40 k$$    (k is constant)

=> $$90a - 90b = 90 (a - b) = 40 k$$

=> k = $$\frac{9\left(a-b\right)}{4}$$

Since k is a natural number (a-b) should be a multiple of 4

If a = 9 , the values of b that satisfies the given equation are 1,5

If a = 8 , the value of b that satisfies the given equation is 4

If a = 7 , the values of b that satisfies the given equation is 3

If a = 6 , the values of b that satisfies the given equation is 2

If a = 5 , the values of b that satisfies the given equation is  1

The number could be = 510,620,730,840,950, 910

Thus, there are 6 numbers that satisfy these conditions.

None of the answers given are correct. The reasoning is as given below.

999000 is a multiple of 8 but not of 16. If N! is a multiple of 16, M! would also be a multiple of 16 and hence M!-N! would be a multiple of 16.

Hence, as M!-N! = 999000, it would imply that N! is a multiple of 8 and not of 16. Therefore, N is either 4 or 5. So, N! is either 24 or 120. So, it would imply that M! is either 999024 or 999120. Both of which are not factorials for any natural number. 

Hence, the given question is wrong. 

The terms $$x, 17, 3x - y^{2} - 2$$ and $$3x + y^{2} - 30$$ are in A.P.

=> Common difference = $$d = 17 - x$$ ----------Eqn(I)

$$d = 3x - y^2 - 19$$ ----------Eqn(II)

$$d = 2y^2 - 28$$ ----------Eqn(III)

From eqn(I) & (II), => $$17 - x = 3x - y^2 - 19$$

=> $$4x - y^2 = 36$$ -------Eqn(IV)

From eqn(II) & (III), => $$3x - y^2 - 19 = 2y^2 - 28$$

=> $$x - y^2 = -3$$ ---------Eqn(V)

Solving eqn(IV) & (V), we get : 

$$x = 13 , y^2 = 16$$

=> Terms are = $$13,17,21,25$$

$$\therefore$$ Sum = $$13+17+21+25 = 76$$, which is divided by 2.   (among the given options)

Natural numbers = $$x , y , (x+y) , (x-y)$$

Statement I : As all the numbers are prime, therefore, either x or y has to be 2 because otherwise (x+y) cannot be prime.

Case 1 : If x = 2, then (x-y) cannot be prime

Case 2 : If y = 2, numbers = $$(x-2) , x , (x+2)$$

These numbers are prime, hence all possibility = 3,5,7

$$\therefore$$ Sum = 2+3+5+7 = 17

Using statement II, we cannot find the required sum, as no specific value of mean is given.

Thus, statement I alone is sufficient.

In Base B, $$297_B = 2B^2 + 9B + 7$$

and $$792_B = 7B^2 + 9B + 2$$

It is given that $$297_{B}$$ is a factor of $$792_{B}$$

=> $$\frac{7B^2 + 9B + 2}{2B^2 + 9B + 7}$$ must be an integer    

=> $$\frac{(2B^2 + 9B + 7) + (5B^2 - 5)}{2B^2 + 9B + 7}$$

=> $$\frac{5B^2 - 5}{2B^2 + 9B + 7} + 1 = k$$

=> $$5B^2 - 5 = (2B^2 + 9B + 7) k$$      (where $$k$$ is factor)

Put $$k = 1$$

=> $$5B^2 - 5 = 2B^2 + 9B + 7$$

=> $$B^2 - 3B - 4 = 0$$

=> $$(B - 4) (B + 1) = 0$$

=> $$B = 4 , -1$$

Since, B is a base,so B must be greater than 9. Hence, it is not possible

Put $$k = 2$$

=> $$5B^2 - 5 = 4B^2 + 18B + 14$$

=> $$B^2 - 18B - 19 = 0$$

=> $$(B - 19) (B + 1) = 0$$

=> $$B = 19 , -1$$

$$\therefore B = 19$$

Expression : $$\sum_{n=1}^{13}\frac{1}{n} = \frac{x}{13!}$$

=> $$\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + ...... + \frac{1}{13} = \frac{x}{13!}$$

=> $$x = \frac{13!}{1} + \frac{13!}{2} + \frac{13!}{3} + ......... + \frac{13!}{13}$$

Now, if $$x$$ is divided by 11

=> $$\frac{13! + \frac{13!}{2} + \frac{13!}{3} + ......... + \frac{13!}{13}}{11}$$

All terms are divisible by 11 except $$\frac{13!}{11}$$

$$\therefore$$ Remainder if x is divided by 11 = Remainder of $$\frac{13!}{11 \times 11}$$

= $$(10! \times 12 \times 13) \% 11$$

= $$(10 \times 1 \times 2) \% 11 = 20 \% 11 = 9$$

Let us analyse all numbers from 100 to 200 (both excluded).

Total numbers when unit digit is 2 = 10 {102,112,122, ..., 192}

Total numbers when tens digit is 2 = 10 {120,121,121, ..., 129}

We can see that the there is exactly one number{122} which have both tens and unit digit as 1. So total unique numbers from 100 to 200 (both excluded) which contain at least one 2 digit = 10 + 10  - 1 = 19

From 200 to 299 (both included), all numbers have at least one digit as 2. Total such number = 100.

From 300 to 400, 400 to 500, 500 to 600, 600 to 700 and 700 to 800 we will have 19 number for each pair with at least one 2 digit.

Therefore, total whole numbers between 100 and 800 contain the digit 2 = 100 + 19*6 = 214.

As the number is between 10 and 100 and 100 cannot be the number we are looking for, we can assume the number to be of two-digits.
Let the number be xy.
According to the question, for the number to be interesting
x + y + xy = 10x + y
On solving, we get 
xy = 9x
or, x (9 - y) = 0
x cannot be 0, because we need a number greater than or equal to 10.
So, 9 - y = 0
=> y = 9
For all the numbers whose unit digit is 9 will be an interesting number.
So, the numbers are 19, 29, 39, 49, .....99
There are 9 such numbers out of 91 total numbers between 10 and 100 including both.
Required fraction = $$\dfrac{9}{91}$$ = 0.0989
As this is not given in any of the options, the answer will be "none of the above".
Hence, option E is the correct answer.

$$p^q = q^p$$.
It has been given that $$q = 9p$$.

Substituting, we get, 
$$p^{9p}=(9p)^p$$
$$(p^p)^9 = 9^p*p^p$$
=> $$(p^p)^8 = 9^p$$
$$p^{8p}=9^p$$
Raising the power to $$\frac{1}{p}$$ on both sides, we get, 
$$p^8=9$$
$$p=\sqrt[8]{9}$$.
Therefore, option D is the right answer. 

$$(xxx)_{b}=x^3$$
=> $$xb^2+xb+x = x^3$$
=> $$b^2+b+1=x^2$$
On substituting b=1,and b=2, we get $$x^2$$ as $$3$$, and $$7$$. Since $$3$$ and $$7$$ are not perfect squares, we can infer that no number satisfies the given condition. Therefore, option E is the right answer.

Let us evaluate the statements one by one:
I: 103 and 7 are the only prime factors of 1000027
On successively dividing 1000027 by 103 and 7, we get 1387 as the answer. 
1387 is divisible by 19. 
Therefore, statement I is false. 
II: $$\sqrt[6]{6!}>\sqrt[7]{7!}$$
Raising the power to 42 on both sides, we get,
$$[6!]^7>[7!]^6$$
$$6!*[6!]^6 > 7^6*[6!]^6$$
$$7^6$$ is greater than $$6!$$.
Therefore, statement II is false. 
III:It has been given that the person travels one-half of the journey at x kmph. 
Let us assume the distance to be '2d'. 
Let us assume the average speed to be 2d and check for the feasibility.
Let the speed at which the person travels the other half of the journey be y.
d/x + d/y = 2d/2x
d/x + d/y = d/x
=> d/y = 0 or y tends to infinity.
Therefore, such a scenario is not possible and hence, statement III is true.
Only statement III is true. Therefore, option C is the right answer.

Little Pika can add   (1000, 1001), (1001, 1002), (1002, 1003), (1003, 1004), (1004, 1005) and (1009, 1010).

Similarly, he can add (1010, 1011), (1011, 1012), (1012, 1013), (1013, 1014), (1014, 1015) and (1019, 1020).

Similarly, he can add (1020, 1021), (1021, 1022), (1022, 1023), (1023, 1024), (1024, 1025) and (1029, 1030).

Similarly, he can add (1030, 1031), (1031, 1032), (1032, 1033), (1033, 1034), (1034, 1035) and (1039, 1040).

Similarly, he can add (1040, 1041), (1041, 1042), (1042, 1043), (1043, 1044) , (1044, 1045) and (1049, 1050).

We can see that there are 30 cases when we have changed unit and tens digit. Now the hundreds digit can be anything from {0, 1, 2, 3, 4}. 

Hence, total number of such pairs which Pika can add = 5*30 = 150. 

He can also number of form, (1099, 1100), (1199, 1200), (1299, 1300), (1399, 1400) (1499, 1500) and (1999, 2000) 

Therefore, we can say that Pika can add 150+6 = 156 numbers. Hence, option C is the correct answer. 

The meter skips all the numbers in which there is a 5

From, (1 to 99) -> 5 occurs 10 times in tens place and 10 times in units place (which also includes 55)

=> Total occurrence = $$10 + 10 - 1 = 19$$

Similarly from (100 to 199), from (200 to 299)..., from (400 to 499) , from (600 to 699),...from (900 to 999)

It occurs = $$8 \times 19 = 152$$

Now, from (500 to 599), there are 100 numbers, => micromanometer reading can change from 499 to 600.

Thus, total numbers skipped from (1 to 999) = $$19 + 152 + 100 = 271$$

Similarly, from (1000 to 1999) = 271

and from (2000 to 2999) = 271

Also, 3005 and 3015 are also skipped

=> Total number of skips = $$271 + 271 + 271 + 2 = 815$$

$$\therefore$$ Actual pressure = $$3016 - 815 = 2201$$

Account number = $$2 x_1x_2x_3x_4x_5x_6x_7$$

Case 1 : $$x_1x_2x_3$$ is exactly same as $$x_4x_5x_6$$

=> $$x_1x_2x_3 = x_4x_5x_6 = 000$$ and $$x_7$$ = 1 to 9   (as '20000000' is not valid)

=> 9 possibilities

Now, $$x_1x_2x_3 = x_4x_5x_6$$ = 001 to 999 and $$x_7$$ = 0 to 9

=> $$999 \times 10 = 9990$$ possibilities.

Case 2 : $$x_1x_2x_3$$ is exactly same as $$x_5x_6x_7$$

=> $$x_1x_2x_3 = x_5x_6x_7 = 000$$ and $$x_4$$ = 1 to 9   (as '20000000' is not valid)

=> 9 possibilities

Now, $$x_1x_2x_3 = x_5x_6x_7$$ = 001 to 999 and $$x_4$$ = 0 to 9

=> $$999 \times 10 = 9990$$ possibilities.

Subtracting common possibilities in above cases.

=> $$x_4x_5x_6 = x_5x_6x_7$$

=> $$x_4 = x_5 = x_6 = x_7$$

Except '0', the possibilities are 1111,2222,....,9999 => 9 possibilities.

$$\therefore$$ Maximum possible number of customers having a ‘magic’ account number

= $$(9 + 9990) + (9 + 9990) - (9)$$

= $$19989$$

Expression : $$x^{2}+ 4xy+ 6y^{2}-4y+ 4$$

= $$(x^2 + 4xy + 4y^2) + (2y^2 - 4y + 4)$$

= $$(x + 2y)^2 + \frac{1}{2} (4y^2 - 8y + 8)$$

= $$(x + 2y)^2 + \frac{1}{2} [(2y)^2 - 2 (2y) (2) + (2)^2 + 4]$$

= $$(x + 2y)^2 + \frac{1}{2} [(2y - 2)^2 + 4]$$

= $$(x + 2y)^2 + \frac{1}{2} (2y - 2)^2 + 2$$

Since, $$x$$ and $$y$$ are real, => Min value of $$(x + 2y)^2 = 0$$

Minimum value of $$(2y - 2)^2 = 0$$

$$\therefore$$ Minimum value of expression = $$0 + 0 + 2 = 2$$

Given that unit digit of $$X$$ and product of all 4 digits of $$X$$ are prime.

The product of two numbers to be prime is possible only when one of the numbers is prime and the other is ‘1’.

The possibilities for the prime unit digits are - 2,3,5,7     (Note that 1 is not a prime number)

Hence the possibility of remaining 3 digits, considering the product of all 4 digits to be prime is ‘111’ only.

Hence all the possible numbers are 1112,1113,1115,1117

$$\therefore$$ Total 4 integers are possible.