CAT Number Systems Questions

We are given that average of three distinct integers is 28, that means the sum of these three integers is 28x3=84

Let us write $$x+y+z=84$$
x, y, z being the three distinct integers in ascending order. 

If the smallest number is increased by 7 and the largest number is reduced by 10
$$(x+7)+(y)+(z-10)=81$$

New arithmetic mean will be $$\frac{81}{3}=27$$
And this is said to be 2 more than the middle number, meaning
$$27-2=y=25$$

$$x+z=59$$

We are given that difference between the largest and the smallest numbers becomes 64, 
$$(z-10)-(x+7)=64$$
$$z-x=81$$

Adding the two equations we get, $$2z=140$$
$$z=70$$

There are multiple ways of solving such questions involving remainders; one easy way is to look for a power of numerator that leaves a remainder of 1 or 01 when divided by the denominator. 

In this instance, 1000, when divided by 13, leaves a remainder of -1

We can rewrite the numerator as $$\frac{10^{66}\times\ 100}{13}$$

The remainder would be $$\left[\frac{10^{66}}{13}\right]_R\times\ \left[\frac{100}{13}\right]_R$$
$$\left(-1\right)^{22}\times\ 9$$
9

Therefore, Option C is the correct answer. 

To find the value of $$10^{100}mod\left(7\right)$$
When 10 is divided by 7, it leaves a remainder 3, so the above equation can be written as, 
$$3^{100}mod\left(7\right)$$

Now looking at the cyclicality of powers of 3 when divided by 7, 
$$3^1mod  7=3$$

$$3^2mod  7=2$$

$$3^3mod  7=6$$

$$3^4mod  7=4$$

$$3^5mod  7=5$$

$$3^6mod  7=1$$

From this calculation, it is evident that the powers of 3 modulo 7 repeat every 6 steps. This forms a cycle: 3, 2, 6, 4, 5, 1

$$3^{100}=\left(3^6\right)^{16}\times\ \left(3^4\right)$$

Since $$3^6mod 7=1$$

We just need to consider $$3^4mod 7$$ which equals 4

Hence the answer is 4. 

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, $$(10^{n-1}+10^{n-2}+10^{n-3}+...) \times ((n-1)!)\times(Sum\ of\ numbers)$$

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. 

There are multiple ways of solving these sorts of questions. One method is to look for powers of the term in the numerator that leave a remainder of 1 or -1 when divided by the denominator. 

Noting down the powers of 3, 3, 9, 27, 81, 243

243 is one such number, 242 is multiple of 11 (11 times 22), hence 243 will leave a remainder of 1 when divided by 11. 

243 is 3 raised to power 5; we can rewrite the given term as $$\dfrac{3^{330}\times\ 3^3}{11}$$
The overall remainder will be $$\left[\dfrac{3^{330}}{11}\right]_R\times\ \left[\dfrac{3^3}{11}\right]_R$$

$$\left[\dfrac{3^{5\times\ 66}}{11}\right]_R\times\ \left[\dfrac{3^3}{11}\right]_R$$
$$\left[\dfrac{243^{66}}{11}\right]_R\times\ \left[\dfrac{3^3}{11}\right]_R$$
$$1^{66}\times\ \left[\dfrac{27}{11}\right]_R$$
$$1\times\ 5$$
$$5$$

Therefore, Option A 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,e a total of 288 numbers

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

Therefore, 378 is the correct answer. 

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 D

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 => the factor of 2 should be atleast 3, for 168 to be a factor => n = 3.

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

=> So, m + n = 15.

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 C

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: $$N\ =\ p_1\times\ p_2$$ (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 $$(p+1)\cdot(q+1)$$(Where, $$N=a^p\cdot b^q,$$ 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 $$N=2^4\times\ 3^2\ =144$$, and the second lowest value of N is $$N=2^2\times\ 3^4\ =324$$.

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.

1-digit numbers => We have 1 to 9 => 9

2-digit numbers => x y, we have 9 ways to choose x from 1 to 9 => 9 ways and 9 ways to choose y (0 to 9 except x) => 9*9 = 81

3-digit numbers => x y z, we have 9 ways to choose x, 9 ways to choose y and 8 ways to choose z => 9*9*8 = 648.

Total numbers till 1000 without digits repeated in them is 9 + 81 + 648 = 738.

Let the six numbers be a, b, c, d, e, f in ascending order

a+b = 28

e+f =  56

If we want to maximise the average then we have to maximise the value of c and d and maximise e and minimise f

e+f = 56

As e and f are distinct natural numbers so possible values are 27 and 29

Therefore c and d will be 25 and 26 respecitively

So average = $$\frac{\left(a+b+c+d+e+f\right)}{6}=\frac{\left(28+25+26+56\right)}{6}=\frac{135}{6}=22.5$$

A is the HCF of $$3^k+4^k+5^k$$ for different values of k.

For k = 1, value is 12

For k = 2, value is 50

For k = 3, value is 216

HCF is 2. Therefore, A = 2

$$4^k+3\left(4^k\right)+4^{k+2}=4^k\left(1+3\right)+4^{k+2}=4^{k+1}+4^{k+2}=4^{k+1}\left(1+4\right)=5\cdot4^{k+1}$$

HCF of the values is when k = 1, i.e. 5*16 = 80

Therefore, B = 80

A + B = 82

To find the largest possible value of n, we need to find the value of n such that n! is less than 15000.

7! = 5040

8! = 40320 > 15000

This implies 15000! is not divisible by 40320!

Therefore, maximum value n can take is 7.

The answer is option B.

Since the total number of students, when divided by either 9 or 10 or 12 or 25 each, gives a remainder of 4, the number will be in the form of LCM(9,10,12,25)k + 4 = 900k + 4.

It is given that the value of 900k + 4 is less than 5000.

Also, it is given that 900k + 4 is divided by 11.

It is only possible when k = 2 and total students = 1804.

So, the number of 12 students group = 1800/12 = 150.

$$2.25\leq2+2^{n+2}\leq202$$

$$2.25-2\le2+2^{n+2}-2\le202-2$$

$$0.25\le2^{n+2}\le200$$

$$\log_20.25\le n+2\le\log_2200$$

$$-2\le n+2\le7.xx$$

$$-4\le n\le7.xx-2$$

$$-4\le n\le5.xx$$

Possible integers = -4, -3, -2, -1, 0, 1, 2, 3, 4, 5

If we see the second expression that is provided, i.e 

$$3+3^{n+1}$$, it can be implied that n should be at least -1 for this expression to be an integer.

So, n = -1, 0, 1, 2, 3, 4, 5.

Hence, there are a total of 7 values.

Given the 4 digit number :

Considering the number in thousands digit is a number in the hundredth digit is b, number in tens digit is c, number in the units digit is d.

Let the number be abcd.

Given that a+b+c = 14. (1)

                 b+c+d = 15. (2)

                 c = d+4. (3).

In order to find the maximum number which satisfies the condition, we need to have abcd such that a is maximum which is the digit in thousands place in order to maximize the value of the number. b, c, and d are less than 9 each as they are single-digit numbers.

Substituting (3) in (2) we have b+d+4+d = 15, b+2*d = 11.  (4)

Subtracting (2) and (1) : (2) - (1) = d = a+1.   (5)

Since c cannot be greater than 9 considering c to be the maximum value 9 the value of d is 5.

If d = 5, using d = a+1, a = 4.

Hence the maximum value of a = 4 when c = 9, d = 5.

Substituting b+2*d = 11. b = 1.

The highest four-digit number satisfying the condition is 4195

Let the number be 'abc'. Then, $$2<a\times\ b\times\ c<7$$. The product can be 3,4,5,6.

We can obtain each of these as products with the combination 1,1, x where x = 3,4,5,6. Each number can be arranged in 3 ways, and we have 4 such numbers: hence, a total of 12 numbers fulfilling the criteria.

We can factories 4 as 2*2 and the combination 2,2,1 can be used to form 3 more distinct numbers. 

We can factorize 6 as 2*3 and the combination 1,2,3 can be used to form 6 additional distinct numbers. 

Thus a total of 12 + 3 + 6 = 21 such numbers can be formed.

Here there are two cases possible

Case 1: When 7 is at the left extreme

In that case 3 can occupy any of the three remaining places and the remaining two places can be taken by (0,1,2,4,5,6,8,9)

So total ways 3(8)(7)= 168

Case 2: When 7 is not at the extremes

Here there are 3 cases possible. And the remaining two places can be filled in 7(7) ways.(Remember 0 can't come on the extreme left)

Hence in total 3(7)(7)=147 ways

Total ways 168+147=315 ways

$$0.2<\frac{n}{11}<0.5$$ 

=> 2.2<n<5.5

Since n is an even natural number, the value of n = 4

$$0.2<\frac{m}{20}<0.5$$  => 4< m<10. Possible values of m = 5,6,7,8,9

Since $$0.2<\frac{n}{m}<0.5$$, the only possible value of m is 9

Hence m-2n = 9-8 = 1

Total number of numbers from 100 to 999 = 900

The number of three digits numbers with unique digits:

_ _ _

The hundredth's place can be filled in 9 ways ( Number 0 cannot be selected)

Ten's place can be filled in 9 ways

One's place can be filled in 8 ways

Total number of numbers = 9*9*8 = 648

Number of integers in the set {100, 101, 102, ..., 999} have at least one digit repeated = 900 - 648 = 252

Possible values of x = 3,4,5,6,7,8,9

When x = 3, there is no possible value of y

When x = 4, the possible values of y = 22

When x = 5, the possible values of y=21,22

When x = 6, the possible values of y = 20.21,22

When x = 7, the possible values of y = 19,20,21,22

When x = 8, the possible values of y=18,19,20,21,22

When x = 9, the possible values of y=17,18,19,20,21,22

The unique values of N = 26,27,28,29,30,31

The number of multiples of 2 between 1 and 120 = 60

The number of multiples of 5 between 1 and 120 which are not multiples of 2 = 12

The number of multiples of 7 between 1 and 120 which are not multiples of 2 and 5 = 7

Hence, number of the integers 1, 2, … , 120, are divisible by none of 2, 5 and 7 = 120 - 60 - 12 - 7 = 41

$$ab\ =\ 4^{2017}=2^{4034}$$

The total number of factors = 4035.

out of these 4035 factors, we can choose two numbers a,b such that a<b in [4035/2] = 2017.

And since the given number is a perfect square we have one set of two equal factors.

.'. many pairs(a, b) of positive integers are there such that $$a\leq b$$ and $$ab=4^{2017}$$ = 2018. 

The four digit even numbers will be of form:

1100, 1122, 1144 ... 1188, 2200, 2222, 2244 ... 9900, 9922, 9944, 9966, 9988

Their sum 'S' will be (1100+1100+22+1100+44+1100+66+1100+88)+(2200+2200+22+2200+44+...)....+(9900+9900+22+9900+44+9900+66+9900+88)

=> S=1100*5+(22+44+66+88)+2200*5+(22+44+66+88)....+9900*5+(22+44+66+88)

=> S=5*1100(1+2+3+...9)+9(22+44+66+88)

=>S=5*1100*9*10/2 + 9*11*20

Total number of numbers are 9*5=45

.'. Mean will be S/45 = 5*1100+44=5544.

Option D

Since $$c<9$$, we can have the following viable combinations for $$b\times\ c\ =96$$ (given our objective is to minimize the sum):

$$48\times\ 2$$ ; $$32\times3$$ ; $$24\times\ 4$$ ; $$16\times6$$ ; $$12\times8$$ 

Similarly, we can factorize $$a\times\ b\ = 432$$ into its factors. On close observation, we notice that $$18\times24\ and\ 24\ \times\ 4\ $$ corresponding to $$a\times b\ and\ b\times\ c\ $$ respectively together render us with the least value of the sum of $$a+b\ +\ c\ \ =\ 18+24+4\ =46$$

Hence, Option D is the correct answer.

$$\ \frac{\ n^2+3n+4n+12}{n^2-4n+3n-12}$$

=$$\ \frac{\ n^{ }\left(n+3\right)+4\left(n+3\right)}{n^{ }\left(n-4\right)+3\left(n-4\right)}$$

=$$\ \frac{\left(\ n+4\right)\left(n+3\right)}{\left(n-4\right)\left(n+3\right)}$$

=$$\ \frac{\left(\ n+4\right)}{\left(n-4\right)}$$

=$$\ \frac{\left(\ n-4\right)+8}{\left(n-4\right)}$$

=$$\ 1+\frac{8}{\left(n-4\right)}$$ which will be maximum when n-4 =8

n=12

D is the correct answer.

$$n^2-m^2=105$$

(n-m)(n+m) = 1*105, 3*35, 5*21, 7*15, 15*7, 21*5, 35*3, 105*1.

n-m=1, n+m=105  ==> n=53, m=52

n-m=3, n+m=35 ==> n=19, m=16

n-m=5, n+m=21  ==> n=13, m=8

n-m=7, n+m=15 ==> n=11, m=4

n-m=15, n+m=7 ==> n=11, m=-4

n-m=21, n+m=5 ==> n=13, m=-8

n-m=35, n+m=3 ==> n=19, m=-16

n-m=105, n+m=1 ==> n=53, m=-52

Since only positive integer values of m and n are required. There are 4 possible solutions.

Assume the numbers are a and b, then ab=616

We have, $$\ \ \frac{\ a^3-b^3}{\left(a-b\right)^3}$$ = $$\ \frac{\ 157}{3}$$

=> $$\ 3\left(a^3-b^3\right)\ =\ 157\left(a^3-b^3+3ab\left(b-a\right)\right)$$

=> $$154\left(a^3-b^3\right)+3*157*ab\left(b-a\right)$$ = 0

=> $$154\left(a^3-b^3\right)+3*616*157\left(b-a\right)$$ = 0        (ab=616)

=>$$a^3-b^3+\left(3\times\ 4\times\ 157\left(b-a\right)\right)$$    (154*4=616)

=> $$\left(a-b\right)\left(a^2+b^2+ab\right)\ =\ 3\times\ 4\times\ 157\left(a-b\right)$$

=> $$a^2+b^2+ab\ =\ 3\times\ 4\times\ 157$$

Adding ab=616 on both sides, we get

$$a^2+b^2+ab\ +ab=\ 3\times\ 4\times\ 157+616$$

=> $$\left(a+b\right)^2=\ 3\times\ 4\times\ 157+616$$ = 2500

=> a+b=50

$$2^4 \times 3^5 \times 10^4$$

=$$2^4 \times 3^5 \times 2^4*5^4$$

=$$2^8 \times 3^5 \times 5^4$$

For the factor to be a perfect square, the factor should be even power of the number.

In $$2^8$$, the factors which are perfect squares are $$2^0, 2^2, 2^4, 2^6, 2^8$$ = 5

Similarly, in $$3^5$$, the factors which are perfect squares are $$3^0, 3^2, 3^4$$ = 3

In $$5^4$$, the factors which are perfect squares are $$5^0, 5^2, 5^4$$ = 3

Number of perfect squares greater than 1 = 5*3*3-1

=44

Let the six-digit number be ABCDEF

F = A+B+C, E= A+B, C=A, B= 2A, D= E+F.

Therefore D = 2A+2B+C = 2A + 4A + A= 7A.

A cannot be 0 as the number is a 6 digit number.

A cannot be 2 as D would become 2 digit number.

Therefore A is 1 and D is 7.

We know that one of the 3 numbers is 37.
Let the product of the other 2 numbers be x.
It has been given that 73x-37x = 720
36x = 720
x = 20

Product of 2 real numbers is 20. 
We have to find the minimum possible value of the sum of the squares of the 2 numbers. 
Let x=a*b
It has been given that a*b=20

The least possible sum for a given product is obtained when the numbers are as close to each other as possible. 
Therefore, when a=b, the value of a and b will be $$\sqrt{20}$$. ($$\sqrt{20}$$ is irrational number, but it is a real number)

Sum of the squares of the 2 numbers = 20 + 20 = 40.

Therefore, 40 is the correct answer. 

Let 'a' and 'b' are those two numbers. 

$$\Rightarrow$$ $$a^2+b^2 = 97$$

$$\Rightarrow$$ $$a^2+b^2-2ab = 97-2ab$$

$$\Rightarrow$$ $$(a-b)^2 = 97-2ab$$

We know that $$(a-b)^2$$ $$\geq$$ 0

$$\Rightarrow$$ 97-2ab $$\geq$$ 0

$$\Rightarrow$$ ab $$\leq$$ 48.5

Hence, ab $$\neq$$ 64. Therefore, option D is the correct answer.

$$4^{n} > 17^{19}$$

$$\Rightarrow$$ $$16^{n/2} > 17^{19}$$

Therefore, we can say that n/2 > 19

n > 38

Hence, option D is the correct answer. 

At $$x = 0, 2^x = 1$$ which is in the given range [0.25, 200]

$$2^x + 2$$ = 1 + 2 = 3 Which is divisible by 3. Hence, x = 0 is one possible solution. 

At $$x = 1, 2^x = 2$$ which is in the given range [0.25, 200]

$$2^x + 2$$ = 2 + 2 = 3 Which is divisible by 4. Hence, x = 1 is one possible solution. 

At $$x = 2, 2^x = 4$$ which is in the given range [0.25, 200]

$$2^x + 2$$ = 4 + 2 = 6 Which is divisible by 3. Hence, x = 2 is one possible solution. 

At $$x = 3, 2^x = 8$$ which is in the given range [0.25, 200]

$$2^x + 2$$ = 8 + 2 = 3 Which is not divisible by 3 or 4. Hence, x = 3 can't be a solution. 

At $$x = 4, 2^x = 16$$ which is in the given range [0.25, 200]

$$2^x + 2$$ = 16 + 2 = 18 Which is divisible by 3. Hence, x = 4 is one possible solution. 

At $$x = 5, 2^x = 32$$ which is in the given range [0.25, 200]

$$2^x + 2$$ = 32 + 2 = 34 Which is not divisible by 3 or 4. Hence, x = 5 can't be a solution. 

At $$x = 6, 2^x = 64$$ which is in the given range [0.25, 200]

$$2^x + 2$$ = 64 + 2 = 66 Which is divisible by 3. Hence, x = 6 is one possible solution. 

At $$x = 7, 2^x = 128$$ which is in the given range [0.25, 200]

$$2^x + 2$$ = 128 + 2 = 130 Which is not divisible by 3 or 4. Hence, x = 7 can't be a solution. 

At $$x = 8, 2^x = 256$$ which is not in the given range [0.25, 200]. Hence, x can't take any value greater than 7.

Therefore, all possible values of x = {0,1,2,4,6}. Hence, we can say that 'x' can take 5 different integer values. 

$$(x -1)x(x+1) = 15600$$
=> $$x^3 - x= 15600 $$
The nearest cube to 15600 is 15625 = $$25^3$$
We can verify that x = 25 satisfies the equation above.
Hence the three numbers are 24, 25, 26. Sum of their squares = 1877

For the value of given expression to be minimum, the values of $$a, b, c$$ and $$d$$ should be as close as possible. 30/4 = 7.5. Since each one of these are integers so values must be 8, 8, 7, 7. On putting these values in the given expression, we get
 $$(8 - 8)^{2} + (8 - 7)^{2} + (8 - 7)^{2}$$
=> 1 + 1 = 2

Let us assume that three positive consecutive integers are x, x+1, x+2. They are raised to first, second and third powers respectively. 

$$x^{1} + (x+1)^{2} + (x+2)^{3} = (x + (x+1) +(x+2))^{2}$$

$$x^{1} + (x+1)^{2} + (x+2)^{3}$$ = $$(3x + 3)^{2}$$

$$x^{3} + 7x^{2} + 15x + 9$$ = $$9x^{2} + 9 + 18x$$

After simplifying you get,

$$x^{3} - 2x^{2} - 3x = 0$$

=> x=0,3,-1

Since x is a positive integer, it can only be 3.

So, the minimum of the three integers is 3. Option a) is the correct answer.

Let the first operation be (1+40-1) = 40, the second operation be (2+39-1) = 40 and so on

So, after 20 operations, all the numbers are 40. After 10 more operations, all the numbers are 79

Proceeding this way, the last remaining number will be 781

$$7^4$$ = 2401 = 2400+1
So, any multiple of $$7^4$$ will always end in 01
Since 2008 is a multiple of 4, $$7^{2008}$$ will also end in 01

After the first sale, the remaining quantity would be (x/2)-0.5 and after the second sale, the remaining quantity is 0.25x-0.75

After the last sale, the remaining quantity is 0.125x-(7/8) which will be equal to 0

So 0.125x-(7/8) = 0 => x = 7

1/m + 4/n = 1/12
So, 1/m = 1/12 - 4/n
So, m = 12n/(n-48)
Since m is positive, n should be greater than 48
Also, since n is an odd number, it can take only 49, 51, 53, 55, 57 and 59
If n = 49, 51, 57 then m is an integer, else it is not an integer
So, there are 3 pairs of values for which the equation is satisfied

The number of players in all the teams put together = k * n

The number of players that are common is 1*n = n

So, the number of players in the tournament = kn - n = n(k-1)

Let the number be xxyy
xxyy = 1000x + 100x + 10y +y = 1100x+11y = 11(100x+y)
Since xxyy is a perfect square, and 11 is one of the factors, it should be a multiple of 121
So, xxyy = 121k, where k is also a perfect square.
For k = 4, xxyy is a 3 digit number and for k > 82, xxyy is a five digit number
Between 4 and 82, only for k = 64, the number is of the form xxyy
121*64 = 7744
So, there is only 1 number 7744 which is of the form xxyy and a perfect square.

Alternatively:

The number should be definitely more than 32 and less than 100 as the square is a two digit number.

A number of such form can be written as $$(50 \pm a)$$ and $$100 - a$$ where $$0 \leq a \leq 100$$
So, the square would be of form $$(50 \pm a)^2 = 2500 + a^2 \pm 100a$$ or $$ (100 - a)^2$$ i.e. $$10000 + a^2 + 200 a$$

In both cases, only $$a^2$$ contributes to the tens and ones digit. Among squares from 0 to 25, only 12 square i.e. 144 has repeating tens and ones digit. So, the number can be 38, 62, or 88. Checking these squares only 88 square is in the form of xxyy i.e. 7744.

Maximum sum of the four numbers <= 384=99+97+95+93
384/10 = 38.4
So, the perfect square is a number less than 38.4
The possibilities are 36, 25, 16 and 9
For the sum to be 360, the numbers can be 87, 89, 91 and 93
The sum of four consecutive odd numbers cannot be 250
For the sum to be 160, the numbers can be 37,39,41 and 43
The sum of 4 consecutive odd numbers cannot be 90
So, from the options, the answer is 41.

The addition of numerator and denominatpr should give a prime no. Only option E gives that. 

3 is a factor of 189 and 183 => A and D eliminated

17 is a factor of 187 and 221 => B and C eliminated

181 is prime.

a/d = a/b * b/c * c/d = 1/3 * 2 * 1/2 = 1/3

Similarly, b/e and c/f are 3 and 3/8 respectively.

b/e = b/c*c/d*d/e = 3

c/f = c/d*d/e*e/f = 3/8

=> Value of abc/def = 1/3 * 3 * 3/8 = 3/8

According to given condiiton we have p = (1 × 1!) + (2 × 2!) + (3 × 3!) + (4 × 4!) + … + (10 × 10!) . So n × n! = [(n + 1) - 1] × n! = (n + 1)! - n!. So equation becomes p = 2! - 1! + 3! - 2! + 4! - 3! + 5! - 4! +… + 11! - 10!.  So p = 11! - 1! = 11! - 1.  p + 2 = 11! + 1 .So when it is  divided by 11! gives a remainder of 1. Hence, option 4.

 Let A = 100x + 10y + z  and B = 100z + 10y + x .According to given condition B - A = 99(z - x) As (B - A) is divisible by 7 . So clearly  (z - x) should be  divisible by 7.  z and x can have values 8,1 or 9,2 , such that 8-2=9-2=7 and  y can have  value from 0 to 9.
So Lowest possible value of A lowest x,y and z which is  is 108 and the highest possible value of A is 299. 

Rightmost non-zero digit of $$30^{2720}$$ is same as rightmost non-zero digit of $$3^{272}$$.

272 is of the form 4k.

All $$3^{4k}$$ end in 1.

=> Right most non-zero digit is 1.

Let n can be a 2 digit or a 3 digit number.

First let n be a 2 digit number.
So n = 10x + y and Pn = xy and Sn = x + y
Now, Pn + Sn = n
Therefore, xy + x + y = 10x + y , we have y = 9 .
Hence there are 9 numbers 19, 29,.. ,99, so 9 cases .  

Now if n is a 3 digit number.
Let n = 100x + 10y + z
So Pn = xyz and Sn = x + y + z
Now, for Pn + Sn = n ;  xyz + x + y + z = 100x + 10y + z ; so. xyz = 99x + 9y .
For above equation there is no value for which the above equation have an integer (single digit) value.

Hence option D. 

The no. has all the digits as odd no. and is divisible by 3. So the possibilities are 

1113
1119
1131
1137
1155
1173
1179
1191
1197
Hence 9 possibilities . 

$$\frac{(30^{65}-29^{65})}{(30^{64}+29^{64})} = ((30-29)*\frac{(30^{64}+30^{63}*29+....+29^{64})}{(30^{64}+29^{64})}$$ , which is greater than 1 . Hence option D.

We know that x = $$16^3 + 17^3 + 18^3 + 19^3 = (16^3 + 19^3) + (17^3 + 18^3)$$

= $$(16 + 19)(16^2 - 16 * 19 + 19^2) + (17 + 18)(17^2 - 17 * 18 + 18^2)$$ = 35 × odd + 35 × odd = 35 × even = 35 × (2k)

=> x = 70k

=> Remainder when divided by 70 is 0.

The remainder when $$15^{23}$$ is divided by 19 equals $$(-4)^{23}$$
The remainder when $$23^{23}$$ is divided by 19 equals $$4^{23}$$
So, the sum of the two equals$$(-4)^{23}+(4)^{23}=0$$

From the options a, c and d all can possibly occur. Hence option b. Besides, if all people have different number of acquaintances, then first person will have 26 acquaintance, second person will have 25 acquaintance, third person will have 24 and so on till 27 th person will have 0 acquaintance. 0 acquaintance is practically not possible.

Between 100 and 200 both included there are 51 even nos. There are 7 even nos which are divisible by 7 and 6 nos which are divisible by 9 and 1 no divisible by both. hence in total 51 - (7+6-1) = 39 

There is one more method through which we can find the answer. Since we have to find even numbers, consider the numbers which are divisible by 14, 18 and 126 between 100 and 200. These are 7, 6 and 1 respectively.

Since  in all three cases the last digit is 1, the number should give remainder 1 when divided individually by 2,3,5 . So the no. may be 31 or 91 . Now 31 in base 2,3 and 5 give first digit as 1 in all the 3 cases while 91 gives exactly two out of the three cases the leading digit as 1. Hence option D.

positive integers n in the range $$12 \leq n \leq 40$$ such that the product (n -1)*(n - 2)*…*3*2*1 is not divisible by n, implies that n should be a prime no. So there are 7 prime nos. in given range. Hence option B.

No. of terms in series T , 3+(n-1)*8 = 467 i.e. n=59.

Now S will have atleast have of 59 terms i.e 29 .

Also the sum of 29th term and 30th term is less than 470.

Hence, maximum possible elements in S is 30. 

Consider the options:

24: (Square of sum of digits - the number) = 36 - 24 = 12

54: (Square of sum of digits - the number) = 81 - 54 = 27

34: (Square of sum of digits - the number) = 49 - 34 = 15

45: (Square of sum of digits - the number) = 81 - 45 = 36

So, option b) is the correct answer.

Let x = y+z such that z > y.

We know that $$48*(z - y)^2 = z^2 - y^2$$ 

Solving the above equation, we get z + y = 48

So, option d) is the correct answer.

$$2^4 = 16 = -1$$ (mod $$17$$)
So, $$2^{256} = (-1)^{64} $$(mod $$17$$)
$$= 1$$ (mod $$17$$)
Hence, the answer is 1. Option a).

In this problem, the lights are flashed at the intervals 2.5, 4.25 and 5.125 seconds and put off after one second each.

The total duration of intervals of these lights are (2.5+1) = 3.5 s, (4.25+1) = 5.25 s and (5.125+1) = 6.125 s.

We have to find the minimum duration. It would be the LCM of thes three numbers.
Since each word is put after a second. So LCM [$$(\frac{5}{2}+1 )(\frac{17}{4}+1)(\frac{41}{8}+1)$$] = LCM of numerator / HCF of denominator = 49*3/2 = 73.5. Hence they will glow for full one second after 73.5-1 =72.5 sec.

HCF of [(9/2), (27/4), (36/5)] = HCF of numerators / LCM of denominators = 9/20

Total weight = 18.45 lb

So no. of parts = 18.45 / (9/20) = 18.45*20/9 = 41 

Hence option d) is the correct answer.

Since after division of a number successively by 3, 4 and 7, the remainders obtained are 2, 1 and 4 respectively, the number is of form ((((4*4)+1)*3)+2)k = 53K 

Let k = 1; the number becomes 53

If it is divided by 84, the remainder is 53.

Option d) is the correct answer.

Alternative Solution.

Consider only for 3 and 4 and the remainders are 2 and 1 respectively.

So 5 is the first number to satisfy both the conditions. The number will be of the form 12k+5. Put different integral values of k to find whether it will leave remainder 5 when divided by 7. So the first number to satisfy such condition is 48x4+5= 53

Consider n=1 we have $$7^{6} - 6^{6}$$ which is = $$(7^{3} + 6^{3})(7^{3} - 6^{3})$$ = 13 * 127 * 43 which is divisible by all the 3 options.

Option d) is the correct answer.

Let x and y be the age of older and younger boy respectively(both single digit). According to given condition we know that $$y^2 = x$$.

Also Father's age = 10y + x and Mother's age = (10x+y)/2.

Only value which satisfies above equations is x=4 and y=2 .

Each box contains at least 120 and at most 144 oranges.

So boxes may contain 25 different numbers of oranges among 120, 121, 122, .... 144.

Lets start counting.

1st 25 boxes contain different numbers of oranges and this is repeated till 5 sets as 25*5=125.

Now we have accounted for 125 boxes. Still 3 boxes are remaining. These 3 boxes can have any number of oranges from 120 to 144.

Already every number is in 5 boxes. Even if these 3 boxes have different number of oranges, some number of oranges will be in 6 boxes.

Hence the number of boxes containing the same number of oranges is at least 6.

Let the 4 digit no. be xyzw.
 According to given conditions we have x+y=z+w --- Eq 1 , x+w=z --- Eq 2,y+w=2x+2z --- Eq 3
Eq 2 - Eq 3 : x-y = -2x-z --- Eq 4
Eq 1+ Eq 4 :2x = -2x+w
4x=w  --- Eq 5
Substitute w = 4x inEq2
5x = z
Substitute w = 4x inEq3
y+4x=2x+10x
y = 8x
Now the minimum value x can take is 1 so z =5 and the no. is 1854, which satisfies all the conditions.
 Hence option A .

Let the number be X.
From the given information, 53x - 35x = 540 => 18x = 540 => x = 30

So, new product = 35*30 = 1050

The product of 44 and 11 in decimal is 484.
If base is x , then 3*x^3+4*x^2+x+4 = 484 .
Hence, the given base system is of number 5.
Now, we have to convert 3111 (in base 5) to decimal number system.

3111 in base 5 equals $$1*5^0 + 1*5^1 + 1*5^2 + 3*5^3 = 1+5+25+375=406$$

If the number of pages is 44, the sum will be 44*45/2 = 22*45 = 990
So, the number 10 was added twice

We know that a=$$b^2-b$$.

So$$a^2-a$$ = b($$b^3-2b^2-b+2$$) . = (b - 2)(b - 1)( b)(b + 1)

The above given is a product of 4 consecutive numbers with the lowest number of the product being 2(given b >= 4)

In any set of four consecutive numbers, one of the numbers would be divisible by 3 and there would be two even numbers with the minimum value of the pair being (2,4).

Thus, for any value of b >=4, $$a^2-4$$ would be divisible by 3 x 2 x 4 = 24.

Thus, option C is the right choice. Options A and B are definitely wrong as a set of four consecutive numbers need not always include a multiple of 5 eg:(6,7,8,9)

The possible arrangements are 1,multiples of 2 ,remaining. So we have 1+2+4+8+16+32+64+31 = 158. Hence minimum no. of bags required is 8. 

We have a +c=e so possible summation 6+4=10 or 4+2 = 6.
Also b=2d so possible values  4=2*2 or 10=5*2.
So considering both we have b=10 , d=5, a=4 ,c=2, e=6.
Hence the correct option is B .

To be divisible by 4 , last 2 digits of the 5 digit no. should be divisible by 4 . So possibilities are 12,16,32,64,24,36,52,56 which are 8 in number. Remaining 3 digits out of 4 can be selected in $$^4C_3 $$ ways and further can be arranged in 3! ways . So in total = 8*4*6 = 192

Take x=3 , z=2 , y=5.

$$y(x-z)^2 = 5(3-2)^2 = 5$$

Option A gives 5 which is odd.

A red light flashes three times per minute and a green light flashes five times in 2 min at regular intervals. So red light fashes after every 1/3 min and green light flashes every 2/5 min. LCM of both the fractions is 2 min . 

Hence they flash together after every 2 min. So in an hour they flash together 30 times .

Case 1: $$a_1=0$$
So, D equals $$0.0a_20a_20a_2...$$
So, 100D equals $$a_2.0a_20a_2...$$
So, 99D equals $$a_2$$

Case 2: $$a_2=0$$
So, D equals $$0.a_10a_10a_1...$$
So, 100D equals $$a_10.a_10a_1....$$
So, 99D equals $$a_10$$

So, in both the cases, 99D is an integer. From the given options, only option C satisfies this condition (198=2*99) and hence the correct answer is C. 

Number of multiples of 3 between 100 and 200 = 66 - 33 = 33
Number of odd multiples = 16
Number of odd multiples of 21 = 3 (105, 147, 189)
So, the required number = 13

For number of zeroes we must count number of 2 and 5 in prime numbers below 100.

We have just 1 such pair of 2 and 5.

Hence we have only 1 zero.

In option d), x-y is even. So, the product of the three terms is even. So, d) cannot be true.

The numbers 1421, 1423 and 1425 when divided by 12 give remainder 5, 7 and 9 respectively.

5*7*9 mod 12 = 11 * 9 mod 12 = 99 mod 12 = 3

The difference of the numbers = 34041 - 32506 = 1535

The number that divides both these numbers must be a factor of 1535.

307 is the only 3 digit integer that divides 1535.

$$55^3 + 17^3 - 72^3$$ = $$(55-72)k + 17^3$$. This is divisible by 17

Remainder when $$55^3$$ is divided by 3 = 1

Remainder when $$17^3$$ is divided by 3 = -1

Remainder when $$72^3$$ is divided by 3 = 0

So, $$55^3 + 17^3 - 72^3$$ is divisible by 3

So, the answer is d) 3 and 17

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

$$0.04 = 0.1 - 2xy => xy = 0.03$$

So, |xy| = 0.03

$$(|x| + |y|)^2 = x^2 + y^2 + 2|xy| = 0.1 + 0.06 = 0.16$$

So, |x|+|y| = 0.4

Quotient of 1982/12 = 165, remainder = 2
Quotient of 165/12 = 13, remainder = 9
Quotient of 13/12 = 1, remainder = 1
Remainder of 1/12 = 1
So, the required number in base 12 = 1192

Let the four consecutive positive integers be $$a,a+1,a+2$$ and $$a+3$$.
Therefore, $$n=1+a(a+1)(a+2)(a+3)$$
Or, $$n = 1+(a^2+3a)*(a^2+3a+2)$$
Or, $$n = (a^2+3a)^2 + 2*(a^2+3a)+1 = (a^2+3a+1)^2$$
Hence, n is a perfect square and therefore not a prime.

The product of four consecutive positive integers is always even. Hence, n is always odd.
Therefore, from the given statements, only A and C are true.

Observe the pattern given below.
$$11^2 = 121$$
$$111^2 = 12321$$
$$1111^2 = 1234321$$ and so on.

So, $$11111111^2 = 123456787654321$$

$$(ab)^2$$ = ccb

ccb > 300

The last digit of the number ab must be same as that of the square of ab.

So, b can be 0, 1, 5 or 6.

$$20^2$$=400 and $$30^2$$=900 are three digit numbers and greater than 300. But the first 2 digits are not same. Hence, b is not 0.

If b is 5, then the ten's digit of ab's square will be 2 => c = 2. But if c is 2, then ccb is not greater than 300. Hence, b is not 5.

If b is 6, then $$26^2$$ = 576 is the only three digit number that is greater than 300. But, it is not in the form of ccb => b is not 6.

If b is 1, then $$21^2$$=441 satisfies all the given conditions => b is 1.

$$7^3$$ = 343

$$7^{84}$$ = $$(7^3)^{28}$$ = $$343^{28}$$

$$343^{28}$$ mod 342 = $$1^{28}$$ mod 342 = 1

if $$n^3$$ is odd then $$n$$ will be odd. let's say it is $$2k+1$$
then $$n^2$$ will be = $$(4k^2 + 4k + 1)$$ which will be odd
Hence answer will be C.

A digit number when squared produces a 3 digit number. This means that the number ranges from [10, 31].
First digit of $$BE^2$$ should be unit digit of $$E^2$$. But unit digit of $$E^2$$ is B. Look at the numbers and the unit digit of their square.

0-0, 1-1, 2-4, 3-9, 4-6, 5-5, 6-6, 7-9, 8-4, 9-1. Only 2-4, 3-9, 4-6, 7-9, 8-4 and 9-1 are kind of pairs we are looking after.But all the pairs except 9-1 produce a number greater than 31. Now, the number we can form from 9-1 is 19 whose square is 361 which satisfies all the condition we are looking for. This is the only such number.

Maximum five digit number which can be formed by using numbers is 43210
And minimum five digit number = 10234
Difference = 43210 - 10234 = 32976

Possible numbers with unit's place as 5 = $$4 \times 3 \times 2 \times 1 = 24$$

Possible numbers with unit's place as 4 and ten's place 3,2,1 = $$3 \times 3 \times 2 \times 1 = 18$$

Possible numbers with unit's place as 3 and ten's place 2,1 = $$2 \times 3 \times 2 \times 1 = 12$$

Possible numbers with unit's place as 3 and ten's place 1 = $$1 \times 3 \times 2 \times 1 = 6$$

Total possible values = 24+18+12+6 = 60

Let's say N is our number
N = (899K + 63) or N = ($$29 \times 31$$K) + 63
So when it is divided by 29, remainder will be $$\frac{63}{29}$$ = 5

Let the number 'n' belong to the set A.
Hence, the remainder when n is divided by 2 is 1
The remainder when n is divided by 3 is 2
The remainder when n is divided by 4 is 3
The remainder when n is divided by 5 is 4 and 
The remainder when n is divided by 6 is 5

So, when (n+1) is divisible by 2,3,4,5 and 6.
Hence, (n+1) is of the form 60k for some natural number k.
And n is of the form 60k-1

Between numbers 0 and 100, only 59 is of the form above and hence the correct answer is 1

As we can see here that total number of students are = 60+84+108 = 252

Now given condition is that in one room only the students of the same subject can be there and the number of rooms should be minimum that means the number of students in a particular room will be maximum.

This Maximum number of students will be HCF (Highest common factor) of 60, 84 and 108 and that will be 12

Hence, number of rooms will be = 252/12 = 21

As we know for a number to be divisible by 125, its last three digits should be divisible by 125
So for a five digit number, with digits 2,3,8,7,5 its last three digits should be 875 and 375
Hence only 4 numbers are possible with its three digits as 875 and 375
I.e. 23875, 32875, 28375, 82375
 

The last digit of powers of 2 follow a pattern as given below.

The last digit of $$2^1$$ is 2
The last digit of $$2^2$$ is 4
The last digit of $$2^3$$ is 8
The last digit of $$2^4$$ is 6

The last digit of $$2^5$$ is 2
The last digit of $$2^6$$ is 4
The last digit of $$2^7$$ is 8
The last digit of $$2^8$$ is 6

Hence, the last digit of $$2^{51}$$ is 8

For a number to be divisible by 8, last 3 digits must be divisible by 8.

Last 3 digits of this number are 354.

354 mod 8 = 2

Hence, 2 is the remainder.

Expression can be reduced to 16n + 7 + $$\frac{6}{n}$$
Now to make above value  an integer n can be 1,2,3,6,-1,-2,-3,-6
Hence answer will be D).

let's say number is x.

So $$\frac{7x}{8} - \frac{7x}{18} = 770$$

or $$x = 1584$$

When PQ = 64

Possible values of P and Q are as follows:

64 and 1 ; 32 and 2 ; 16 and 4 ; 8 and 8

So possible sums are 65,34,20 and 16

Hence answer would be D)

Let's say m=5k and n=5t
So m-n = 5(k-t) will be divisible by 5.
$$m^2 - n^2 = 25(k^2 - t^2)$$ will be divisible by 5.
$$m+n = 5(k+t)$$ will be divisible by 5 but not necessarily with 10.

Let's say three numbers are (a-2),a,(a+2)

So 3(a-2) = 2(a+2) - 3

a=7 and a+2 = R = 9

The values of 1!, 2!, 3!, 4!, 5!, 6! and 7! are 1, 2, 6, 24, 120, 720 and 5040 respectively.

So, the digits must lie from 1 to 6 only to satisfy the conditions.

6 cannot be one of the digits as the at least one digit in the final number is more than 6.

145 = 1! + 4! + 5!

Given expressions can be reduced as follows

A = $$\frac{1}{4.000004}$$

B = $$\frac{1}{6.000003}$$

C = $$\frac{1}{6.000002}$$

Among all of them B is smallest.

Let's say n=2k+1 where k=1,2,3....
So $$n(n^2-1) = 4k(k+1)(2k+1)$$
As above expression is already divisible by 4 simultaneously $$k(k+1)(2k+1)$$ will always be divisible by 6.
Hence complete term will be divisible by 24.

According to the divisible rule of 9, the sum of all digits should be divisible by 9.
i.e. 55+A+B = 9k
So sum can be either 63 or 72.
For 63, A+B should be 8.
In given options, option B has values of A and B whose sum is 8 and by putting them we are having a number which divisible by both 9 and 8.
Hence answer will be B.

Total sales value $$= 1148 = 4\cdot7\cdot41$$
Let AB be the actual price of the product and CD be the actual quantity.

Since the quantity sold reduced by 54 upon reversing
$$CD-DC\ =\ (10\cdot C+D)-\left(10\cdot D+C\right)=9\cdot\left(C-D\right)$$
wkt, $$9\cdot\left(C-D\right)=54\ \ \longrightarrow\ \ C-D=6$$

Now, wkt, $$AB\cdot CD=1148\ \ \&\ \ BA\cdot DC=1148$$
The possible values of CD are $$93$$, $$82$$, & $$71$$
Since only 82 divides 1148, $$CD=82$$, Therefore, $$AB=14$$
Actual price and Actual quantity are 14 and 82 respectively.

Alternatively,
Since final value $$= 1148 = 4\cdot7\cdot41$$ remains the same,
1148 can be represented as product of interchangeable numbers i.e. $$41 \times 28$$ and $$82 \times 14$$
As inventory sold reduced by 54, so, the entry of quantity sold should be 28 and price entry will be 41, and hence, actual price has to be 14 and actual sales volume has to be 82.

Total sales value $$= 1148 = 4\cdot7\cdot41$$
Let AB be the actual price of the product and CD be the actual quantity.

Since the quantity/inventory increased by 54 upon reversing
$$CD-DC\ =\ (10\cdot C+D)-\left(10\cdot D+C\right)=9\cdot\left(C-D\right)$$
wkt, $$9\cdot\left(C-D\right)=54\ \ \longrightarrow\ \ C-D=6$$

Now, wkt, $$AB\cdot CD=1148\ \ \&\ \ BA\cdot DC=1148$$
The possible values of CD are $$93, 82 $$ & $$ 71$$
Since only 82 divides 1148, $$CD=82$$, Therefore, $$AB=14$$
Actual price and Actual quantity are 14 and 82 respectively.

Alternatively,
Since final value $$= 1148 = 4\cdot7\cdot41$$ remains the same,
1148 can be represented as product of interchangeable numbers i.e. $$41 \times 28$$ and $$82 \times 14$$
As inventory decreased by 54, so, the quantity sold should be 82 and actual price will be 14.

The highest power of 5 in 80! = [80/5] + [$$80/5^2$$] = 16 + 3 = 19

So, the highest power of 5 which divides 80! exactly = 19

13*62 = 806

31*26 = 806

Hence the answer is option b

As 55 and 124 don't have any common factor, and n has to be a minimum integer, Hence, it should be 124 only. So that given equation won't have a remainder.

let's say k+4 = 7m
k = 7m-4
Now for k+2n or 7m+(2n-4) is also  multiple of 7.
or 2n-4 should be a multiple of 7 
So 2n-4 = 7p
or 2n = 7p+4
For p=2; n=9 (p cannot be 1 as n is an integer )

If 2x+12 is perfectly divisible by x, then 12 must be divisible by x.
Hence, there are six possible values of x : (1,2,3,4,6,12)

Let the Prime number be 6n+1.

So $$(p^2 - 1)$$ = 6n(6n+2) = 12n(3n+1)

For any value of n , n(3n+1) will have a factor of 2

Hence given equation will be always be divisible by 24

For 203 :
first step = $$2\times 3^2 + 0 \times 3^1 + 3 \times 3^0$$ = 21
second step = $$2 \times 3^1 + 1 \times 3^0$$ = 7
So two steps needed to reduce it to 7

8+12 =( 20 = 2+0) = 2
7+14 =( 21 = 2+1) = 3
10+18= (28= 2+8) = 10
 

$$2^{60}$$ or $$4^{30}$$ when divided by 5
So according to remainder theorem 
remainder will be $$(-1)^{30}$$ = 1.

Take 64 for example.

$$\sqrt{64}$$ = 8

$$\sqrt{8}$$ = 2.828

$$\sqrt{2.828}$$ is close to 1.7

So, we can see that the result is tending towards 1.

Putting value of x=0 or 1 and solving all four options,
We will find that only option A satisfies the equation with both values, hence answer will be A.

$$n^3-n$$ can be written as:
$$(n-1)n(n+1)$$ (where n is a positive integer)
i.e. product of three consecutive integers.
Hence for any number n=2 or >2 , product will have a factor of 6 in it.
When two numbers are prime in product, then third number will always be divisible by 6 
Or product will always have a factor of $$3\times2$$ into it.