We are aske dto find the number of numbers that are divisible by none of 2, 5 and 7. Instead, let us find the number of numbers that are divisible by atleast one of 2, 5 and 7.

Recall that n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(C ∩ A) + n(A ∩ B ∩ C).
Therefore, Number of Multiples of (2 or 5 or 7) = Number of Multiples of (2) + Number of Multiples of (5) + Number of Multiples of (7) – Number of Multiples of (2 and 5) – Number of Multiples of (5 and 7) – Number of Multiples of (7 and 2) + Number of Multiples of (2 and 5 and 7)
Number of Multiples of (2 or 5 or 7) = Number of Multiples of (2) + Number of Multiples of (5) + Number of Multiples of (7) – Number of Multiples of (10) – Number of Multiples of (35) – Number of Multiples of (14) + Number of Multiples of (70)
Number of Multiples of (2 or 5 or 7) = 60 + 24 + 17 – 12 – 3 – 8 + 1 = 79

Since we know the number of integers that can be divided by atleast one of 2, 5 and 7 is 79. We can arrive at the number of integers that can be divided by none of 2, 5 and 7, by subtracting from the whole of 120.

No of integers of 1, 2, … , 120, that are divisible by none of 2, 5 and 7 = 120 – 79 = 41

2⁴ x 3⁵ x 10⁴
2⁴ x 3⁵ x 2⁴ x 5⁴
2⁸ x 3⁵ x 5⁴
Now, for any number to be a perfect square, it must have an even number of factors.
So, if we consider 2⁸, only powers of 0, 2, 4, 6, 8 can lead to perfect squares – 5 ways
If we consider 3⁵, only powers of 0, 2, 4 can lead to perfect squares – 3 ways
If we consider 5⁴, only powers of 0. 2, 4 can lead to perfect squares – 3 ways
So, total number of possibilities = 5 x 3 x 3 ways = 45 ways
Since we need to find the number of factors greater than 1,
Required number of ways = 45 -1 = 44 ways

Given that N and X are positive integers such that N^N = 2¹⁶⁰ and N² + 2^N is an integral multiple of 2^x
Now we have to find the largest possible value for x
N^N = 2¹⁶⁰
We can rewrite this 2¹⁶⁰ as (2^p)^q
such that 2^p = N = q
pq = 160

We can try by substituting p and q with 2 and 80 respectively
(2²)⁸⁰ = 4⁸⁰
This doesn’t work so we can try by substituting p and q with 4 and 40 respectively
(2⁴)⁴⁰ = 16⁴⁰
This also doesn’t work so now we can substitute p and q with 5 and 32 respectively
Such that (2⁵)³² = 32³²

Now we can say that N = 32
N² + 2^N ⟹ 32² + 2³² = 2^x × k (where k is the integral multiple)
N² + 2^N ⟹ 2¹⁰ + 2³² = 2^x × k since 32² = (2⁵)²
N² + 2^N ⟹ 2¹⁰ (1 + 2²²) = 2^x × k
This (1 + 2²²) is an odd number so the number is going to be 2¹⁰ × multiple of the odd number
Hence the largest power of 2 can be 10 i.e. 2¹⁰
X_max = 10
The largest possible x is 10

Given 0.25 ≤ 2x
0.25 = 1414 =(1212)² = 2^-2
Similarly, 2⁶ = 64, 2⁷ = 128, 2⁸ > 200
Substituting the values and checking for divisibility
Number of Integers x = 5

We will select the 4 digits first and arrange them later.
Out of the 4 digits, one of them should be 2 and one of them should be 3.
2, 3, , .
So, we just need to select the other two digits…
The two digits could be (1,1), (2, 2), (3, 3), (1, 2), (1, 3), or (2, 3).
So, the selection of numbers could be…
2, 3, 1, 1
2, 3, 2, 2
2, 3, 3, 3
2, 3, 1, 2
2, 3, 1, 3
2, 3, 2, 3
Each of these selections could be re-arranged in a number of ways.
2, 3, 1, 1 (4!2!4!2! = 12 ways)
2, 3, 2, 2 (4!3!4!3! = 4 ways)
2, 3, 3, 3 (4!3!4!3! = 4 ways)
2, 3, 1, 2 (4!2!4!2! = 12 ways)
2, 3, 1, 3 (4!2!4!2! = 12 ways)
2, 3, 2, 3 (4!2!∗2!4!2!∗2! = 6 ways)
So total number of possibilities = (12 + 4 + 4 + 12 + 12 + 6) = 50 ways.
Alternate method:
(Arrangements with at least one 2 and one 3) = (All possible arrangements) – (Arrangements with either 1 or 2) – (Arrangements with either 1 or 3) + (Arrangements with only 1)
Think why we need to add (Arrangements with only 1)!
_, _, _, _
(All possible arrangements) = 3⁴
Each blank could be any one of 1, 2 or 3.
(Arrangements with either 1 or 2) = 2⁴
Each blank could be any one of 1 or 2.
(Arrangements with either 1 or 3) = 2⁴
Each blank could be any one of 1 or 3.
(Arrangements with only 1) = 1
Each blank is filled with 1.
(Arrangements with at least one 2 and one 3) = (All possible arrangements) – (Arrangements with either 1 or 2) – (Arrangements with either 1 or 3) + (Arrangements with only 1)
(Arrangements with at least one 2 and one 3) = 3⁴ – 2⁴ – 2⁴ + 1
(Arrangements with at least one 2 and one 3) = 81 – 16 – 16 + 1
(Arrangements with at least one 2 and one 3) = 82 – 32 = 50

For a 4-digit number we can make 4 placeholders to represent the four digits
__ __ __ __
Now, we know that the sum of the thousands, hundreds and the tens place digit is 14, and the sum of the hundreds, tens and the units place digit is 15.

So we can conclude that the unit’s digit is one unit greater than the thousandth place.

To maximize the 4-digit number the value of a should be as large as possible. And the value of the tens digit can be 9 at most. So the maximum value of a is 4.

To satisfy the sum of the thousands, hundreds and the tens place digit, the value of the hundreds place should be equal to 1.
Hence the largest possible number is 4195

The natural numbers are divided into groups as (1), (2, 3, 4), (5, 6, 7, 8, 9), ….. and so on.
If you observe that the first group ends with 1², the second group ends with 2²and the third group ends with 3²
Hence, the 15th group ends with 15² = 225.
The 14th group ends with 14² = 196.
Therefore, the 15th group contains (197, 198, . . . . . . . . . . ,225)
The sum of the numbers in the 15th group =292292(197 + 225) = 6119.

Let’s consider a three-digit number to be ‘abc’.
Given that, three-digit numbers increase by 198 when the three digits are arranged in the reverse order.
100c + 10b + a – 100a – 10b – c = 198
99c – 99a = 198
c – a = 2
So, the difference between the hundreds place digit and the units place digit is 2.
The possible combinations are:
1 _ 3, 2 _ 4, 3 _ 5, 4 _ 6, 5 _ 7, 6 _ 8, 7 _ 9.
We have 10 numbers for each combination.
Hence, the total numbers are 70.

2 < x < 10
x can take any of the values from the set {3, 4, 5, 6, 7, 8, 9}
14 < y < 23
y can take any of the values from the set {15, 16, 17, 18, 19, 20, 21, 22}

The highest value N (i.e x+y) can take = 9+22 = 31. (at x = 9; y = 22)
30 can be obtained at x = 9; y = 21
29 can be obtained at x = 9; y = 20
28 can be obtained at x = 9; y = 19
27 can be obtained at x = 9; y = 18
26 can be obtained at x = 9; y = 17
25 can be obtained at x = 9; y = 16
But, x+y=25 is not the desired sum, hence the different values of x+y are {31,30,29,28,27,26}.
Hence, x+y, and thereby N can take 6 distinct values.

 _ _ _ on the 4 digit 7 has to come before 3
So, ⁴C₂ possibilities for 7 coming before 3 into the spaces
7 ___ 3 ____, 2 numbers are already there and 2 remaining spots we have to fill in
From 8 digits available (Excluding 7 and 3)
So, ⁸C₂ ways of choosing
After choosing two from 8, that two can be placed in any way
For example : 1 and 2 can be arranged as 12 and 21
So, ⁴C₂ × ⁸C₂ × 2
Now we need to subtract the possibility where 0 comes in the first position
0753, So these are in the form 0__ __ __
In this 3 places 7 before 3 can be placed in ³C₂ ways and remaining 1 digit
Can be chosen from 7 (Excluding 0,3,7) digits.
³C₂ × 7 = 21 numbers
Should subtract 21 from ⁴C₂ × ⁸C₂ × 2
= (6 × 28 × 2) – 21
= 315

Let the two numbers be a and b
From the question, we know
a x b = 616
and = 
= 
= 
= 
= – 1 = 
= 
= 
= 3 x 3 x 4 = 36
a – b = 6, ab = 616
616 can be rewritten as 22 x 28, where 28 – 22= 6 using trial and error
So, a + b = 28 + 22 = 50

Given that the sum of squares of two numbers is 97 i.e. a² + b² = 97
From the given options we have to find which one cannot be their product i.e. ab
A. 64 ⟹ 2ab = 128
B. −32 ⟹ 2ab = -64
C. 16 ⟹ 2ab = 32
D. 48 ⟹ 2ab = 96
2ab is found because we know that
a² + b² + 2ab ≥ 0
a² + b² – 2ab ≥ 0
By this we can know that 97 + 128 works but 97 – 128 doesn’t works so we can understand option A cannot be the product and the rest can be.
a² + b² ≥ |2ab|
a² + b² ≥ 2ab
a² + b² ≥ -2ab
⟹ a2+b2+2 ≥ |ab|
So here 2ab should lie between +97 and -97 or ab should be less than 97/2 or greater than −97/2, so except option A all the other options works so option A 64 cannot be the product

Let the two-digit number be xy, which can be expressed as 10x + y
Given that the two-digit number is more than thrice the number obtained by interchanging the digits
So, 10x + y > 3 × (10y + x)
=> 10x + y > 30y + 3x
=> 7x > 29y
=> x > 297297 × y
Approximately, x > 4y
Let us fix values for y and check for conditions,
For y = 1, x can take the values of 5 , 6 , 7 , 8 , 9 (As, 51 > (3 × 15), 61 > (3 ×16), 71> (3 × 17), 81 > (3 × 18), 91 > (3 × 19))
Similarly, for y = 2, the condition satisfies only for x = 9 (As, 92 > (3 × 29))
The remaining values of y does not satisfy the given conditions.
So, Total = 5 + 1 = 6 numbers

Let the three real numbers be a, b, c. c is taken as 73 instead of 37
So, 73ab = 37ab + 720
36 ab = 720
ab = 720/36 = 20
We know, AM ≥ GM
(a2+b2)/2 ≥ ab
(a² + b²) ≥ 2ab
(a² + b²) ≥ 40
(Or)
ab = 20
(a,b) = (√20, √20)
(a² + b²) = 20 + 20 = 40

Given that the product of three consecutive positive integers is 15600 and we have to find the sum of the squares of the integers
As we have two zeroes in 15600 we should have some number ending in zero or in five
156 = 12 × 13
15600 = 12 × 13 × 100 = 12 × 13 × 25 × 4
15600 = 24 × 25 × 26
The product of n (n+1)(n+2) should have a multiple of 5²
The sum of the squares of these integers 24 , 25 , 26 is
24² + 26² + 25² = 576 + 625 + 676
24² + 26² + 25² = 1877