CAT Previous Year Questions: Number Series- 2 (June 6) - CAT MCQ

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

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

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

which will be maximum when n - 4 = 8
n = 12

n² - m² = 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.

2⁴ × 3⁵ × 10⁴ 
=2⁴ × 3⁵ × 2⁴ × 5⁴ 
= 2⁸ × 3⁵ × 5⁴ 
For the factor to be a perfect square, the factor should be even power of the number.
In 2⁸, the factors which are perfect squares are 2⁰, 2², 2⁴, 2⁶, 2⁸ = 5
Similarly, in 3⁵, the factors which are perfect squares are 3⁰, 3², 3⁴ = 3
In 5⁴, the factors which are perfect squares are 5⁰, 5², 5⁴ = 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.

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


Adding ab = 616 on both sides, we get
a² + b² + ab + ab = 3 × 4 × 157 + 616
=> (a + b)² = 3 × 4 × 157 + 616 = 2500
=> a + b = 50

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 √20.
Sum of the squares of the 2 numbers 20 + 20 = 40.
Therefore, 40 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.