CAT Previous Year Questions: Number Series- 1 (June 5) - CAT MCQ

The prime factorizations of 168 and 1134 are as follows:
168 = 2³ × 3 × 7
1134 = 2 × 3⁴ × 7
Clearly, the smallest positive integral value of n, such that 168 is a factor of 1134ⁿ is 3.
1134ⁿ = 1134³ = 2³ × 3¹² × 7³
Clearly, the smallest positive integral value of m, such that 1134³ is a factor of 168^m is 12.
Therefore, m + n = 12 + 3 = 15

Single digit numbers with non-repeating digits = 9
(The unit’s digit is non-zero)
Two digit numbers with non-repeating digits = 9 × 9
(The tenth’s digit is non-zero and the unit digit can be any digit except the tenth’s digit.)
Three digit numbers with non-repeating digits = 9 × 9 × 8
(The hundred’s digit is non-zero and the tenth’s digit can be any digit except the hundred’s digit and the unit digit can be any digit except the tenth’s digit.)
So, totally there are (9 + 9 × 9 + 9 × 9 × 8) = 738 natural numbers up to 1000 with non-repeating digits.

 k divides m + 2n
So, k also divides 2(m + 2n) = 2m + 4n
It is given that k divides 3m + 4n
Which means, k should also divide (3m + 4n) – (2m + 4n)
∴k divides m
Since k divides m and m + 2n
k should also divide (m + 2n) – m = 2n
Therefore, k divides m and 2n.

A positive integer less than 50, having exactly two distinct factors other than 1 and itself, is either a perfect cube below 50 or an integer that is a product of exactly two distinct primes.
Case i)
Perfect cubes below 50 are 2³ and 3³. So, two numbers here
Case ii)
For the product of two primes to be below 50, the individual primes should be below 25.
(Because, the smallest prime is 2 and multiplying 2 with anything greater than or equal to 25 yields a number greater than or equal to 50.)
2, 3, 5, 7, 11, 13, 17, 19, 23 are prime numbers less than 25. 
2, 3, 5, 7 are the primes less than √50
, any product of two numbers among them yields a product less than 50.
So, there are 4C₂ = 6
 pairs here.
11, 13, 17, 19, 23 are the primes greater than √50
, any product of two numbers among them yields a product greater than 50.
So, there are 0 pairs here.
Between the two lists 11 and 13 can pair with 2 and 3, while 17, 19, and 23 can only pair with 2.
So, there are 7 pairs here.
So, totally, there are 2 + 6 + 0 + 7 = 15 such numbers.

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.

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 = 

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.

Let the number be ‘abc’. Then, 2 < a × b × 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 x 2 and the combination 2,2,1 can be used to form 3 more distinct numbers.
We can factorize 6 as 2 x 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.

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.

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