CAT Previous Year Questions: Number System (June 3) - 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.

It is given,

For n = 1 to n = 19, value of function is zero.
For n = 20 to n = 44, value of function will be 1.
44 = 20 + n - 1
n = 25 which is equal to given value.
This implies N = 44

Let the number of 100 cheques, 250 cheques and 500 cheques be x, y and z respectively.
We need to find the maximum value of z.
x + y + z = 100 ...... (1)
100x + 250y + 500z = 15250
2x + 5y + 10z = 305 ...... (2)
2x + 2y + 2z = 200 ....... (1)
(2) - (1), we get
3y + 8z = 105
At z = 12, x = 3
Therefore, maximum value z can take is 12.

Let the number of 100 cheques, 250 cheques and 500 cheques be x, y and z respectively.
We need to find the maximum value of z.
x + y + z = 100 ...... (1)
100x + 250y + 500z = 15250
2x + 5y + 10z = 305 ...... (2)
2x + 2y + 2z = 200 ....... (1)
(2) - (1), we get
3y + 8z = 105
At z = 12, x = 3
Therefore, maximum value z can take is 12.

Let the numbers be of the form 100a + 10b + c, where a, b, and c represent single digits.
Then (100c + 10b + a) - (100a+ 10b + c) = 198
99c - 99a = 198
c - a = 2.
Now, a can take the values 1-7. a cannot be zero as the initial number has 3 digits and cannot be 8 or 9 as then c would not be a single-digit number.
Thus, there can be 7 cases.
B can take the value of any digit from 0-9, as it does not affect the answer. Hence, the total cases will be 7× 10 = 70.

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 x 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 x d = 11. b = 1.
The highest four-digit number satisfying the condition is 4195

Given a, b, c are rational numbers.
Hence a, b, c are three numbers that can be written in the form of p/Q
Hence if one both the root is 2 + √3 and considering the other root to be x.
The sum of the roots and the product of the two roots must be rational numbers.
For this to happen the other root must be the conjugate of 2 + √3  so the product and the sum of the roots are rational numbers which are represented by: b/a, c/a
Hence the sum of the roots is 2 + √3 + 2 - √3 = 4
The product of the roots is 2 + √3 x  2 - √3 =1
b/a = 4, c/a = 1.
b = 4 x a, c= a.
Since b = c3
4 x a = a3 
a² = 4.
a = 2 or -2.
|a| = 2