CAT Questions on Factors of a Number

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CAT Questions on Factors of a Number
CAT Questions on Factors of a Number

CAT Questions on Factors of a Number:

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Question 1: A “tragic number” is a number which can be expressed as the sum of three of its factors. For example, 6 can be expressed as the sum of 1, 2 and 3. How many tragic numbers are there that are less than 50?

a) 6
b) 7
c) 8
d) 9

Question 2: If N = 1980, Find the number and sum of its even factors.

a) 28, 6552
b) 24, 5616
c) 24, 4630
d) 28, 5672

Question 3: How many integers are both multiples of 1253124 and factors of 1253127?

a) 3
b) 10
c) 4
d) 11

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Question 4: What is the number of even factors of 36000 which are divisible by 9 but not by 36?

a) 20
b) 4
c) 10
d) 12

Question 5: How many factors of 36288 are perfect cubes?

a) 9
b) 4
c) 6
d) 8

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Solutions for CAT Questions on Factors of a Number:

Solutions:

1) Answer (C)

The factors of a number ‘x’ can be of the form x2 , x3 , x5 , ….
If 2 is not a factor of the number then the highest three factors of the number can be x3, x5, x7. The sum of these three is less than x. So 2 has to be a factor of the number.
If 3 is not a factor of the number the the highest three factors of the number can be x2, x4, x5. The sum of these three is less than x. So 3 has to be a factor of the number.
If 2 and 3 are factors, 6 is also a factor.
Also the sum of x2, x3, x6 is exactly equal to x.
So all numbers which are multiples of 6 are tragic numbers.
There are 8 such numbers which are below 50.

2) Answer (B)

1980 = 2232115
Number of even factors = Total number of factors – Number of odd factors.
= (2+1)(2+1)(1+1)(1+1) – (2+1)(1+1)(1+1) = 24
Sum of even factors = Sum of all the factors – sum of odd factors
= (2(2+1)1(21)) x (3(2+1)1(31)) x (11(1+1)1(111)) x (5(1+1)1(51))(3(2+1)1(31)) x (11(1+1)1(111)) x (5(1+1)1(51))
= 7*13*12*6 – 13*12*6 = 5616

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3) Answer (B)

1253124 = 59372
1253127 = 59381
Thus, the factors of 59381 which are multiples of 59372 are 59372,5937359381. Thus, there are 10 such integers

4) Answer (B)

36000=253253
Since we are talking of even factors, there must be at least one 2 in the required factors.
Since the number is divisible by 9, we must have both the threes.
We cannot have more than 1 two as it will make the number divisible by 36.
So we have 1 way of choosing 2, 1 way of choosing 3, 4 ways of choosing 5.
Thus the required number of factors are
1*1*4 = 4

5) Answer (C)

36288=26347
For any perfect cube, all the powers of its prime numbers have to be multiples of 3.
So, if the factor is of form 2a3b7c, a can take values 0, 3, 6
And b can take values 0, 3
And c can take value 0.
==> There are 3*2*1 = 6 possibilities.

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