# CAT Questions on Factors of a Number

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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 $125^{3124}$ and factors of $125^{3127}$?

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

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:

The factors of a number ‘x’ can be of the form $\frac{x}{2}$ , $\frac{x}{3}$ , $\frac{x}{5}$ , ….
If 2 is not a factor of the number then the highest three factors of the number can be $\frac{x}{3}$, $\frac{x}{5}$, $\frac{x}{7}$. 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 $\frac{x}{2}$, $\frac{x}{4}$, $\frac{x}{5}$. 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 $\frac{x}{2}$, $\frac{x}{3}$, $\frac{x}{6}$ 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.

1980 = $2^2 * 3^2 * 11 * 5$
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
= $(\frac {2^{(2+1)} -1} {(2-1)})$ x $(\frac {3^{(2+1)} -1} {(3-1)})$ x $(\frac {11^{(1+1)} -1} {(11-1)})$ x $(\frac {5^{(1+1)} -1} {(5-1)})$ – $(\frac {3^{(2+1)} -1} {(3-1)})$ x $(\frac {11^{(1+1)} -1} {(11-1)})$ x $(\frac {5^{(1+1)} -1} {(5-1)})$
= 7*13*12*6 – 13*12*6 = 5616

$125^{3124}$ = $5^{9372}$
$125^{3127}$ = $5^{9381}$
Thus, the factors of $5^{9381}$ which are multiples of $5^{9372}$ are $5^{9372}$,$5^{9373}$ … $5^{9381}$. Thus, there are 10 such integers

$36000 = 2^{5}*3^{2}*5^{3}$
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

$36288 = 2^6 * 3^4 * 7$
So, if the factor is of form $2^a * 3^b * 7^c$, a can take values 0, 3, 6