### Number Theory - Factors Questions

Number Theory is one of the heavily
tested topics in CAT and is probably best to practice a lot of questions in
this topic.

This is a very interesting topic to
prepare for and a fun topic if you go about it the right way. I have given two
questions on Number Theory – on the topic of factors below.

Questions

- A number N^2 has 15 factors. How many factors can N have?
- If a three digit number ‘abc’ has 2 factors (where a, b, c are digits), how many factors does the 6-digit number ‘abcabc’ have?

Correct Answer

Question 1: 6 or 8 factors

Question 2: 16 factors

Explanatory
Answer

Qn:
A number N^2 has 15 factors. How many factors can N have?

Any
number of the form p

^{a}q^{b}r^{c}will have (a+1) (b+1)(c+1) factors, where p, q, r are prime. (This is a very important idea)
N

^{2}has 15 factors.
Now,
15 can be written as 1 * 15 or 3 * 5.

If
we take the underlying prime factorization of N2 to be p

^{a}q^{b}, then it should have (a + 1) (b+1) factors. So, N can be of the form
p

^{14}or p^{2}q^{4}
p

^{14}will have (14+1) = 15 factors
p

^{2}q^{4}will have (2+1) * (4 +1) = 15 factors.
Importantly,
these are the

**possible prime factorizations that can result in a number having 15 factors.**__only two__
Qn:
If a three digit number ‘abc’ has 2 factors (where a, b, c are digits), how
many factors does the 6-digit number ‘abcabc’ have?

To
start with ‘abcabc’ = ‘abc’ *1001 or abc * 7*11*13 (This is a critical idea to
remember)

‘abc’
has only two factors. Or, ‘abc’ has to be prime. Only a prime number can have
exactly two factors. (This is in fact the definition of a prime number)

So,
‘abcabc’ is a number like 101101 or 103103.

’abcabc’
can be broken as ‘abc’ * 7 * 11 * 13. Or, a p * 7 * 11 * 13 where p is a prime.

As
we have already seen, any number of the form p

^{a}q^{b}r^{c}will have (a+1) (b+1)(c+1) factors, where p, q, r are prime.
So,
p * 7 * 11 * 13 will have = (1+1)*(1+1)*(1+1)*(1+1) = 16 factors

Labels: CAT, CAT number theory, CAT questions, Factors

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