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
  1. A number N^2 has 15 factors. How many factors can N have?
  2. 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 paqbrc will have (a+1) (b+1)(c+1) factors, where p, q, r are prime. (This is a very important idea)
N2has 15 factors.
Now, 15 can be written as 1 * 15 or 3 * 5.
If we take the underlying prime factorization of N2 to be paqb, then it should have (a + 1) (b+1) factors. So, N can be of the form
p14or p2q4
p14will have (14+1) = 15 factors
p2q4will have (2+1) * (4 +1) = 15 factors.
Importantly, these are the only two possible prime factorizations that can result in a number having 15 factors.
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 paqbrcwill 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

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