### Number Theory Questions - Factors

Questions:

- How many numbers are there less than 100 that cannot be written as a multiple of a perfect square greater than 1?
- Find the smallest number that has exactly 18 factors?

Correct Answer:

Qn 1: 61 numbers

Qn 2: 180

Explanatory answers

Qn1 :How many numbers are
there less than 100 that cannot be written as a multiple of a perfect square?

To begin with, all prime numbers will be part of this list. There are 25 primes less than 100. (That is a nugget that can come in handy)

To begin with, all prime numbers will be part of this list. There are 25 primes less than 100. (That is a nugget that can come in handy)

Apart from this, any
number that can be written as a product of two or more primes will be there on
this list. That is, any number of the form pq, or pqr, or pqrs will be there on
this list (where p, q, r, s are primes). A number of the form p

This is a brute-force question.

^{n}q cannot be a part of this list if n is greater than 1, as then the number will be a multiple of p^{2}.This is a brute-force question.

First let us think of
all multiples of 2 * prime number. This includes 2*3, 2*5, 2*7, 2 * 11 all the
way up to 2*47 (14 numbers)

The, we move on to all
numbers of the type 3 * prime number 3*5, 3*7 all the way up to 3*31 (9
numbers)

Then, all numbers of
the type 5 * prime number – 5*7, 5*11, 5*13, 5*17, 5*19 (5 numbers),

Then, all numbers of
the type 7 * prime number and then 7*11,
7*13 (2 numbers).

There are no numbers
of the form 11 * prime number which have not been counted earlier.

Post this, we need to count all numbers of the form p*q*r, where p, q, r are all prime.

In this list, we have
2*3*5, 2*3*7, 2*3*11, 2*3*13 and 2*5*7. Adding 1 to this list, we get totally 36
different composite numbers.

Along with the 25
prime numbers, we get 61 numbers that cannot be written as a product of a
perfect square greater than 1

Alternative Method

There is another
method of solving this question.

We can list all multiples of perfect squares (without repeating any number) and subtract this from 99

We can list all multiples of perfect squares (without repeating any number) and subtract this from 99

4 - there are 24
multiples of 4 { 4, 8, 12, …96}

9 - There are 11
multiples, 2 are common with 4 (36 and 72), so let us add 9 new numbers to the
list{ 9, 18, 27, ….99}

16 - 0 new multiples

25 - 3 new multiples { 25, 50, 75

36 – 0 new ones

49 – 2 { 49, 98}

64 - 0

81 - 0

So, total multiples of perfect squares are 38. There are 99 numbers totally. So, there are 61 numbers that are not multiples of perfect squares

This is a difficult and time-consuming question. But a question that once solved, helps practice brute-force counting. Another takeaway is the fact that there are 25 primes less than 100. There is a function called pi(x) that gives the number of primes less than or equal to x. pi(10) = 4, pi(100) = 25

4. Find the smallest number that has exactly 18 factors?

Any number of the form p

16 - 0 new multiples

25 - 3 new multiples { 25, 50, 75

36 – 0 new ones

49 – 2 { 49, 98}

64 - 0

81 - 0

So, total multiples of perfect squares are 38. There are 99 numbers totally. So, there are 61 numbers that are not multiples of perfect squares

This is a difficult and time-consuming question. But a question that once solved, helps practice brute-force counting. Another takeaway is the fact that there are 25 primes less than 100. There is a function called pi(x) that gives the number of primes less than or equal to x. pi(10) = 4, pi(100) = 25

4. Find the smallest number that has exactly 18 factors?

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)
Now, the number we are
looking for has 18 factors. It can comprise one prime, two primes or three
primes.

Now, 18 can be written
as 1 * 18 or 3 * 6 or 9 * 2 or 2 * 3 * 3

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

^{a}q^{b}, then it can be of the form
p

^{1}q^{8}or p^{2}q^{5}
If we take the
underlying prime factorization of N to be p

^{a}, then it can be of the form
p

^{17}
If we take the
underlying prime factorization of N to be p

^{a}q^{b}r^{c}, then it can be of the form
p

^{1}q^{1}r^{2}
So, N can be of the
form p

^{17}, p^{2}q^{5}, p^{1}q^{8}or p^{1}q^{2}r^{2}
Importantly, these are
the

**prime factorizations that can result in a number having 18 factors.**__only possible__
Now, let us think of
the smallest possible number in each scenario

p

^{17}- Smallest number = 2^{17}
p

^{2}q^{5 }– 3^{2}* 2^{5}
p

^{1}q^{8}– 3^{1}* 2^{8}
p

^{1}q^{1}r^{2}– 5^{1}* 3^{1}* 2^{2}
The smallest of these numbers is 5

^{1}* 3^{1}* 2^{2}= 180Labels: CAT number theory, CAT Solutions

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