__Product of Factors__

__Perfect square as a product of two factors__

In case of perfect square number we have odd number of factors i.e. the number of factors are odd hence in that case required number of ways in which we can write perfect square number as a product of its two factors are (n – 1)/2 if we do not include the square root of the number and (n + 1 )/2 if we include the square root of the number. So number of ways to express a perfect square as product of two different factors (that means excluding its square root) is

½ {(p + 1)(q + I)(r + I) … – 1)}. And if we include the square root then required numbs 1/2 {(p+1)(q+1)(r+1) … +1}

**Let’s take one example to understand this. **

**Example 2**: **In how many ways you can express 36 as the product of two of its factors?**

**Solution : ****Step 1**: Prime factorization of 36 i.e. we write 36 = 2^{2} x 3^{2}

**Step 2**: Number of factors of 36 will be (2+1)(2+1)=9 (i.e. factors are 1,2,3,4,6,9,12,18,36)

**Step 3**: Since we are asked total number of ways hence we include square root of 36 i.e. 6 as well. Thus number of ways you can express 36 as the product of two of its factors is (9+1)/2=5

__Product of two co-prime numbers __

To express the number as a product of co-prime factors we will use the following steps:

**Step 1:**Write Prime factorisation of given number i.e. convert the number in the form where p_{1}, p_{2},p_{3}…..p_{n }are prime numbers and a,b,c….. are natural numbers as their respective powers.

**Step 2:** In the above step we have n prime numbers then the number of ways to express the number as the product of two co prime numbers =2^{n-1}

Because two primes are always co-prime and after we pick 1 prime the other prime can be picked in 2^{n-1}ways.Hence number of ways in which we can write given number as a product of two co prime factors =2^{n-1 }

**Example 1: In how many ways you can write 315 as product of two of its co-prime factors. Solution: **

Step 1: Prime factorization of 315 i.e. we write 315 = 3^{2} x 5^{1} x 7^{1}

Step2: In above step 3 prime numbers are used. Hence number of ways are 2^{3-1 }= 4.Infact we can mention these cases as well 9×35, 5 x 63, 7 x 45, 15 x 21.

__Remember:__ The number of numbers, which are less than N= **p ^{a} q^{b} r^{c}(where p, q, r are prime numbers and the a,b,c are natural numbers as **

**their respective powers)and are co-prime to N is given by \[N\left\{ \left( 1-\frac{1}{p} \right)\left( 1-\frac{1}{q} \right)\left( 1-\frac{1}{r} \right) \right\}\]**

__Product of all the factors __

**To find product of all the factors we follow the following steps:**

Step1: Prime factorization N= a^{p}b^{q}c^{r}

Step2: Number of factors (say X)

Step3: Product of all the factors is given by = N^{X/2}

**Example 2: Find the product of all the factors of 120 ?**

**Solution: **

**Step 1:** Prime factorization : = 2^{3}X5^{1}X3^{1}

**Step2: **Number of factors are ( 3+1)(1+1)(1+1)= 16

**Step 3:** Product of all the factors =( 2^{3}X5^{1}X3^{1})^{16/2} =(120)^{8}

__Number of Co-Prime Numbers that are less than Given Number__

When number n is written in the form of n = a^{p}b^{q}c^{r} ….. where a, b, c …. are the numbers. The number of co-primes to number n and also less than *n* is given by . This is also denoted by $n\,\left( 1-\frac{1}{a} \right)\,\left( 1-\frac{1}{b} \right)\,\left( 1-\frac{1}{c} \right)$

*N(o) Euler’s totient.*

And sum of all these co-primes are given by N × N(Ø)=$\frac{{{n}^{2}}}{2}\left( 1-\frac{1}{a} \right)\,\left( 1-\frac{1}{b} \right)\,\left( 1-\frac{1}{c} \right)$ ……..

**Example 1:** Find the number of co-primes to 90 that are less than 90.

**Solution: **

Since 90 = 21 × 32 × 51 so number of numbers that are less than 90 and co-prime to 90

is 90(Ø) $=90\,\left( 1-\frac{1}{2} \right)\,\left( 1-\frac{1}{3} \right)\,\left( 1-\frac{1}{5} \right)=24$

**Example 2:** Find the sum of all the co-prime numbers to 90 and less than 90.

**Solution:**

Sum of all the co-prime numbers less than 90 is 90 × 90(Ø) = 90 × 24 = 2160

__Product of all the Factors__

Write the number in prime factorization format *N = a ^{p}b^{q}c^{r}* …..

We know that number of factors of

*N*is given by

*n*(

*F*)= (

*p*+ 1)(

*q*+ 1)(

*r*+ 1).

Then product of all the factors of given by

Let us understand this by taking one example.

Example 3: Find the product of all the factors of 60.

Solution:

1^{st} we will list down all the factors of 60 these are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60.

Now we will list down all these factors in the form of prime factorization form

1 = 2^{0}3^{0}5^{0}

2 = 2^{1}3^{0}5^{0}

3 = 2^{0}3^{1}5^{0}

and so on then product of all these factors is

(2^{0}3^{0}5^{0})(2^{1})(3^{1})(2^{2})(5^{1})(2^{1}3^{1})(2^{1}5^{1})(2^{2}3^{1})(3^{1}5^{1})(2^{2}5^{1})(2^{1}3^{1}5^{1})(2^{2}3^{1}5^{1})

= (2^{1+2+1+1+2+2+1+2})(3^{1+1+1+1+1+1})(5^{1+1+1+1+1+1})

= (2^{2}3^{1}5^{1})^{6} = 60^{6}

**Example 4:** find the product of all the factors of 100.

**Solution:**

Product of all the factors is calculated as we have done in previous example –

(1)(2^{1})(2^{2})(5^{1})(2^{1}5^{1})(2^{2}5^{1})(5^{2})(2^{1}5^{2})(2^{2}5^{2}) = (2^{1+2+1+2+1+2})(5^{1+1+1+2+2+2}) = (2^{2}5^{2})^{9/2} = 2^{9}5^{9}

__Pair of Factors that are Co-Prime to each other__

Write the given number in prime factorization format *n = a ^{p}b^{q}* (If number has two prime factors) then number of pairs of factors which are co-prime to each other

= (

*p*+1)(

*q*+ 1) +

*pq*=2

*pq*+(

*p*+

*q*) +1

Let us understand the logic behind this with an example:

**Example 5:** How many pairs of factors of number 108 such that they are co-prime to each other?

**Solution:**

We know that 108 = 2^{2}3^{3} let two factors are *a* and *b*, it is given that they are co-prime to each other.

**Case (i)** when *A* = 1, then *B* can take any factor of 108 so total (2+1)(3+1) such pairs exist.

**Case (ii)** when *A* = 2^{1} then we have *B* = 3^{1} or 3^{2} or 3^{3 }= 3 such numbers

**Case (iii)** when *A* = 2^{2} then we have *B* = 3^{1} or 3^{2} or 3^{3} = 3 such numbers

Hence total such numbers is (2+1)(3+1) +2 ×3 = 12 + 6 = 18

**Now let us generalize if number is in the form of N = a^{p}b^{q}**

Let the above number has two factors as (*A*, *B*) and they are Co-prime to each other now to find such pairs let us see different cases: –

**Case (i):** When *A* = 1 then *B* can be any factor of number *N* hence total such values are (*p* + 1)(*q* + 1)

**Case (ii):** When *A* = *a ^{x}* then

*B*can take any value from 31, or 32 or ….. 3

*q*for one value of ‘

*x*’ there exist ‘

*q*’ such values of

*B*and number of possible values of ‘

*x*’ is ‘

*p*’ hence total such pairs are ‘

*pq*’

**Hence total number of such pairs are (**

*p*+ 1)(*q*+ 1)+*pq*= 1+ (*p*+*q*) + 2*pq***Example 6:** How many pairs of factors of number 196 such that they are co-prime to each other?

**Solution:**

We know that 196 = 2^{2} × 7^{2} in this case we have only 2 prime factors hence number of such pairs is given by 1 +4 + 2 × 4 = 13 pairs

If three prime factors then Number *n = a ^{p}b^{q}c^{r}*

Let the above number has two factors as (

*A*,

*B*) and they are co-prime to each other now to find such pairs let us see different cases:

**Case (i):** When *A* = 1 then *B* can assume any value from all the factors of number n and total such values of *B* is (*p *+ 1)(*q*+ 1)(*r* +1) = *pqr* + *pq* + *pr* + *qr* + 1

**Case (ii):** When *A* is in the form *a ^{x}* (Total

*p*such value of

*x*exist) then

*B*can assume any factor of number

*b*except 0, and total number of factors of

^{q}c^{r}*b*except 1 is

^{q}c^{r}*qr*hence total such pairs is ‘

*pqr*’.

Similarly when

*A*is in the form of

*b*and

^{y}*c*then also we will get ‘

^{z}*pqr*’ such pairs. Hence total such pairs in this format is 3

*pqr*.

Case (iii): When a is in the form of *a ^{x}* (Total

*q*such values of

*x*exist) and

*B*is in the form of

*b*(

^{y}*y*can assume

*q*values) total such pairs are

*pq*, similarly for the combination of

*a*&

^{x}*c*and

^{z}*b*&

^{y}*c*we will get

^{z}*pr*and

*qr*such pairs respectively. So total number of such pairs is

*pq + pr + qr*

**Hence total number of pairs is 1+ (**

*p+q + r*)+ 2(*pq +pr + qr*) + 4*pqr***Example 7:** How many pairs of factors of number 300 such that they are co-prime to each other?

**Solution:**

Since 300 = 2^{2} × 3^{1} × 5^{2} here we have three prime factors hence number of such pairs is given by the formula 1 + (*p +q +r*)+2(*pq +pr +qr*)+4*pqr* here *p* = 2,* q* = 1 and *r* = 2

Hence number of such pairs is 1 +5 + 2 (2 + 2 + 4) = 22 pairs.

**Example 8:** How many pairs of factors of number 2700 such that they are co-prime to each other?

**Solution:**

Since 2700 = 2^{2} × 3^{3} × 5^{2} here we have three prime factors hence number of such pairs is given by the formula 1 + (*p +q +r*) +2(*pq +pr +qr*) +4*pqr* here *p* = 2,* q* = 3 and *r* = 2

Hence number of such pairs is 1 + 7+ 2 (6 + 6 + 4) = 1+ 7 + 32 = 40 pairs.

**Pairs of Factors that are co-prime to Each Other**

**In all cases let n has total number of factors as F then**

*N***=**then number of such pairs = 1 + (

*a*^{p}b^{q}*p*+

*q*) + 2

*pq*

*N***=**then number of such pairs 1 + (

*a*^{p}b^{q}c^{r}*p*+

*q*+

*r*) +2(

*pq*+

*pr*+

*qr*) + 4

*pqr*

If

**then number of such pairs 1 + (**

*N*=*a*^{p}b^{q}c^{r}d^{s}*p +q +r +s*) +2 (

*pq +pr + ps +qr +qs +rs*)+ 4(

*pqr +pqs +prs + qrs*)+ 8

*pqrs*

If

*N*= $({{x}_{1}}^{n1})\,({{x}_{2}}^{n2})…..({{x}_{k}}^{nk})$ then number of such pairs are $=\sum\limits_{n=1}^{k}{\left( Tn \right)}$ here$Tn=\left( {{2}^{n-1}} \right)\,\{\,\sum\limits_{n=1}^{n}{x1.x2.xn}\}$

Or to further generalize it number of such pairs is $\sum\limits_{n=1}^{n}{\left[ \left( 2n-1 \right)\left\{ \sum\limits_{n=1}^{n}{x1.x2.xn} \right\} \right]}$

*Assignment: *

__Questions:__

**1**: In how many ways can you express 216 as a product of two of its factors?

a) 4

b) 6

c) 8

d) 9

Answer: c

Solution:

Step 1: 216 = 2^{3}3^{3}

Step 2: Number of factors are (3+1)(3+1) = 4 x 4= 16

Step 3: So number of ways ½ (4)(4) = 8

**2**: Find the sum of factors of 216.

a) 950

b) 850

c) 600

d) 1000

Answer: c

Solution:

Step 1: 216 = 2^{3}3^{3}

Step 2: Sum of factors = (2^{0}+2^{1}+2^{2}+2^{3})(3^{0}+3^{1}+3^{2}+3^{3})= 15 x 40 = 600

**3**:**In how many ways you can write 200 as product of two of its co-prime factors. a) 1 b) 2 c) 4 d) 8**

Answer: b

Step1: 200 = 2^{3}5^{2}

Hence number of ways are 2^{2-1 }= 2.Infact we can mention these cases as well 8 x 25,

1 x 100.

**4**: In how many ways you can write 10890 as product of two of its co-prime factors.

a) 7

b) 3

c) 31

d) 15

Answer: d

*Solution: *We will do the solution step by step

Step1: first prime factorization i.e.2^{1} x 3^{2} x 5^{1}x 11^{2}

Step2: 4 prime number are used in this so the number of ways are 2^{4}-1=15