Number Theory : Factor Theorem  Vikas Saini

If N is a composite number = a^p x b^q x c^r
Here a,b and c are prime numbers and p,q and r are their power respectively.
Number of factors = (p+1)(q+1)(r+1) Let find the number of factors of 60.
60 = 2^2 x 3 x 5.
No of factors = (2+1)(1+1)(1+1) = 3 x 2 x 2 = 12.
Factors of 60 are 1,2,3,4,5,6,10,12,15,20,30,60.
No. of factors of a prime no = 2 Find the number of factors of 97.
97 = 97^1
No of factors = 1+1 = 2.
If any number has total number of factors are 3 only.
It means that number is a square of a prime number.
Q. If 11 < n < 600.
For how many values of n, the no of factors are 3 ?
Solution :
We know 24^2 = 576 and 25^2 = 625.
3^2 = 9 and 4^2 = 16.
3^2 < n < 24^2.
We need to work on prime numbers between 3 & 24.
The numbers are 5,7,11,13,17,19,23.
Total 7 numbers.
Even factors & odd factors of any number
If any composite number N = 2^a x p^b x q^c
No of Even factors = a x (b+1) x (c+1)
No of Odd factors = (b+1) (c+1)
Q. Find the even & odd factors of 432.
432 = 2^4 x 3^3.
Total factors = (4+1) (3+1) = 20.
No of Even factors = 4 x (3+1) = 16.
No of Odd factors = 3+1 = 4.
Factors of 432 = 1,2,3,4,6,8,9,12,16,18,24,27,36,48,54,72,108,144,216,432.
How to find even factors
(2,4,8,16) (1,3,9,27)
Firstly multiply by 2
2(1,3,9,27) = 2,6,18,54
Then by 4
4(1,3,9,27)=4,12,36,108.
Then by 8
8(1,3,9,27)=8,24,72,216.
Then by 16
16(1,3,9,27)=16,48,144,432.
Even factors are 2,6,18,54,4,12,36,108,8,24,72,216,16,48,144,432.
How to find odd factors
1,3,9,27 itself.
Q. Find total number of factors,even factors and odd factors of 300.
Solution :
300 = 2^2 x 3 x 5^2.
Total factors = (2+1) (1+1) (2+1) = 3 x 2 x 3 = 18.
Even factors = 2 (1+1) (2+1) = 12.
Odd factors = (1+1) (2+1) = 6.
Multiple of 2 (1,2,4)
Multiple of 3 (1,3)
Multiple of 5 (1,5,25)
To find Even factors :
(2,4)(1,3)(1,5,25)
(2,4)(1,3)(1,5,25)
(2,4)(1,5,25,3,15,75)
(2,10,50,6,30,150,4,20,100,12,60,300)
To find odd factors
(1,3)(1,5,25)
(1,5,25,3,15,75)
Q. If 4n has 15 factors, 5n has 12 factors. Find no of factors in 3n.
Solution :
Important part here is multiplication of 4 & 5.
4 = 2^2.
Let n = 2^a x 5^b.
4n = 2^(a+2) x 5^b.
Total factors = (a+2+1) (b+1)
15 = (a+3) (b+1)…………………..(i)
5n = 2^a x 5^(b+1)
Total factors = (a+1) (b+1+1)
12 = (a+1) (b+2)…………………….(ii)
From equation (i) & (ii)
a = 2 , b =2.
n = 2^2 x 5^2
3n = 2^2 x 3 x 5^2
Total factors = 3 x 2 x 3 = 18.
Q. For how many positive values of N, it must be an integer
(N + 102) / (N +2).
Solution :
For such questions we need to work on factors only.
(N+2+100) / (N+2)
1 + 100 / N+2.
100 = 2^2 x 5^2.
Total factors = (2+1) (2+1) = 9.
But here N+2 > 2.( N must be positive)
We need to remove 1 & 2 from total factors.
Required values = 9 – 2 = 7.
Check :
Factors of 100 are 1,5,25,2,10,50,4,20,100.
After removing 1,2 it can be checked by putting N+2 = 5,25,10,50,4,20,100.
N=3,23,8,48,2,18,98.
(N+102) / (N+2) = 21,5,11,3,26,6,2. (Total 7 values)
Q. For how many total values of N, this must be an integer.
(N + 28 ) / (N + 8 ).
Solution :
We need to work on factors.
(N+8+20) / (N+8 )
1 + (20 / N+8 )
60 = 2^2 x 5.
Total factors = 3 x 2 = 6.
But in question it is given that for total values, so factors can be negative too.
Total = 2 x 6 = 12 .
Check :
Factors of 20: – (1,2,4,5,10,20).
Negative factors : (1,2,4,5,10,20).
N + 8 = 1,2,4,5,10,20,1,2,4,5,10,20.
N = 9,10,12,13,18,28,7,6,4,3,2,12.
(N+28 ) / (N+8 ) = 19,9,4,3,1,0,21,11,6,5,3,2.
Total 12 integer values.
Q. Find the smallest number with 12 divisors.
Solution :
12 = 2 x 6, 3x 4.
To get smallest number we need to work on 2 & 3 only.
N = 2^a x 3^b.
Total divisors = (a+1)(b+1)
(a,b) = (1,5)(5,1)(2,3)(3,2)
N = 2 x 3^5 = 243.
N = 2^5 x 3 = 96.
N = 2^2 x 3^3 = 108.
N = 2^3 x 3^2 = 72.
Smallest number here is 72.
Q. A composite number is given 150.
(a) Find the sum of divisors.
(b) Find the sum of even divisors.
(c) Find the sum of odd divisors.
Solution :
150 = 2 x 3 x 5^2.
(a)Sum of divisors
(2^0 + 2) (3^0 + 3)(5^0 + 5 + 5^2)
= (1+2)(1+3)(1+5+25)
=3 x 4 x 31
= 372.
Check :
1+150+2+75+3+50+5+30+6+25+10+15 = 372.
(b) sum of even divisors
2 (1+3)(1+5+25)
=2 x 4 x 31
=248.
Check :
150+2+50+30+6+10 =248.
(c)sum of odd divisors
(1+3)(1+5+25)
= 4 x 31
= 124.
Check :
1+75+3+5+25+15=124.
Q. How many divisors of 1296 are perfect square ?
Solution =
1296 = 36^2 = (2^2 x 3^2)^2
1296 = 2^4 x 3^4
Perfect square = [ (4/2) + 1] [(4/2) + 1] = 3 x 3 = 9.
To find perfect square :
(1,2,4,8,16,32)(1,3,9,27,81)
(1,4,16)(1,9,81)
(1,9,81,4,36,324,16,144,1296)
Total 9.
Q. How many divisors of 36^36 are perfect cubes ?
Solution :
36 = 2^2 x 3^2.
36^36 = 2^72 x 3^72.
Perfect cubes = [ (72/3) + 1] [(72/3) + 1]
= 25 x 25 = 625.
Q . How many of the first 1200 natural numbers are not divisible by any 2,3,5 ?
Solution :
Solve it by Euler’s theorem
If N = p^a x q^b x r^c
E(N) = N (1 – 1/p)(1 – 1/q) (1 – 1/r)
1200 = 2^4 x 3 x 5^2.
Numbers not divisible by 2,3,5
= 1200 (1 – 1/2)(1 – 1/3) (1 – 1/5)
= 1200 (1/2)(2/3)(4/5)
= 320.
Q. How many divisors of 112 are divisible by 4,but not 8.
Solution :
112 = 2^4 x 7.
Factors = 5 x 2 = 10.
(1,2,4,8,16)(1,7)
(1,2,4,7,8,14,16,28,56,112)
4 = 2^2,8=2^3.
No divisible by 4 = (52) x 2 = 6.
No divisible by 8 = (53) x 2 = 4.
Divisors = (6 – 4) = 2.
But in 8, 4 is already counted.
Hence 1.
Q. How many divisors of 243 are divisible by 3, but not 27.
Solution :
243 = 3^5.
(1,3,9,27,81,243)
3=3^1, 27=3^3
No divisible by 3 = (61) = 5.
No divisible by 27 = (63)= 3.
Divisors = (5 – 3)= 2.