Formula To Find Ordered & Unordered Pairs Possible For A Given LCM  Hemant Malhotra

hemant_malhotra
Director at ElitesGrid  CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
First let us understand some basic things about LCM
We know LCM of 2^{3 }* 3^{2} and 2^{2 }* 3^{3 }is 2^{3} * 3^{3}
Here we look for highest power of 2 and highest power 3. Now if we want to find Lcm of 2^{3} * 3^{3 }and 2^{3} * 3^{3} still LCM will be 2^{3} * 3^{3} so if highest power is same in both the number there will be no effect in LCM
so highest power should be distributed in atleast one of the number.
If LCM of a and b is 2^{3} * 3^{3} then we will say that 2^{3} will be in a or will be in b or will be in a and b both , same for 3^{3}
Question  The LCM of three positive integers a, b, c is 119^{2}. Find the total number of ordered triplets (a ,b ,c)
119^{2} = 17^{2} * 7^{2}.
 > LCM is 17^{2} * 7^{2}
So, powers of 17 till the maximum power available (17^{0}, 17^{1}, 17^{2}) and powers of 7 till the maximum power available 7 (7^{0}, 7^{1}, 7^{2}) will be distributed among a, b, c such that at least one of a, b, c has the highest power of 17 and at least one of them has the highest power of 7.
Each of a, b, c can have any one of three powers of 17 in 3 ways. = > a, b and c can have powers of 17 in 3 * 3 * 3= 27 ways.
But this will also contain cases when a, b, c will not have highest power of 17 so we have to remove those cases.
Cases where highest power is in none of a, b and c is in 2 ways for each of a, b and c = > a, b and c cannot have highest power of 17 in 2 * 2 * 2 = 8 ways.
at least one = total cases  cases when none of them has highest power
a, b and c can have powers of 17, such that at least one of them has the highest power of 17, in 27 – 8 = 19 ways.
Similarly, a, b and c can have powers of 7, such that at least one of them has the highest power of 7, in 19 ways.
so total ways = 19 * 19 = 361
Case1 for ordered when we want to find number of pairs when LCM is given
let two numbers x and y and their lcm is 2^{2} * 3^{3} * 5^{3}
now atleast one of x and y , highest power of 2 is 2. highest power of 3 is 3 and highest power of 5 is 3
now let's take 2^{2 }so x can be 2^{0 }or 2^{1 }or 2^{2}.. same in case of y
so 3 * 3 = 3^{2 }ways for them
but there will be a case when neither x nor y have the power 2^{2} so we have to remove that case because lcm won't be 2^{2}
so we will remove cases when x and y have only 20 or 2^1 so 2*2 = 2^{2} cases we have to remove
so total cases = 3^{2 } 2^{2 }= 5 = 2 * 2 + 1
Similarly when we choose 3^{3} then total cases = 4^{2}  3^{2} = 7 = 2 * 3 + 1
For case of 5^{3}, it is 4^{2 } 3^{2 }= 7 = 2 * 3 + 1
so Total case = (3^{2 } 2^{2}) * ( 4^{2 } 3^{2}) * ( 4^{2}  3^{2})
= (2 * 2 + 1) (2 * 3 +1) (2 * 3+1)
See the pattern ? so the formula is
Number of ordered pairs possible for LCM = N = P_{1}^{a} x P_{2}^{b} x P_{3}^{c}
= [ (a+1)^{2} a^{2}] [b+1)^{2} b^{2}] [ (c+1)^{2} c^{2}]
= (2a + 1 ) ( 2b +1) (2c+1)
Where, a, b and c are power of prime factors
If LCM of two numbers is 360, how many such ordered pairs are possible?
360 = 2^{3} * 3^{2 }* 5
so our formula is (2a+1)(2b+1)(2c+1)
where a , b and c are power of prime
So [(2*3+1)(2*2+1)(2*1+1)
7 * 5 * 3 = 105
The LCM of three positive integers X, Y and Z is 119^2. Find the total number of ordered triplets (X, Y and Z) and unordered triplets
total ordered =361, now How to Find Unordered
Ordered=total cases (with arrangement) and Unordered (Without arrangement)
Case1  when all are different
Case 2  when 2 are same and one is different
Case 3  when all are same
a) case1 when all are different : let those cases are x and number of ways to arrange them =3!=6 so 6x
b) cases when two are same and one is different :
means basically we want to find ordered pair of 2 numbers
those will be (2a + 1) (2b + 1) = (2 * 2 + 1) (2 * 2 + 1) = 25 but this will also contain case when all three are equal so remove that
so 24 cases now ways to arrange them 3!/2!=3
so total 24 * 3=72
Case3 when all are same .. only 1 case will be there
so ordered =361
ordered = 6x + 3y + z, 361 = 6x + 72 + 1
so (36173)/6=x
so unordered=x+y+z = (36173)/6 + 24 + 1