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


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    First let us understand some basic things about LCM

    We know LCM of 23 * 32 and 22 * 33 is 23 * 33

    Here we look for highest power of 2 and highest power 3. Now if we want to find Lcm of 23 * 33 and 23 * 33 still LCM will be 23 * 33 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 23 * 33 then we will say that 23 will be in a or will be in b or will be in a and b both , same for 33

    Question - The LCM of three positive integers a, b, c is 1192. Find the total number of ordered triplets (a ,b ,c)

    1192 = 172 * 72.

    -- > LCM is 172 * 72

    So, powers of 17 till the maximum power available (170, 171, 172) and powers of 7 till the maximum power available 7 (70, 71, 72) 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 22 * 33 * 53

    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 2so x can be 20 or 21 or 22.. same in case of y

    so 3 * 3 = 3ways for them

    but there will be a case when neither x nor y have the power 22 so we have to remove that case because lcm won't be 22

    so we will remove cases when x and y have only  20 or 2^1 so 2*2 = 22 cases we have to remove

    so total cases = 32 - 22 = 5 = 2 * 2 + 1

    Similarly when we choose 33 then total cases = 42 - 32 = 7 = 2 * 3 + 1

    For case of 53, it is 42 - 32 = 7 = 2 * 3 + 1

    so Total case = (32 - 22) * ( 42 - 32) * ( 42 - 32)

    = (2 * 2 + 1) (2 * 3 +1) (2 * 3+1)

    See the pattern ? so the formula is

    Number of ordered pairs possible for LCM = N = P1a x P2b x P3c

    = [ (a+1)2 -a2] [b+1)2 -b2] [ (c+1)2 -c2]

    = (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 = 23 * 32 * 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 (361-73)/6=x

    so unordered=x+y+z = (361-73)/6 + 24 + 1


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