# CAT Question Bank - Ordered & Unordered Solutions

• @Kushal-Khandelwal

Funda: If N= P1^a * P2^b * P3^c then ordered solution such that LCM = N is (2a + 1) * (2b + 1) * (2c + 1)
Unordered will be [(2a + 1) * (2b + 1) * (2c + 1) + 1] /2

LCM (a, b) = 144 = 2^4 x 3^2
Ordered pairs = 9 * 5 = 45 pairs
Unordered pairs = (45 + 1)/2 = 23 pairs

Explanation: @Soumya-Chakraborty

How many pairs of 2 numbers are there whose LCM is 400?

Let us generalise this process. First of all we will try to find the ordered pairs of numbers. Then we will remove the order from it.
LCM (a,b) = 400 = 2^4 ∗ 5^2
Let us think from the perspective of each prime.
First of all, both ‘a’ and ‘b’ must be factors of 2^4
So we have 5 possibilities of both ‘a’ and ‘b’. Hence 5^2 possibilities.
But from these possibilities we must remove those cases where ‘a’ and ‘b’ are neither 2^4. This will happen when both ‘a’ and ‘b’ are factors of 2^3. So that is 4^2 cases that must be removed.
So, we have (5^2 − 4^2) cases where the LCM will be 24

Similarly, for 5^2, we will have (3^2 − 2^2) cases where the LCM of ‘a’ and ‘b’ will be 5^2

So, we have a total of (5^2−4^2) ∗ (3^2 − 2^2) = 45 cases where LCM (a,b) = 400

These are the ordered solutions. In order to convert to unordered solution, we must first understand that except for the case (400,400), all cases have been counted twice. So, the number of unordered solutions will be (45+1)/2=23 cases

• @rowdy-rathore ordered 30

• @rowdy-rathore 361
LCM(a,b,c) = 7^2 * 17^2
Pair = (3^3 - 2^3) * (3^3 - 2^3)
= 19 * 19 = 361

solutions from Nitin Gupta sir (AplhaNumeric) shared below (for both ordered and unordered)

• @rowdy-rathore
(3^10-3)/3! +1=9842

• @zabeer it will be HCF ( a + b + 1) = 221

13*17
hcf can be 13 or 17
a+b+1 = 17 or 13
we will find out co prime pairs right ??