CAT Question Bank  Ordered & Unordered Solutions

Funda: Number of ordered pairs possible for LCM of N = P1^a x P2^b x P3^c
= (2a + 1) (2b + 1) (2c + 1)Here 360 = 2^3 * 3^2 * 5
So number of ordered pairs possible = (2 * 3 + 1) (2 * 2 + 1)(2 * 1 + 1) = 105More on this concept can be read @ Formula To Find Ordered & Unordered Pairs Possible For A Given LCM  Hemant Malhotra

Funda: Number of Triangles with Integer sides for a given perimeter.
 If the perimeter p is even then, total triangles is [p^2]/48.
 If the perimeter p is odd then, total triangles is [(p+3)^2]/48
 If it asks for number of scalene triangle with a given perimeter P, then subtract 6 and apply the same formula . For even [(p6)^2]/48 and for odd [(p3)^2]/48.
Here P = 2009 (odd)
So total triangles possible = (2012)^2/48 = 84336

AM ≥ GM
(p/q + q/r + r/p)/3 ≥ 1
(p/q + q/r + r/p) ≥ 3
So minimum value possible is 3.
No triplets exist.
[ @shashank_prabhu ]

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] /2LCM (a, b) = 144 = 2^4 x 3^2
Ordered pairs = 9 * 5 = 45 pairs
Unordered pairs = (45 + 1)/2 = 23 pairsExplanation: @SoumyaChakraborty
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 24Similarly, 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

@rowdyrathore ordered 30

@rowdyrathore 361
LCM(a,b,c) = 7^2 * 17^2
Pair = (3^3  2^3) * (3^3  2^3)
= 19 * 19 = 361

@badalravi correct.
solutions from Nitin Gupta sir (AplhaNumeric) shared below (for both ordered and unordered)

@rowdyrathore
(3^103)/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 ??is the answer = 6
