If are prime numbers and are the positive integers such that LCM of is , then the number of ordered pairs is
- A.252
- B.254
- C.225
- D.224
Show answer and solution
Correct option: C
Solution
For each prime factor, count pairs (a,b) where max(a,b) equals the required exponent.
For prime r with max exponent 2: pairs (a,b) with max(a,b)=2, where a,b ∈ {0,1,2}. Total pairs with a,b ≤ 2 is 9, minus pairs with max ≤ 1 which is 4, giving 5. Formula: 2(2)+1 = 5.
For prime s with max exponent 4: 2(4)+1 = 9.
For prime t with max exponent 2: 2(2)+1 = 5.
Total ordered pairs = 5 × 9 × 5 = 225.
For prime r with max exponent 2: pairs (a,b) with max(a,b)=2, where a,b ∈ {0,1,2}. Total pairs with a,b ≤ 2 is 9, minus pairs with max ≤ 1 which is 4, giving 5. Formula: 2(2)+1 = 5.
For prime s with max exponent 4: 2(4)+1 = 9.
For prime t with max exponent 2: 2(2)+1 = 5.
Total ordered pairs = 5 × 9 × 5 = 225.
[AIEEE 2006]