LCM/HCF

Question

Find the number of combinations of (a, b, c) if LCM (a, b) = 1000, LCM (b, c) = 2000, LCM (c, a) = 2000.

I am getting total 96 combinations

Comments

Although @Shivam Chauhan ji has provided correct answer. But for detailed analysis, Please refer my answer.

@ reena_kandari ji. Your answer is coming wrong because you are doing double counting. 

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I think its 100 not sure

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you are counting extra combinations IN powers of 2 you are counting (3,3) twice and you can figure out for powers of 5.

Answer 1

LCM (a, b) = 2³×5³

LCM (b, c) = 2⁴×5³

LCM (c, a) = 2⁴×5³

Therefore prime factorization shows that only 2 and 5 are involved in combination of numbers.

Let a = 2^p×5^q

b = 2^r×5^s

c = 2^t×5^u

Calculation for powers of 2:

Above LCM's shows that 2⁴ came from c thus t = 4.  Now either p or r will take 3 and other will take 0, 1, 2, 3.

Total options for (p,r) = (0,3) (1,3) (2,3) (3,3) (3,0) (3,1) (3,2) = 7

Calculation for powers of 5: 
Maximum power of 5 is 3.  So any two of q, s, u have maximum power 3.  and other will take 0, 1, 2, 3.

Total options for (q,s,u) = (0,3,3) (1,3,3) (2,3,3) (3,0,3) (3,1,3) (3,2,3) (3,3,0) (3,3,1) (3,3,2) (3,3,3) = 10

Total combinations = 7 x 10 = 70

Comments

Done @Chhotu ji

Answer 2

LCM (a, b) = $2^{3}×5^{3}$, LCM (b, c) = $2^{4}×5^{3}$ and LCM (c, a) = $2^{4}×5^{3}$

Therefore prime factorization shows that only 2 and 5 are involved in combination of numbers.

Let a = $2^{P}×5^{Q}$ , b = $2^{R}×5^{S}$ and a = $2^{T}×5^{U}$. then  answer is 70. For very detailed explanation, Please refer attached image.

Comments

Above LCM's shows that 2⁴ came from c thus t = 4.  Now either p or r will take 3 and other will take 0, 1, 2, 3. I didn't get this logic. can you please explain....