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Answer will be option 3 i.e. 4 pairs. Explanation: Let first no=51x , other no = 51y then, 51x*51y=1530*51 => x*y= 30
So there are 4 pairs i.e. (5,6),(15,2),(10,3),(30,1).
The LCM and HCF of two numbers are 1530 and 51. Find how many such pairs are possible.
A
$$2$$B
$$3$$C
D
$$1$$
Let the numbers be $$51a$$ and $$51b$$, where $$a$$ and $$b$$ are co-primes.
We know that LCM$$\times$$HCF=Product of two numbers, therefore, we have,
$$51a\times 51b=(51\times 1530)\\ \Rightarrow 2601ab=51\times 1530\\ \Rightarrow ab=\dfrac { 51\times 1530 }{ 2601 } \\ \Rightarrow ab=30$$
The co-primes with product $$30$$ are $$(1,30),(2,15),(3,10)$$ and $$(5,6)$$.
Therefore, the required numbers are $$(51,1530),(102,765),(153,510)$$ and $$(255,306)$$
Hence, the number of possible pairs are $$4$$.