Question 59

HCF of 3240, 3600 and a third number is 36 and their LCM is $$2^4 \times 3^5 \times 5^2 \times 7^2$$. The third number is

Solution

Let the third number be $$x$$. H.C.F. = 36. Prime factorization of :

3240 = $$2^3\times3^4\times5$$

3600 = $$2^4\times3^2\times5^2$$

=> $$x=2^2\times3^{n}\times k$$, where $$n\geq2$$ and $$k$$ is any prime number and $$k\neq5$$

Also, LCM is $$2^4 \times 3^5 \times 5^2 \times 7^2$$

=> $$n=5$$ and $$k=7^2$$

$$\therefore$$ $$x=2^2\times3^5\times7^2$$

=> Ans - (B)

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