# HCF of 3240, 3600, and a third number is 36 and their LCM is 2^{4} × 3^{5} × 5^{2} × 7^{2} . The third number is = ?

A. 2^{2} × 3^{5} × 7^{2}

B. 2^{2} × 5^{3} × 7^{2}

C. 2^{5} × 5^{2} × 7^{2}

D. 2^{3} × 3^{5} × 7^{2}

**Answer: Option A **

__Solution(By Examveda Team)__

3240 = 2^{3}× 3

^{4}× 5

3600 = 2

^{4}× 3

^{2}× 5

^{2}

HCF = 36 = 2

^{2}× 3

^{2}

Since H.C.F. is the product of lowest powers of common factors, so the third number must have (2

^{2}× 3

^{2}) as its factors.

Since L.C.M. is the product of highest powers of common prime factors, so the third number must have 3

^{5}and 7

^{2}as its factors.

∴ Third number = 2

^{2}× 3

^{5}× 7

^{2}

Related Questions on Problems on H.C.F and L.C.M

## Join The Discussion