Subject: General Aptitude · Chapter: H.C.F. and L.C.M. of Numbers · Exam: 2005 · Marks: · Difficulty:
H.C.F. of 3240, 3600 and a third number is 36 and their L.C.M. is $2^{4} \times 3^{5} \times 5^{2} \times 7^{2}$. The third number is
(a)$2^{2} \times 3^{5} \times 7^{2}$
(b)$2^{2} \times 5^{3} \times 7^{2}$
(c)$2^{5} \times 5^{2} \times 7^{2}$
(d)$2^{3} \times 3^{5} \times 7^{2}$
Answer
Answer (as printed): A
Explanation
$3240=2^{3} \times 3^{4} \times 5 ; 3600=2^{4} \times 3^{2} \times 5^{2}$; $$\text { H.C.F. }=36=2^{2} \times 3^{2} .$$ Since H.C.F. is the product of lowest powers of common factors, so the third number must have $\left(2^{2} \times 3^{2}\right)$ as its factor. 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. $$\therefore \quad \text { Third number }=2^{2} \times 3^{5} \times 7^{2} .$$
Explanation as extracted from the printed page; notation may be imperfect.