Question

How many factors of
$N={12}^{12}\times {14}^{14}\times {15}^{15}$
are multiple of
$K={12}^{10}\times {14}^{10}\times {15}^{10}$

• A. $2\times 4\times 5$
• B. $3\times 5\times 6$
• C. $8\times 7\times 4\times 5$
• D. $9\times 8\times 6\times 5$

Answer 1

The correct option is D $9\times 8\times 6\times 5$

$N={12}^{12}\times {14}^{14}\times {15}^{15}$

$=(2^2\times 3{)}^{12}\times (7\times 2{)}^{14}\times (5\times 3{)}^{15}$

$=2^{24}\times 3^{12}\times 7^{14}\times 2^{14}\times 5^{15}\times 3^{15}$

$=2^{38}\times 3^{27}\times 5^{15}\times 7^{14}$

$K=(12{)}^{10}\times (14{)}^{10}\times (15{)}^{10}$

$=2^{20}\times 3^{10}\times 7^{10}\times 2^{10}\times 3^{10}\times 5^{10}$

$=2^{30}\times 3^{20}\times 5^{10}\times 7^{10}$

Thus factors of N which are multiple of k are
$2^{30}\times 3^{20}\times 5^{10}\times 7^{10}(2^8\times 3^7\times 5^5\times 7^4)$

$willbe(8+1)\times (7+1)\times (5+1)\times (4+1)9\times 8\times 6\times 5$