Let y be the product of all the divisors of the number 720 . If y is divisible by 15p, then maximum value of ((p/3)) equals

Question

Let y be the product of all the divisors of the number 720 . If y is divisible by 15p, then maximum value of ((p/3)) equals (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

720=24× 32× 51 Total number of cases <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-thxealab6j86sca2.jpg" /> Similarly 3 more cases with powers 23, 24, 20. Power of 3=5× (2 + 4)=30 Power of 5=5× 3=15 15P=5P× 3P Max. value of P=15

Q. Let $y$ be the product of all the divisors of the number $720$ . If $y$ is divisible by $1 5^{p},$ then maximum value of $(\frac{p}{3})$ equals

$720 = 2^{4} \times 3^{2} \times 5^{1}$
Total number of cases

Similarly $3$ more cases with powers $2^{3}, 2^{4}, 2^{0} .$
Power of $3 = 5 \times (2 + 4) = 30$
Power of $5 = 5 \times 3 = 15$
$1 5^{P} = 5^{P} \times 3^{P}$
Max. value of $P = 15$

Q. Let y be the product of all the divisors of the number 720 . If y is divisible by 15p, then maximum value of (3p​) equals

720=24×32×51 Total number of cases Similarly 3 more cases with powers 23,24,20. Power of 3=5×(2+4)=30 Power of 5=5×3=15 15P=5P×3P Max. value of P=15

Q. Let $y$ be the product of all the divisors of the number $720$ . If $y$ is divisible by $1 5^{p},$ then maximum value of $(\frac{p}{3})$ equals

$720 = 2^{4} \times 3^{2} \times 5^{1}$
Total number of cases

Similarly $3$ more cases with powers $2^{3}, 2^{4}, 2^{0} .$
Power of $3 = 5 \times (2 + 4) = 30$
Power of $5 = 5 \times 3 = 15$
$1 5^{P} = 5^{P} \times 3^{P}$
Max. value of $P = 15$