We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.More informationAgree
asked 2016-05-06 21:04:53 
If '$A$' is the smallest positive integer that is a multiple of $147$ and has exactly $147$ positive integral divisors,including $1$ and itself,the value of $\dfrac{A}{147}$ is ?
like 2 dislike 0 comment
profileJay Patel
answered 2016-05-06 21:28:57 
Ans: $15552$

$147=3×7^2$ (prime factorization)

If a number can be written as $x^p×y^q×z^r$ where $x,y,z$ are prime factors of that number, then, total number of factors of that number is $(p+1)(q+1)(r+1)$. We will use the same concept here.

Since $A$ is a multiple of $147$ and the smallest, let's take
$A=3^a×7^b×2^c$ where $a\ge 1$ and $b \ge 2$
(Note that we have taken the third factor as 2 to minimize the value of $A$)

Then $(a+1)(b+1)(c+1)=147$ because $A$ has $147$ positive integral divisors.

Since $147=3×7×7$, 
=> $c=6,~a=6,~b=2$

Hence, $A=3^6×7^2×2^6$

like 1 dislike 0 comment

Answer This Question