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Avyay

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 ?

Jay 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 $147=3×7×7$,

$(c+1)=7,~(a+1)=7,~(b+1)=3$

=> $c=6,~a=6,~b=2$

Hence, $A=3^6×7^2×2^6$

$\dfrac{A}{147}=\dfrac{3^6×7^2×2^6}{3×7^2}=3^5×2^6=15552$

$147=3×7^2$ (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 $b \ge 2$

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

Since $147=3×7×7$,

$(c+1)=7,~(a+1)=7,~(b+1)=3$

=> $c=6,~a=6,~b=2$

Hence, $A=3^6×7^2×2^6$

$\dfrac{A}{147}=\dfrac{3^6×7^2×2^6}{3×7^2}=3^5×2^6=15552$

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