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Q.

What is the maximum value of $m$ such that $7^m$ divides into 14! evenly?

 A.

1

 B.

2

 C.

3

 D.

4

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Solution:
Option(B) is correct

The term 14! equals the product of the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13$ and $14$.

Only two of these numbers are divisible by 7.

The numbers are 7 and 14.

Hence, 14! can be expressed as the product of $k×7×14$, where $k$ is not divisible by 7.

Now, since there are two 7s in 14!, the numbers 7 and $7^2$ divide 14! evenly.

$7^3$ and further powers of 7 leave a remainder when divided into 14!.

Hence, the maximum value of $m$ is 2.


(2) Comment(s)


Anjali Agrawal
 ()

Please elaborate the solution .. I m not getting it . How do we know that 49 divide 14! Evenly ?



Shobia
 ()

$\frac{14}{7}$ will give 2 with remainder 0,,

so the ans: 2