Answer

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Hint: Determining the divisibility of a dividend is easy if the divisor is broken into its factors. If we try to check for divisibility by $45$ itself, it might lead to a complex calculation. So, it is advised to break $45$ into its prime factors.

Complete step-by-step answer:

Here, we have to check the divisibility of $25110$ by $45$.

And it is easy to determine divisibility with divisor’s prime factors, thus prime factors of divisor $45$ is

$45={{3}^{2}}\times 5\times 1$

Thus, prime factors of 45 are: $3,5,1...\text{ }\left( 1 \right)$

Now, to determine the divisibility of $25110$ by $45$, we have to check its divisibility with $3,5,1$, i.e.,

Divisibility rules of $3$ and $5$ will be checked on $25110$, as it is already divisible by $1$.

Now,

Divisibility rule of $3$ is, if sum of all the digits of dividend is divisible by $3$ or not, if it is divisible by $3$ then the whole dividend is said to be divisible by $3...\text{ }\left( 2 \right)$

Similarly, divisibility rule of $5$ is, if the last digit of dividend is either $0$ or $5$, then dividend is said to be divisible by $5...\text{ }\left( 3 \right)$

From combining equation (2) and (3), we can say that

A dividend is said to be divisible by $45$, only if it is divisible by $3$ and $5$.

Thus, applying divisibility rule of $3$ on $25110$ from equation (2), we get

$25110\to \left( 2+5+1+1+0 \right)=9$, which is divisible by 3

And applying divisibility rule of $5$ from equation (3), we get

$25110\to $ last digit is $0$, hence it is also divisible by $5$

Thus, we can say that $25110$ is divisible by $3$ and $5$.

Hence, $25110$ is divisible by $45$.

Note: An easiest way to solve such problems is direct division of dividend through divisor. If the remainder comes out to be $0$ then it can be said that dividend is divisible by divisor.

Complete step-by-step answer:

Here, we have to check the divisibility of $25110$ by $45$.

And it is easy to determine divisibility with divisor’s prime factors, thus prime factors of divisor $45$ is

$45={{3}^{2}}\times 5\times 1$

Thus, prime factors of 45 are: $3,5,1...\text{ }\left( 1 \right)$

Now, to determine the divisibility of $25110$ by $45$, we have to check its divisibility with $3,5,1$, i.e.,

Divisibility rules of $3$ and $5$ will be checked on $25110$, as it is already divisible by $1$.

Now,

Divisibility rule of $3$ is, if sum of all the digits of dividend is divisible by $3$ or not, if it is divisible by $3$ then the whole dividend is said to be divisible by $3...\text{ }\left( 2 \right)$

Similarly, divisibility rule of $5$ is, if the last digit of dividend is either $0$ or $5$, then dividend is said to be divisible by $5...\text{ }\left( 3 \right)$

From combining equation (2) and (3), we can say that

A dividend is said to be divisible by $45$, only if it is divisible by $3$ and $5$.

Thus, applying divisibility rule of $3$ on $25110$ from equation (2), we get

$25110\to \left( 2+5+1+1+0 \right)=9$, which is divisible by 3

And applying divisibility rule of $5$ from equation (3), we get

$25110\to $ last digit is $0$, hence it is also divisible by $5$

Thus, we can say that $25110$ is divisible by $3$ and $5$.

Hence, $25110$ is divisible by $45$.

Note: An easiest way to solve such problems is direct division of dividend through divisor. If the remainder comes out to be $0$ then it can be said that dividend is divisible by divisor.

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