A four-digit number $'abcd'$ is divisible by $3$. The number is rearranged as $'bcda'$. Is the second number divisible by $3$?

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Hint: A number in the form $'abcd'$ means every digit is shown with a variable separately. Use the divisibility rule of $3$, which says that if the sum of every digit of a number is divisible by $3$ then the number itself is divisible by $3$.

It is given in the question that the number $'abcd'$ is completely divisible by $3$. This means that the number $'abcd'$ satisfies the divisibility rule of $3$.
According to the divisibility rule of $3$, if the sum of all the digits in a number is divisible by $3$ then the number itself is divisible by $3$.
This implies that, $\left( {a + b + c + d} \right)$ is also divisible by $3$.
So, when the number is rearranged as $'bcda'$, the sum of the digits of the number would be $\left( {b + c + d + a} \right)$ which is the same in value as $\left( {a + b + c + d} \right)$.
Therefore, the second number $'bcda'$ is also divisible by $3$ as it satisfies the divisibility rule of $3$.
Note: To establish a better understanding of this concept let us take another case. In this case, if a three-digit number $'abc'$ is divisible by $3$ then check divisibility of $100 \times bca$ with $3$.
Similarly, here we can say using the divisibility rule that if $'abc'$ is divisible by $3$, and then $\left( {a + b + c} \right)$ is necessarily divisible by $3$.
So, for $100 \times bca$ by using divisibility rule you get $\left( {a + b + c + 0 + 0} \right) = \left( {a + b + c} \right)$ , which is already proven divisible by $3$.
In this type of question, getting handy with divisibility rules helps a lot. Try to remember divisibility rules from $1 - 10$. Don’t be confused by seeing a four-digit number in this form $'abcd'$. Here every digit is represented with separate variable, i.e. $abcd = 1000a + 100b + 10c + d$