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A number being successively divided by 3,5 and 8 leaves remainder 1,4 and 7 respectively. Find the respective remainders if the orders of divisors be reversed.

Answer
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Hint: Assume the quotient and the remainder as some variable and then according to the given conditions of the question we obtain an equation and solve it to get the required result.

Complete step-by-step answer:
Let us say that a number n, when successively divided by a, b, and c leaves a remainder of p, q, and r.
Let us suppose there is a number k and it is divided by c and the divide leaves a number as remainder r then it can be written in the form as ck+r.
Same logic can be extended to give the value of n, when it is divided by three different numbers as 3, 5 and 8 leaving the remainders as 1, 4 and 7 respectively.
This all implies that the value of n becomes as,
n=8q+7
Now dividing the above number by 5 which leaves remainder as 4 is given as,
n= 5(8q+7)+4
Similarly, applying the last divisor and remainder n becomes,
n= 3(5(8q+7)+4)+1
Solving the above,
\[\begin{align}
  & \Rightarrow n=3(5(8q+7)+4)+1 \\
 & \Rightarrow n=3(40q+39)+1 \\
 & \Rightarrow n=120q+117+1 \\
 & \Rightarrow n=120q+118. \\
 & \\
\end{align}\]
Now, if it is divided by 8.
We have \[n=120q+118=8(15q+14)+6\], remainder 6.
And if \[15q+14\] is divided by 5, we get remainder as 4.
And if \[3q+2\] is divided by 3, we get remainder 2.
Hence, this all gives us the remainders as 6, 4 and 2, which is our required result.

Note: The possibility of error in this question can be taking different variables for different quotients and different remainders which would get the wrong result because then the equation will become complex and would not be easy to solve.