A number when successively divided by 5, 2, 3 gives remainder 0, 1, 2…

Given:
A number when successively divided by 5, 2, 3 gives remainders 0, 1, 2 respectively.
We must find the remainders when the same number is successively divided by 2, 3, 5.

Step 1: Work backward using successive division

Let the number be N.

1. Dividing by 3 gives remainder 2

   N=3a+2

2. This result when divided by 2 gives remainder 1

   3a+2 = 2b+1 ⇒ 3a = 2b−1

   Smallest solution: a=1, b=2

3. This result when divided by 5 gives remainder 0

   2b+1=5k

   Smallest solution: b=2 ⇒ N=25

So, the smallest number satisfying the given condition is 25.

Step 2: Divide 25 successively by 2, 3, 5

1. 25÷2 → remainder 1, quotient = 12

2. 12÷3 → remainder 0, quotient = 4

3. 4÷5 → remainder 4
   
   Final Answer: Remainders = 1, 0, 4