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The least number when successively divided by 3, 5, 7 and 3 leaves remainders 1, 2, 4 and 1 respectively. What are the respective remainders if the order of the divisor is reversed?

The least number when successively divided by 3, 5, 7 and 3 leaves remainders 1, 2, 4 and 1 respectively. What are the respective remainders if the order of the divisor is reversed? Correct Answer 1, 1, 3, 1

The least number A, in this case, will be determined as follows:

3

A

 

5

B

1

7

C

2

3

D

4

 

1

1

 

We know that, Dividend = Divisor × Quotient + Remainder

D = 3 × 1 + 1 = 4

⇒ C = 7 × D + 4 = (7 × 4) + 4 = 32

⇒ B = 5 × C + 2 = (5 × 32) + 2 = 162

⇒ A = 3 × B + 1 = (3 × 162) + 1 = 487

According to the question,

If the order of divisor is reversed, then, the remainder will be –

3

487

 

7

162

1

5

23

1

3

4

3

 

1

1

 

∴ The remainder will be 1, 1, 3 and 1, respectively when the order of divisor is reversed.

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