Concept: Successive Division
When we divide N successively by 'a' and 'b', it means we first divide N by a, and then divide the quotient of this division by b.
- Dividend = Divisor × Quotient + Remainder
Example: Divide 100 successively by 2 and 5.
Solution:
We first divide 100 by 2. The quotient will be 50
Now, we divide 50 by 5 and the final answer will be 10.
Example: What are the remainder when 100 is successively divided by 3 and 5.
Solution:
We first divide 100 by 3. The quotient will be 33 while the remainder will be 1.
Now, we divide 33 by 5. The quotient will be 6 while the remainder will be 3.
∴ When 100 is successively divided by 3 and 5, the remainders are 1 and 3 respectively.
Example: Find the smallest possible number which when successively divided by 3 and 4 leaves remainder of 1 and 3 respectively.
Solution:
In such examples we need to start from the last of the successive divisions.
When divided by 4 the remainder is 3. Since we need to find the smallest possible such number, we will assume the quotient for this division to be 0.
∴ The smallest possible number which when divided by 4 leaves remainder 3 is 3 itself.
Now, 3 must have been the quotient of previous division.
We need to find the number which when divided by 3 gives quotient as 3 and leaves remainder of 1.
∴ Original number = 3 × 3 + 1 = 10
The smallest number which when successively divided by 3 and 4 leaves remainder of 1 and 3 respectively is 10.
Example: Find the smallest possible number which when successively divided by 2 and 5 leaves remainder of 1 and 4 respectively.
Solution:
In such examples we need to start from the last of the successive divisions.
When divided by 5 the remainder is 4. Since we need to find the smallest possible such number, we will assume the quotient for this division to be 0.
∴ The smallest possible number which when divided by 5 leaves remainder 4 is 4 itself.
Now, 4 must have been the quotient of previous division.
We need to find the number which when divided by 2 gives quotient as 4 and leaves remainder of 1.
∴ Original number = 2 × 4 + 1 = 9
The smallest number which when successively divided by 2 and 5 leaves remainder of 1 and 4 respectively is 9.
Example: Find the smallest possible number which when successively divided by 2, 3 and 5 leaves remainder of 1, 2 and 3 respectively.
Solution:
In such examples we need to start from the last of the successive divisions.
When divided by 5 the remainder is 3. Since we need to find the smallest possible such number, we will assume the quotient for this division to be 0.
∴ The smallest possible number which when divided by 5 leaves remainder 3 is 3 itself.
Now, 3 must have been the quotient of previous division.
We need to find the number which when divided by 3 gives quotient as 3 and leaves remainder of 2.
∴ Number = 3 × 3 + 2 = 11
Now, 11 must have been the quotient of previous division.
We need to find the number which when divided by 2 gives quotient as 11 and leaves remainder of 1.
∴ Initial Number = 2 × 11 + 1 = 23
The smallest number which when successively divided by 2, 3 and 5 leaves remainder of 1, 2 and 3 respectively is 23.
A number (N) when successively divided by a, b and c leaving remainders p, q and r respectively can be written as
N = (a × b × c) × k + n
where k is a whole number and n is the smallest such number.
Example: Find the highest three-digit number possible number which when successively divided by 2, 3 and 5 leaves remainder of 1, 2 and 3 respectively.
Solution:
In such example we need to first find the smallest possible such number.
The smallest number which when successively divided by 2, 3 and 5 leaves remainder of 1, 2 and 3 respectively is 23. (from previous example)
Now, N = 2 × 3 × 5 × k + 23
⇒ N = 30k + 23
Now, N = 30k + 23 < 1000
⇒ k < 977/30
∴ highest possible value of k = 32
∴ N = 30 × 32 + 23 = 983
Example: Find the smallest three-digit number possible number which when successively divided by 3 and 5 leaves remainder of 1 and 2 respectively.
Solution:
In such example we need to first find the smallest possible such number.
The smallest number which when successively divided by 3 and 5 leaves remainder of 1 and 2 respectively is 7.
Now, N = 3 × 5 × k + 7
⇒ N = 15k + 7
Now, N = 15k + 7 > 99
⇒ k > 92/15
∴ least possible value of k = 7
∴ N = 15 × 7 + 7 = 112