*Monday, April 29th, 2019*

If the quotient of the dividend is taken and this is used as the dividend in the next division, such as division is called “successive division”. A successive division process can continue up to any number of steps until the quotient in a division became zero for the first time, i.e., the quotient in the first division is taken and divided in the second division; the quotient in the second division is taken as the dividend in the third division; the quotient in the third division is taken as the dividend in the fourth division and so on.

If we say that 2479 is divided successfully by 3, 5, 7 and 2, then the quotients and remainders are as follows in the successive division.

Dividend |
Divisor |
Quotient |
Remainder |

2479 | 3 | 826 | 1 |

826 | 5 | 165 | 1 |

165 | 7 | 23 | 4 |

23 | 2 | 11 | 1 |

Here we say that when 2479 is successfully divided by 3, 5, 7 and 2, the respective remainders are 1, 1, 4 and 2.

Examples:

**1.** A number when divided successively by 5, and 2 gives respective remainders of 3 and 1. What will be the remainder when largest such two digit number is divided by 12?

**Sol:** We will write down the divisors one after the other and their respective remainder below them:

Then starting from the last remainder, we go diagonally left upwards to the first row multiplying and then directly below adding the figure already obtained. We continue this process until we reach the figure on the extreme left in the second row.

So, we get (1×5) +3=8

So the number is 10k + 8 (10 is the product of divisors).

So the number is of form 10k +8,

for k=0,1,2,3,4…………….

The largest 2-digit number is, 10(9) +8 =98. This, when divided by 12, leaves a remainder of 2.

**2.** A number when successfully divided by 3, 4 and 7 leaves respective remainders of 2, 3 and 1. How many such numbers are there under 1000?

**Sol:** Let us write down all the divisors and their respective remainders as follows:

We start at the bottom right corner 1, and go from 2^{nd} row to 1^{st} row diagonally to the left multiplying. We 1 x 4=4, then we come down to the 2^{nd} row adding, we get, 4 + 3= 7. Again multiplying diagonally left upwards, we get 7 x 3=21. Coming down to 2^{nd} row, adding, we get 21 + 2 =23.

Therefore, the smallest number that satisfies the given condition is 23. The general form of numbers that satisfy the given condition is got by adding the LCM of divisors, which is 84, to 23

Hence, General form is 84k +23.

For k= 0, 1, 2………………..11, the number is less than 1000.

So, there are 12 numbers less than 1000 that satisfy this condition.

**3.** A number when successively divided by 3, 4 and 9 leaves respective remainders of 2, 3 and 7. What will be the remainders if the original number is divided successively by 3, 5 and 7?

**Sol:** Again, we will first find the smallest number that satisfies this condition

The smallest number is

[{(7 x4)+3} x 3 +2] = 95

Now, when 95 is successively divided by 3, 5, and 7, the results are:

Dividend |
Divisor |
Quotient |
Remainder |

95 | 3 | 31 | 2 |

31 | 5 | 6 | 1 |

6 | 7 | 0 | 6 |

The remainders are 2,1 and 6 respectively.

**4.** A number when successively divided by 3,5 and 8 leaves remainders 1,4 and 7 respectively. Find the respective remainders when the order of the divisors is reversed.

**Sol:** We will find the smallest number which is satisfying this condition and then the remainders after reversing the divisors.

The smallest number is

[{(7 x5)+4} x 3 +1] = 118

Now, when 118 is successively divided by 8, 5, and 3, the results are:

Dividend |
Divisor |
Quotient |
Remainder |

118 | 8 | 14 | 6 |

14 | 5 | 2 | 4 |

2 | 3 | 0 | 2 |

The remainders are 6, 4, and 2 respectively.

**5.** A number when successively divided by 3, 4 and 5 leaves remainders 1, 2 and 3 respectively. Find the remainder when it would be divided by:

- 60
- 16
- 15
- 30

**Sol:** Again, we will first find the smallest number that satisfies this condition

The smallest number is

[{(4 x3)+2} x 3 +1] = 43

Hence, the general form of the number satisfying this condition is: 43+60k {k is an integer; 60 is LCM of divisors}.

Now, what would be the remainder when it would be divided by 60.

The answer is 43.

This is according to the Constant Remainder Theorem.

*According to this theorem, if you divided any term of the AP by the common difference you will get the same term i.e. 1 ^{st} term as the remainder.*

And we can find the remainders only when the divisor is a factor of the common difference.

Therefore, if you need to find what would be the remainder when it would be divided by 16.

**The answer is cannot be determined, as 16 is not a factor of 60.**

If it would be divided by 15, the answer would be: 43 mod 15 = 3

If it would be divided by 30, the answer would be: 43 mod 30= 13

Divisibility Rules for CAT Quantitative Aptitude Preparation

Baisc Idea of Remainders

Application of LCM (Lowest Common Multiple) in solving Quantitative Aptitude Problems

How to Find Number of Trailing Zeros in a Factorial or Product

Dealing With Factorials

All questions from CAT Exam Quantitative Aptitude – Number Systems

Quantitative Aptitude – Number Systems – Q1: If the product of three consecutive positive integers is 15600 then the sum of the squares of these integers is

Quantitative Aptitude – Number Systems – Q2: If a, b, c are three positive integers such that a and b are in the ratio 3 : 4 while b and c are in the ratio 2:1, then which one of the following is a possible value of (a + b + c)?

Quantitative Aptitude – Number Systems – Q3: The numbers 1, 2,…,9 are arranged in a 3 X 3 square grid in such a way that each number occurs once and the entries along each column, each row, and each of the two diagonals add up to the same value.

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In 5th question, when no is divided by 15, remainder coming 13. Please check.