A positive integer N when successively divided by 4, 5, and 6 leaves remainders of 2, 3, a… — State PSC Exam

A positive integer $N$ when successively divided by $4$ , $5$ , and $6$ leaves remainders of $2$ , $3$ , and $4$ , respectively. What is the sum of the remainders obtained when the smallest such number $N$ is successively divided by $6$ , $5$ , and $4$ , in that order?

Answer: 7

Answer

The sum of the remainders obtained when the smallest such number is successively divided by 6, 5, and 4 is 7.

The smallest positive integer

N

satisfying the given successive division conditions is 94. Successively dividing 94 by 6, 5, and 4 yields quotients of 15, 3, and 0 with remainders 4, 0, and 3, respectively. The sum of these remainders is

4 + 0 + 3 = 7

Step-by-Step Solution

Formulate the algebraic equations for successive division and calculate the smallest positive value of

N

N = 94

By definition of successive division,

N = 4q_1 + 2

q_1 = 5q_2 + 3

, and

q_2 = 6q_3 + 4

. Setting the non-negative integer quotient

q_3 = 0

yields

q_2 = 4

q_1 = 23

, and

N = 94

Perform successive division of 94 by the divisors 6, 5, and 4 in sequence.

The sequence of remainders is

4

0

, and

3

First stage:

94 \div 6 = 15

remainder

4

. Second stage:

15 \div 5 = 3

remainder

0

. Third stage:

3 \div 4 = 0

remainder

3

Calculate the sum of the three remainders.

4 + 0 + 3 = 7

Adding the individual remainders obtained from each stage of the reverse order division.

Key Concept

Successive Division and Remainder Property

Question

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