A number being successively divided by 3,5,8 leaves remainder 1,4,7 .find the respective remainders if the order of divisor be reveresed
Answers
Answer:
Suppose, the number be A.
When A is divided by 3; B is the quotient and 1 is the remainder. So, A = 3 * B + 1…………(1)
When B is divided by 5; C is the quotient and 4 is the remainder. So, B = 5 * C + 4……………(2)
When C is divided by 8; D is the quotient and 7 is the remainder. So, C = 8 * D + 7…………….(3)
Here, A, B, C and D all are non-zero positive integers.
Combining (1), (2) and (3), we get:
A = 3 * B + 1 = 3 * (5 * C + 4) + 1 = 15 * C + 13 = 15 * (8 * D + 7) + 13 = 120 * D + 118
Thus, A = 120 * D + 118…………………………..(4)
Therefore, A is a sum of two integers (120 * D) and 118.
The first constituent integer (120 * D) is a multiple of 120; thus, (120 * D) will leave no remainder whichever way it may be divided successively by 3, 5 and 8.
But, the case of 118 is not alike!
If 118 is first divided by 8; the quotient is 14 and the remainder is 6.
If 14 is divided by 5; the quotient is 2 and 4 is the remainder.
If 2 is divided by 3; the quotient is 0 and 2 is the remainder.
Therefore, if the same number be successively divided by 8, 5 and 3; it will leave 6, 4 and 2 as remainders respectively.