Question

If a number is successively divided by 3,5 and 8 then the remainders are 1,4 and 7 respectively.
Find the remainders when the same number is divided in reverse order that is 8, 5 and 3.

Answer 1

\[\huge \color{green}{\textbf{Welcome To Openstudy..}}\]

Answer 2

call the number x

Answer 3

x = 3i + 1
i = 5j + 4
j = 8m + 7

Answer 4

next i would try finding x in terms of m and simplify

Answer 5

((x-1)/3 - 4)/5 = 8m + 7

=>

(x-1)/3 = 40m + 39

x = 120m + 118

so x leaves a remainder of 6 when divided by 8

Answer 6

To state the problem more clearly, given an integer N where:

N mod 3 = 1
(N div 3) mod 5 = 4
((N div 3) div 5) mod 8 = 7

Where div means divide and discard the remainder, and mod is the modulo operator, giving the remainder.

Then, find:
N mod 8
(N div 8) mod 5
((N div 8) div 5) mod 3

So, N = 3n + 1 for some integer n. Therefore, N div 3 = n. Thus, n = 5m + 4 for some integer m. Finally, since n div 5 = m, then m = 8p + 7. Let's work backwards to find N in terms of p.

n = 5(8p + 7) + 4 = 40p + 39
N = 3(40p + 39) + 1 = 120p + 118

Now divide N by 8:

N/8 = 15p + 14 + 6/8

So the first remainder is 6. Divide again by 5:

(15p + 14) / 5 = 3p + 10 + 4/5

The second remainder is 4. Divide by 3:

(3p + 10) / 3 = p + 9 + 1/3

The remainder is 1. Thus, the answer is 6, 4, 1.

Answer 7

when divided by 3 the remainder should be 2. not 1 @mayankdevnani