Question

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

Answer 1

Answer:

Please use the method mentioned below

Step-by-step explanation:

Let us say that a number N, when successively divided by a, b, and c leaves a remainder of p, q, and r

=> Before the last division by c, the number must have been of the format of ck + r. Here k is a natural number.

Same logic can be extended to give the value of N.

=> N = a[b(ck + r) + q] + p

In this case:

N = 3[5(8k + 7) + 4] + 1

=> N = 3[40k + 35 + 4] + 1

=> N = 3[40k + 39] + 1

=> N = 120k + 118

Now, we need to calculate the remainders when the number is successively divided by 8, 3 and 5.

The question will be simpler if I just assume some value of 'k'

Let us put k = 0

=> N = 118

=> 118/8 = 14 + remainder of 6

=> 14/3 = 4 + remainder of 2

=> 4/5 = 0 + remainder of 4

Answer 2

Answer:

0,0,0

Step-by-step explanation:

Let number be N

N=K+1

K=5L+4

L=8M+7

so K=5(8M+7)+4

K=40M+39

now N=40M+39+1

=40M+40

now put M=0 we get N=40 which is the least no.

now order of divisor is reversed so it is 8 5 1

first 40 is divided by 8 ,quotient 5 and remainder 0 .now 5 will be divided by 5 then quotient is 1 and remainder is 0 now 1 is divided by 1 so quotient is 1 and remainder will be 0

so order of remainder will 0,0,0