1.

A number when successively divided by 5 3 8 the leaves a remainder of 147 Find the least number​

Answer»

Answer:

6, 2, 4

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

So, the remainders are 6, 2, and 4

Step-by-step EXPLANATION:


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