Math, asked by DKRK1510, 4 months ago

Find the least number which when divided by 8, 10, 12, leaves a remainder 5 but when divided by 13 leaves no remainder.​

Answers

Answered by Anonymous
1

Step-by-step explanation:

Suppose we are given that a number when divided by x, y, and z, leaves a remainder of a, b, and c; then the number will be of the format of LCM(x,y,z)*n + constant

The key in these questions is finding out the value of 'constant'. If all of them leave the same remainder 'r', constant = r. It can also be looked at as the smallest number satisfying the given property.

So, the number N = LCM(3,5,6,8,10,12)*n + 2 = 120n + 2

Now we have to find out a suitable value of n such that 120n + 2 is divisble by 13.

We know that 117 is divisble by 13.

So, we need to find such a value of n such that 3n + 2 is divisble by 13.

That happens when n = 8

=> N = 120*8 + 2 = 962

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