Aptitude - Problems on H.C.F and L.C.M - Discussion

Let x be the greatest possible number such that it leaves the same remainder when it divides 183, 91 or 43.

Since the remainder is the same in each case, the difference of the terms must be exactly divisible by x. Also, x must the greatest possible number that exactly divides the difference between the terms.

Required number, x = HCF of (183 " 91, 91 " 43, 183 " 43) = HCF of (92, 48, 140) = 4.

Best solution, Thanks @Nikita.

Thanks @Komal.

I could finally get it.

Why we taking the difference between these, first explain this.

Hi.

In question, they gave a greatest number, in the sense.

Ex- if you take 4 as a greater num for the given question.

Means this is the greater num which divides all the dividend if you take 5 it won't divide all' the dividends.

If they ask greater num -then take smaller num in the choices and divide all the dividends by that num if you get remainder same then that is the answer.

The answer is 4 and how it is 4 is below,

We can represent any integer number in the form of: D*q + r.
Where D is divisor, q is quotient, r is the remainder.

So each number can be written accordingly:
43 = D*q1 + r1;
91 = D*q2 + r2;
183 = D*q3 + r3;

r1, r2 & r3 will be same in above three equations according to the question.
D is the value that we want to find out. which should be greatest.

On solving three equations we get:

D*(q2-q1)= (91-43)=48
D*(q3-q2)= (183-91)=92
D*(q3-q1)= (183-43)=140
It is obvious that q3>q2>q1.

For the greatest value of D that divide each equation we take the HCF of 48,92,140

THEREFORE ANSWER IS 4.

Good explanation @Prashant.

Easy to understand. Thanks to all the given explanation.

Well done @Saurabh.

I understood the sum now.

Given explanations was good. Thankyou all.