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

Thanks @Saraswati.

Thanks for explaining @Priya.

@Anchit.

We should subtract remainder. i.e 12-2=10.

Let N be the greatest number that will divide 17, 41 and 77, leaving the same remainder in each case. Then the sum of the digits in N is:

It is easy to understand this with a simple example, and the logic behind this is so obvious.
Take number 12 (you can get any number you desire and practice it!).

Dividend = Divisor X Quotient + remainder (when the remainder is a constant!).

12 X 1 + 5 = 17.
12 X 3 + 5 = 41.
12 X 6 + 5 = 77.

Now, take the differences of 17, 41 and 77 in order considering the smallest, the middle and the largest numbers.

(41-17) = 24; (77-17) = 60; (77-41) =36.

Now take 24, 60, and 36 and find the HCF or HCD of these numbers: it is equal to 12.
Look at the "Divisor" used at the beginning. So, the divisor of that arrangement is equal to the HCF of 24, 60 and 36. When the remainder is constant, we can observe this pattern with any set of numbers arranged in this way.

The sum of digits of N = 1+2 =3.

Thanks @Saraswati.

Good explanation, Thanks @Yogesh.

Thanks all for explaining the answer.

Let consider 1305 as N * some value( say X) + remainder.
Similarly N*Y + r = 4665.
N*Z + r =6905,

For removing or subtract each other we get;
N* (X-Y) , N(Y-z), N*(Z-x).

After doing this the number may be changed. But the required HCF (ie N) remains the same. So we can find N. This is the Simplest method for eliminating the remainder.

Easy to understand. Thanks @Suraj.

Good explanation @Yogesh.