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

1st divide 48 with 4 then remainder is 0.

2nd divide 92 with 4 then remainder is 0.

3rd divide 140 with 4 then remainder is 0.

Hence so 4 is a perfect number.

Assume that the required answer is "D" and the remainder that it leaves when dividing 43, 91 and 183.

This means that,

43 = d*q1 + r,

91 = d*q2 + r,

183 = d*q3 + r,

Where q1, q2 and q3 are the respective quotients.

Subtracting the first equation from second, we get:

91 - 43 = (d*q2 + r) - (d*q1 + r).

Hence, 48 = d*(q2 - q1) = d*(an integer).

Similarly, subtracting second equation from first, we get:

183 - 91 = (d*q3 + r) - (d*q2 + r).

Hence, 92 = d*(q3 - q2) = d*(another integer).

Now, 48 = d*(an integer) and 92 = d*(another integer) mean that both 48 and 92 are divisible by d. (In other words, "d" is a common divisor of 48 and 92. )

Now common divisors of 48 and 92 are 1, 2 and 4 only. Out of which, highest is 4. (Alternatively, you can directly find the HCF (or GCD) of 48 and 92). Hence 4 is the answer.

Note:

There is no need to calculate the HCF of 3 numbers, namely (91 - 43), (183 - 91) and (183 - 43). It is sufficient to take any two of these three numbers. The HCF will still be 4.

Yes your right @Sowjanya. But @Prasanth is almost right I think you need more information.

In @Prasanth answer, p is the divisor not dividend and the analysis of sum is correct.

Here in this question greatest number is 7 and it gives the same remainder that is 1 when divided by all. So how is the answer 4?

If the sum of two numbers are 348 and their HCF is 48 then what is the difference of the numbers?

Please tell me the most simple formula.

I want the explanation of the logic.

Is there any quick and efficient method to find HCF and LCM of big numbers such as HCF of 38094, 43762, 84632 for competitive exam if yes then please help me out waiting.

Just divide each option with all figures it is that simple.