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

@All.

Here you need the highest common factor of three numbers those leave the same remainder

Let's do with the formula.

(p)q1 + r = 43.
(p)q2 + r = 91.
(p)q3 + r = 183.
we have used the same 'r' because all three leave the same remainder.
we need to find the 'p' value, which is the highest common factor.
to find the 'p' value, we need to remove 'r' first, ---------> (important point).
it seems that we can do it by making a difference between two equations.

Let's do it.

=> (pq2 +r) - (pq1 +r)
=> pq2+r-pq1-r
=> pq2-pq1 ----> now the reminder is taken out.
=> p(q2-q1) -----> need to find p value ----> (91-43)----> 1

Other numbers after diff will be like;

=> p(q3-q2) --------->(183-91)---------------> 2.
=> p(q3-q1) -------->(183-43) ---------------> 3.

Now we can find HCF of these three numbers to find p-value. Because 'p' is common in all numbers.

Hope you got it.

Good explanation, Thank you all.

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.

Thanks everyone for explaianing.

let the greatest number is H.

Now;

H*Quotient(Q1)+Reminder(R)= 43 -------> (1)
H*Quotient(Q2)+Reminder(R)= 91 -------> (2)
H*Quotient(Q3)+Reminder(R)= 183 -------> (3)
Now, (1)-(2), (3)-(2), (3)-(1)
H*(Q2-Q1)= 48,
H*(Q3-Q2)= 92,
H*(Q3-Q1)= 140.

So 48, 92 and 140 are completely divisible by H, that will be the HCF of these numbers.
So, the HCF of 48, 92 and 140 is 4.

Thanks everyone for explaining the answer.

Thanks everyone for explaining this.

Why do we do the difference of number to each other?

Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.

Explanation: 43, 91, 183.
183 = (183-91) = 92.
91 = (183-43) = 140.
43 = (91-43) = 48.

92, 140, 48.

92 = 2*2*23.
140 = 2*2*35.
48 = 2*2*2*3*2.

Ans 4.

Is there anyone who will clearly explain why one has to take the difference of those numbers?

I mean what is the key note in the question seeing which I will solve the problem in this way?

Please explain someone.