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.

48 = 4 * 12
92 = 4 * 23
140 = 4 * 35.

1,2,3,5 prime numbers so 4 is the correct answer.

Do we know the number no let's assume the number is X therefore 43/X = 4 ithr for 43-4= 40 and similar to the all and by dividing we get reminder as 4.

Thanks for the explanation.

4/43 = remainder is 3.
4/183 =remainder is 3.
4/91 = remainder is 3.
So that' why the answer is 4.

Why we have to use hcf?

@All.

They asked to divide by a number also we need to have the same remainder in all cases.

So, dividing by 4, we get remainder 3 in all cases.

43/4 = 3(remainder).
91/4 = 3(remainder).
183/4 = 3(remainder).

Very good explanation, Thanks @Kuldeep Sharma.

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.