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

Subtract the smallest number from both the highest numbers.
We have three cases:
183 > 43; 183 > 91 and 91 > 43.
183 - 43 = 140,
183 - 91 = 92 and,
91 - 43 = 48.
Why? Please explain.

By division method:

43)183(4
172
-----
11)43(3
33
------
10)11(1
10
-----
1)10(10
10
------
0
Now use the last divisor with the mid-number 1.
You get again the remainder as a 0,
Hence 1 should be the answer.

Take difference:

43 - 91 = 48.
91 - 183 = 92,

Now take the hcf of difference numbers 49, 92.

The HCF Of 48, 92 = 40.

I am not understanding this. Please explain in detail.

Why need to subtract and generate a new number, then take hcf ? Anyone, please explain to me.

Good explanation. Thanks all.

If any one number is prime, then HCF will be 1.

This is good to gain knowledge. Thanks all.

@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.