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

Discussion Forum : Problems on H.C.F and L.C.M - General Questions (Q.No. 1)

Problems on H.C.F and L.C.M

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

Answer: Option

Explanation:

Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43)

= H.C.F. of 48, 92 and 140 = 4.

Discussion:

213 comments Page 10 of 22.

RDbaby said: 1 decade ago

@ Merry.

Pratibhan meant, the common thing coming in prime factor list of all 3 nos is 2*2, i.e 4.

Mozammil anwar said: 6 years ago

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

(1)

Hemalatha said: 1 decade ago

I'm not clear with any one of the above explanation, why we take the difference of those numbers.

Vimal kumar said: 1 decade ago

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

Lovely said: 1 decade ago

Hey prashant your explanation is too good. Thanks.
Is there more short method to do solve this.

Hary said: 2 decades ago

Solving the equations:
91=pq2+r
43=pq1+r
_________________
48=pq2-pq1
thus
48=p(q2-q1)=91-43.

Anup said: 4 months ago

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

(10)

Jayanta atkari said: 1 decade ago

(91 - 43 = 48) = 4*12.

(183 - 91 = 92) = 4*23.

(183 - 43 = 140) = 4*35.

Thus HCF is = 4.

Sweety said: 2 decades ago

I don't understand why we need to find the difference. Please explain the logic behind it.

Niveditaa singh said: 1 decade ago

Is this only method to solve this?

If it can be solve by another way then please explain.

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