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)
1.
Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
4
7
9
13
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:
216 comments Page 4 of 22.

Shreekanth said:   8 years ago
Keep on dividing all the three numbers by 2 so that reminder should be the same ie
2[43 91 183]
[ 21 45 91] remainder will be 1 for all the three.

Divide again by 2.
2[21 45 91]
10 22 45 remainders will be same ie 1.

So after this, if you try to divide all three again by 2 then you get a different value for the reminder.
And hence; 2 * 2 = 4.
(1)

Bala said:   7 years ago
@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.
(1)

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)

Vamsi said:   6 years ago
48 = 4 * 12
92 = 4 * 23
140 = 4 * 35.

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

Anomie said:   5 years ago
P is called the divisor & not dividend.

(Divisor x quotient) + remainder= dividend.
(1)

Vishakha said:   2 months ago
Thanks everyone for explaining this.
(1)

Nagu said:   2 decades ago
Why they took difference between those nos.

Please explain me.

Anu said:   2 decades ago
Explain the logic of this solution.

Sourabh Das said:   2 decades ago
I also want an explanation about the logic of the solution.

Anirudh rai said:   2 decades ago
Is this the only way to this type of problem?

If yes then would you please explain it.


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