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:
210 comments Page 8 of 21.

Ankush said:   7 years ago
@Mrunmay.

Solution they take difference because to make remainder zero. And than that divisor will divine these number Completely.

Davoas said:   7 years ago
What if we have a fixed remainder.

For example take 90, 114 and 230 with remainder 6 what will be the divisor?

Anyone explain me.

Rajesh said:   1 decade ago
1.91-43 = 48, 183-91 = 92, 183-43 = 140.

2.92-48 = 44, 140-92 = 48, 140-48 = 92.

For every time we eliminating uncommon Factors.

Trishana said:   10 years ago
Here in this question greatest number is 7 and it gives the same remainder that is 1 when divided by all. So how is the answer 4?

Shital said:   1 decade ago
Mysterious please tell me how will be that example converted if divide method and factorization method has given in example?

Dhanasekar said:   1 decade ago
Can any one explain how to identify. Whether the problem is based on HCF or least common multiple. Is there any key words?

Sixface said:   7 years ago
It's very simple guys =>

43/4=3,
92/4=3,
183/4=3.

The remainder is 3 for all divide so 4 is the correct answer.

Madhu said:   3 years ago
Take difference:

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

Now take the hcf of difference numbers 49, 92.

The HCF Of 48, 92 = 40.
(19)

Aditya said:   1 decade ago
Find the least number divisible by 12, 32, 42, and 63 and it must be a perfect square?

Please give me the solution.

Phani said:   9 years ago
Is this the common HCF problem if so when I try to solve it using Euclid's method I got the answer '1'.

Am I right?


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