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:

216 comments Page 9 of 22.

Madhu said: 4 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: 10 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?

Samir said: 1 decade ago

Which is the highest five digit number which is equally divisible by 60, 80, and 90 and how to arrive this answer ?

Shital said: 1 decade ago

@Parthiban.

If there is large numbers then it is reliable method to divide and then finding highest common factor?

RAVI VARMA 501 said: 1 decade ago

Swetha and prashanth explanation is good. And abve some are all in kids formulas. . . So dont look others friends.

Sarath said: 1 decade ago

I think the smarter way will be to divide the numbers with options and check. But your speed has to be so high.

Vimalsankar said: 10 years ago

43/4 the remainder is 3. 91/4 the remainder is 3. 183/4 the remainder is 3. So, finally we get the answer 4.

Sakthivel said: 2 decades ago

How to arrive at this equation p(q2-q1)= (91-43) from 43 = pq1 + r;
91 = pq2 + r;
183 = pq3 + r;

Dhanasekar said: 2 decades ago

Simply try divide to divide each by the options, by this which is exactly divided then that is the answer.

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