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

Hi @Ajeeth, very simple for the H.C.F.

For the numbers, you have given (455, 522 & 533).

H.C.F=1, Because the factor of these numbers are 455 = 1 x 5 x 7 x 13,
522 = 1 x 2 x 261 &
533 = 1 x 533.

So except 1, no digit is common in each number. Hence the common digit 1 is the H.C.F of above numbers.
This is the common rule of finding H.C.F. That those digit/digits which will be common in all the numbers.

Eg. You have to find the H.C.F. of 24, 60 & 81. Now factors of each number will be 24 = 2 x 2 x 2 x 3, 60 = 2 x 2 x 3 x 5, & 81 = 3 x 3 x 3 x 3.

Here only the digit '3' is common in all, hence 3 will be H.C.F of these numbers. Also note that if more than 1 digit is common in all the numbers then multiplication of those digits will be the H.C.F (suppose 2 & 5 both are common in all the numbers then the multiplication of both the digit will be 10) hence 10 will be the H.C.F of those numbers.

Thanks.

I don't understand I'm only eleven, please someone help me to get this.

I'm confused by this one, Please explain with division method.

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.

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?

I'm not understanding the logic. Please someone explain me.

I am not understanding please explain me.

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

183 = (183-91) = 92.
91 = (183-43) = 140.
43 = (91-43) = 48.
92, 140, 48.

92 = 2 * 2 * 23.
140 = 2 * 2 * 35.
48 = 2 * 2 * 2 * 3 * 2.
Ans 4.

@Prashant.

In your explanation, 'p' must be a Divisor, not the Dividend.

Here, the remainder is unknown that is why we have only chance to find their difference.