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

Ho we know that 1657 - with 6 and how we know that 2037 is minus with 5 please explain.

Here 6 & 5 are remainders of 1657(say A) & 2037(say B) respectively.

To make A and B(dividends) perfectly divisible by a divisor, the remainders should be subtracted from dividends.

Now if we divide the dividend by a divisor, the remainder will be zero, because we made dividends perfectly divisible.

Now the numbers are 1657-6=1651 and 2037-5=2032.

HCF of 1651,2032.

1651)2032(1
1651
-------
381)1651(4
1524
------
127)381(3
381
------
0
-------
So HCF is 127.

Hope you understand.

THANKS krishna your explanation is too easy.

Can we do 1657+6 or 1657-6 both will give same answer. ?

No .it is not possible .because 1657-6 gives the remainder 0,but 1657+6 gives the remainder 1.

It is not possible .
the term 1657+6 is not a perfet multiplier it gives a remainder 1.the term 1657-6 is a perfect multiplier it gives a remainder 0.that's why they doesnot give same answer.

Thank you mr. M. V. Krishna/Palvoncha.

TO FIND THE HCF OF TWO NUMBERS,

DIVIDE THE BIGGER NUMBER BY SMALLER .

TAKE 2032 AND 1651, ON DIVIDING.

1651)2032( 1
1651
--------------
381
-------------
THEN DIVIDE 1651 BY 381.

381)1651(4
1524
--------------
127
--------------
THEN DIVIDE 381 BY 127.

127)381(3
381
----------
0
----------
THE DIVIDING OF BIGGER NUMBER BY SMALLER NUMBER SHOULD CONTINUE TILL WE GET REMAINDER AS ZERO.

THEN THE LAST DIVISOR IS THE H.C.F OF THE TWO GIVEN NUMBERS.

Answer:

13*127 = 1651+6 = 1657.
16*127 = 2032+5 = 2037.

@Thayagarajan. How we know 127 is the correct answer?, that's we need tell clearly.