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

Why HCF and not LCM?

Lets see,
We have to find the greatest number which on dividing 1657 and 2037 leaves remainder 6 and 5.

Forget about 'greatest'.
we need a number 'which divides 1657' and '2037' ie, if that number is 'x', we have the expression as 1657/x with remainder 6 and 2037/x with remainder 5.

Lets write this in the division form (Dividend = quotient*divisor + remainder)
ie, 1657 = q1 * x + 6.
2037 = q2 * x + 5.

Therefore,
1651 = q1 * x.
2032 = q2 * x.

ie, 'x' (our number) is a common 'FACTOR' of 1651 & 2032 (NOT 'Multiple'). [eg. 24 = 12 *2 or 6*4 or 8*3 ... etc. here, 12 & 2, 6 &4, 8 &3 all are factors of 24 and not multiple].

So, we need common factors of 1651 & 2032 and just select the greatest among them(since the question asks for the greatest).

Division Method: Suppose we have to find the H.C.F. of two given numbers, divide the larger by the smaller one. Now, divide the divisor by the remainder. Repeat the process of dividing the preceding number by the remainder last obtained till zero is obtained as remainder. The last divisor is required H.C.F.

ans...

Required number = H.C.F. of (1657 - 6) and (2037 - 5)

= H.C.F. of 1651 and 2032
let devide the greater no with smaller one means
2030/1651=381
then,1651/381=127
then,381/127=0

so the
The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:127

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.

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.

1st no 1657 - 6 = 1651.
2nd no 2037-5 = 2032 (now, they both get exactly divisible by greatest no.)

the short trick to find HCF: take the difference and see that difference is divisible by these 2 no.(1651 and 2032) or not.

2032 - 1651 = 381.

Factors of 381 = 1,3,127,138.

So, here the ans is 127 which is the greatest divisor of 381.

In these type of questions, just cross-check the answers.

Let the number be "X".

1657/X and 2037/X leaves remainders 6 and 5.
Then, Substitute answers in the place of X and.

1657/127 leaves remainder 6.
2037/127 leaves remainder 5.

Then the answer is 127.

If we divide 7 by 2 it gives a remainder of 1.

Either we add or subtract the remainder 1 from 7 the numbers obtained 8 and 6 which is divisible by 2.

How the answer come in this problem, that we have to add or subtract?

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.

This method of finding HCF of two large no. is called Euclid's division lemma, you can check it further on youtube you will then be able to solve the question.

@Jenifer.

As you see, the question is asking for greatest number which means greatest factor or divisor.

So here we will do HCF/ GCD.