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hcf and lcm

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hcf and lcm

by bblast » Mon Apr 25, 2011 8:50 pm
hcf of 2472,1284 and x is 12.
lcm of these is 2^3 x 3^2 x 5^1 x 103 x 107

a>2^2 x 3 ^2 x 7^1
b>2^2 x 3 ^3 x 103
c>2^2 x 3 ^2 x 5^1
d>none

what is x ?

ans- c

[spoiler]how do we apply lcm x hcf = n1 x n2 here ?[/spoiler]
Last edited by bblast on Tue Apr 26, 2011 5:00 am, edited 1 time in total.
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by Ryandmitri » Mon Apr 25, 2011 9:30 pm
Hey...can you please post the options...
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by bblast » Tue Apr 26, 2011 5:01 am
Ryandmitri wrote:Hey...can you please post the options...
done
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by ankurmit » Wed Apr 27, 2011 2:37 am
--
--------
Ankur mittal
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by sugmomo » Wed Apr 27, 2011 3:43 am
2472 = 2^3 *3 * 103
1284 = 2^2 *3 *107
X =
--------------------------------------
HCF = 2^2 * 3
LCM = 2^3 * 3^2 *5^1 *103 *107

Rule:
HCF contains the common digits among each of the prime factors of 2472, 1284, X
LCM contains the highest powered digits among each of the prime factors of 2472, 1284, X

So, Keeping this in mind.

X = 2^2 *3^2*5. Hence,the answer is C


Reason for choosing 2^2 - HCF rule
Reason for choosing 3^2 - LCM rule
Reason for choosing 5^1 - LCM rule
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by pankajks2010 » Wed Apr 27, 2011 3:59 am
We cannot have a definite value for x. However, considering the options provided and following Sugomo's method the answer is C.

Other possible values for x:
2^2*3^2*5*103
2^2*3^2*5*107
2^2*3^2*5*103*107
2^3*3^2*5*103
2^3*3^2*5*107
2^3*3^2*5*103*107
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by sugmomo » Wed Apr 27, 2011 4:14 am
Agree with pankajks2010...

The above prob. can be confirmed only with the available answer choices.
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by Ryandmitri » Wed Apr 27, 2011 4:51 am
I disagree here....x will not have 103 and 107. As sugmomo mentioned :

2472 = 2^3 *3 * 103
1284 = 2^2 *3 *107
X =
--------------------------------------
HCF = 2^2 * 3
LCM = 2^3 * 3^2 *5^1 *103 *107


LCM will have highest powers of prime numbers in 2472, 1284 and x. LCM has 3^2 but 2472 and 1284 both have 3^1. Therefore x must have 3^2. Similarly 5^1 must be present in x.

So for the time being we have x = 3^2 * 5^1

HCF will have lowest powers of primes numbers in 2472, 1284 and x. HCF does not have 103 and 107 but 2472 has 103 and 1284 has 107...therefore x cannot have either of these numbers.

Therefore till now, x = 3^2 * 5^1 * 103^0 * 107^0 = 3^2 * 5^1

I got tied up cmng to 2.....using the concepts I have used x can have 2^2 or 2^3 but since the given option had 2^2 I went ahead with that...
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by srcc25anu » Wed Apr 27, 2011 10:38 pm
cant we solve this using HCF*LCM = product of numbers
so 12 * 2^3 * 3^2 *5^1 *103 *107 = 2472 * 1284 * x
and x solves out to be 15

AM i missing out something here?
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by pankajks2010 » Wed Apr 27, 2011 11:42 pm
Hi,

The original post was regarding the validity of this method (product of the numbers=HCF*LCM) itself.

While this formula is correct when there are two numbers, however, it is not when the count is greater than that.

For example: Lets consider just 6, 15
The LCM is 30 and the HCF is 3. In this case if we test the above mentioned formula:
6*15=30*3 ----> Very much valid.

However, if we add a third number, say 30 (the numbers now are 6,15 & 30). The LCM and HCF would remain the same 30 & 3 respectively. However, the formula would not hold correct as the count of numbers has gone upto 3 whereas the values of LCM and HCF don't change. While on the LHS of the formula we are multiplying 3 numbers, the RHS remains the same.

Similarly, if we add a fourth number as well, say 3 (the numbers now are 3,6,15 & 30). Again the LCM & HCF would remain the same, while the product of 4 numbers would be significantly larger than the product of the initial two numbers.

Thus, this formula holds only for a count of 2 numbers and not beyond that.
Last edited by pankajks2010 on Thu Apr 28, 2011 9:29 am, edited 1 time in total.
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by srcc25anu » Wed Apr 27, 2011 11:44 pm
thanks pankaj .. for pointing that out.
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by pankajks2010 » Thu Apr 28, 2011 9:42 am
Ryandmitri wrote:I disagree here....x will not have 103 and 107
Hey Ryan,

Lets just check using the following method if x can have both 103 & 107 as its factor or not:

2472 = 2^3 *3 * 103
1284 = 2^2 *3 *107
Let X = 2^2*3^2*5*103*107
--------------------------------------
HCF = 2^2 * 3
LCM = 2^3 * 3^2 *5^1 *103 *107

Well, we see that even for this value of x we get the same LCM & HCF as stated in the question. We would get the same result for all the values of x listed in my post above.

@bblast: Thanks for the question. It has shown up some new concepts.
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by Stendulkar » Fri Apr 29, 2011 5:11 am
Hey Pankaj,

This is 'Ryandmitri'.

You said the following :

2472 = 2^3 *3 * 103
1284 = 2^2 *3 *107
Let X = 2^2*3^2*5*103*107

For HCF you need to take the lowest possible powers of all the factors and for the LCM you need to take the highest possible factors.

For LCM, we will take 2^3 from 2472, 3^2 from X , 5^1 from X, 103^1 from all three and finally 107^1 from all three. Therefore LCM = 2^3*3^2*5^1*103^1*107^1 ( exactly what is given so correct uptil now)

For HCF, we will take lowest possible powers, therefore we take, 2^2 from X and 1284, 3^1 from 1284 and 2472, 103^1 from all three, 107^1 from all three and 5^0 from 1284 and 2472.

HCF = 2^2*5^0*3^1*103^1*107^1 = 2^2*3^1*103^1*107^1. Now if you compare this with what is given in the question then u will see that 103 and 107 are extra. According to what I have explained, X can have a 103 or a 107 but not both. I think the answer to this question is wrong. The answer should be " none of these".
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by Stendulkar » Fri Apr 29, 2011 5:13 am
In my first post I mentioned that X cannot have either 103 or 107. I was wrong there. X can have any one but not both together.
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by pankajks2010 » Sat Apr 30, 2011 4:09 am
Stendulkar wrote:For HCF you need to take the lowest possible powers of all the factors and for the LCM you need to take the highest possible factors.
well HCF is Highest COMMON factor (as textbooks say)...Don't think you are considering COMMON in your definition.

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