### Discussion :: Problems on H.C.F and L.C.M - General Questions (Q.No.9)

Anotny Mathew said: (Oct 15, 2010) | |

Why not 2, 6? |

Priyanka said: (Oct 19, 2010) | |

Yes why not 2, 6 even when they will also give 2028 on (13*2, 13*6) ?. |

Sai said: (Dec 3, 2010) | |

Why (2, 6) ? |

Raj said: (Dec 9, 2010) | |

2,6 r not co-primes.see the notes given above. |

Jaz said: (Dec 12, 2010) | |

Explain this. |

Giri said: (Dec 14, 2010) | |

2 and 6 r not co-primes bcoz 6 has 2 as factor whereas for 3 and 4 have no common factors except 1 |

Nikhil said: (Dec 26, 2010) | |

In that case HCF would become 26. |

Sujay said: (Jan 10, 2011) | |

No of pairs are (1, 12) , (3, 4). This is given in solution. But another pair we can consider here (4, 3) ? please explain this ?. |

Puja said: (Jan 20, 2011) | |

A co-prime no. is that in which no common factor other than 1. Like in the above proble factor will be (3,4) bcoz common factor b/w them is only 1. But if we take factor like (2,6) then common factor b/w them will be 1 and 2. So as these are not common factor hence we will not consider this. @Sujay: you cn take (4,3) instead of (3,4)... ans will be same. |

Raji said: (Apr 22, 2011) | |

x * y = 2028 x * 13 = 2028 x = 156 156 = 13 * 12 and here 13 appears twice and so is the ans!!! |

Vaani said: (Apr 28, 2011) | |

Why should we take co-primes? |

Utsavchavda said: (Jun 26, 2011) | |

Meaning of co-prime ? |

Veena said: (Jun 29, 2011) | |

@vaani In the given problem the H.C.F is 13..so if take the pair other than co-prime(i.e., the pair 2,6)....the H.C.F will become 26...but when we take the pairs 1,12;3,4....there is no common factor other than 1.. so that we have to take the pair of co-primes.... |

Aajush said: (Aug 29, 2011) | |

Its not compulsory to get from prime or co-prime. The thing is to get it not exceeding 2 pair. |

Pallavi said: (Nov 10, 2011) | |

Why not (13*6, 13*2) be another pair? |

Pallavi said: (Nov 10, 2011) | |

Got the point veena, sorry to post the earlier comment. Thank ! |

Rahul Kumar Saraswat said: (Jan 12, 2012) | |

Sir we have a formula that the product of two numbers= LCM * HCM. Why this formula is not used in this question ? |

Samaria Vincent said: (Jan 26, 2012) | |

First of all why not 5, 7 because even these numbers are co-prime numbers. Isn't it? |

S Aditya Gautam said: (May 31, 2012) | |

We take either (1,12) or (4,3) as coprimes and not (2,6) becouse of two reasons: 1)we want to get the product 12 as ab=12 only. 2)we choose coprimes (1,12) or (3,4) not non-coprimes(2,6) because if we choose the non-coprimes(2,6) then we get two nos as (26,78) whose H.C.F is 26 not 13 which against the question. On the other hand if coprime numbers are taken like(1,12) then we get the two nos. as (13,156) and for coprime pairs(3,4) we get the numbers (39,52).In both of the cases u may observe that the H.C.F is 13 only. |

Anup said: (Nov 23, 2012) | |

Is there a simpler method to solve this? This was my solution but I don't know why is it wrong. 13 is the H.C.F. So the numbers should have 13 in them. Also as a*b = 12. We can have these combinations. (13,13*2*2*4) (13*2,13*2*3) (13*3,13*2*2) Please correct me if my logic is not appropriate. |

Sathish Kumar said: (Jan 13, 2013) | |

(a*b)product of number = l.c.m * h.c.f 2028 = x * 13 x = 156. Which means l.c.m = 156 h.c.f = 13 Now 156 = 2 x 6 x 13 Factorize the HCF 13 = 1 * 13 The HCF is the common factor for both numbers whereas the LCM contains all the unique factors of both numbers. So, 13 is a common factor for both numbers. The factors remaining are 2 and 6. So, 2 is a factor of one number and 6 is the factor of the second number. The first number = 13 x 2 = 26 The second number = 13 x 6 = 78 Hope it is clear now. Now Factorize the 26 and 78 26 = 2 * 13 78 = 6 * 13 Here 13 is the common value So h.c.f. is 13 as given in question. And 156 is the l.c.m. And finally the number of pair is 1(13 only) Then how answer as two(2). Tell me clearly, how to solve, its my method is correct. |

Grateful Dead said: (Jun 17, 2013) | |

What is co-prime and why would we use it in this question? |

Priya said: (Aug 9, 2013) | |

Factors of: 6 = 1, 2, 3. 2 = 1, 2. They have highest common factor 2. coprime are those pair that have 1 as hcf. So only (1,12) and (4,3) are such pairs. |

Swati said: (Aug 24, 2013) | |

Hey 13a*13b=2028 then a*b=156 if we take 13 common and divide it by 2028. Then how it is written that a*b=12? |

Aish said: (Sep 12, 2013) | |

Why we can not take pair as (1, 12), (4, 3), (3, 4), (12, 1)? I think it should be 4 pairs. |

Chinnu said: (Sep 18, 2013) | |

(3, 4)4 = 1, 2, 4 why it is coprime number 2 number is also in the factor of 4. |

Arvind said: (Dec 27, 2013) | |

There will be three pairs (26,78), (52,39), (156,13). |

Siva said: (Jan 6, 2014) | |

@Arvind, you must not consider (26, 78) because then the HCF will become 26; as given in the question HCF must be only 13. |

Pragya said: (Jun 20, 2014) | |

Why are we considering coprimes? |

Sriram said: (Jul 11, 2014) | |

We have to consider the co primes with 12. |

Santhosh said: (Jul 19, 2014) | |

1 is neither prime nor composite. In the solution it was given a co prime pair. So can it be co-prime pair ? |

Hemanth said: (Aug 30, 2014) | |

Guys you don't need to know what a coprimes for this sum. Just understand that the HCF given is 13. Now if you take (2, 6) as another pair then the two numbers become 13*2 and 13*6. Now the above two numbers have an HCF of 26 (because in 2 and 6 the common factor is 2) which does not satisfy the condition of the two numbers having HCF of 13 given in the question. Hope you understand. :). |

Naveen said: (Dec 30, 2014) | |

Why ab = 12? |

Dhivya said: (Dec 30, 2014) | |

13a*13b = 2028. ab = 2028/(13*13). ab = 2028/169. ab = 12. |

Sri said: (Jan 2, 2015) | |

@Veena. I too had the doubt of 2,6.... Nice clarification given. Never though to check the HCF formed by the numbers (2, 6). |

Shefali said: (Mar 25, 2015) | |

If we divide 156 by 13 then it is 12. So there are 2 pairs. |

Dhanendra Thakre said: (May 24, 2015) | |

Because in this problem contain highest common factor 13 if you take 2 and 6 then problem condition is not satisfied. Because 2and 6 contain 2 is become highest factor. 13*6 13*2 have HCF 2 not 13. And 13*1 13*12 have HCF 13. |

Dinesh said: (Jul 12, 2015) | |

Why to take the pairs of 12? |

Kanika said: (Jul 28, 2015) | |

When we assume 13a AND 13b as numbers please explain. |

Donald said: (Jul 30, 2015) | |

It's hard to understand not satisfied. |

Yash Srivastava said: (Aug 14, 2015) | |

HCF of 26 and 78 is 2 not 13. |

Deepika said: (Aug 16, 2015) | |

Can anyone explain in a simple way? |

Naheem Sheikh said: (Aug 23, 2015) | |

x*y = 2028. x*13 = 2028. x = 156. Factor = 13*12. Now, please explain? Anyone in details? |

Deepak said: (Aug 26, 2015) | |

4 is definitely not prime number, then why you have taken as co prime? Only one pair can be done i.e. 13, 1. |

Himank said: (Sep 13, 2015) | |

@Naheem. According to the formula. Product = HCF*LCM. Product is given 2028. And HCF is 13. So we get LCM = 2028/13 = 156. Now factors of 156 is calculated which comes out to be 2*2*3*13. |

Priya said: (Sep 22, 2015) | |

I am not getting the concept of ab=12? How it came an some one please explain? |

Sai said: (Dec 20, 2015) | |

13a*13b = 2028. 169ab = 2028. ab = 12. |

Subbu said: (Jan 13, 2016) | |

I am not getting the concept of ab=12. |

Xing said: (Jan 23, 2016) | |

Longer method, divide both side by 13. => 13a * 13b/13 = 2028/13. => 13a * b = 156. => ab = 156/13. => ab = 12. |

Vinay Ambidi said: (Jan 29, 2016) | |

No someone asked why 5, 7 but if we multiply those numbers we will get 5*7 = 35. That's why we have to take 1, 12 or 3, 4. |

Sami said: (May 10, 2016) | |

How to choose the pairs? |

Gogul said: (May 13, 2016) | |

I think the question have its answer. The product of two numbers is 2028. The Number of such pairs is 2. Whatever it is, only 2 pairs, because we have only a and b, if we have a, b, c, d then we may have more pairs. 2 x 6 3 x 4 4 x 3 1 x 12 12 x 1 6 x 2 Correct me if I'm wrong. |

Pradeep said: (May 13, 2016) | |

Why they take the two pair like (1, 12), (3, 4)? Why not (12, 1), (4, 3)? |

Snehesh said: (Jun 15, 2016) | |

I followed this method: 13a * 13b = 2028. 13ab = 2028. ab = 156. Now try to find the unique pairs (Prime factorize) forming 156: 2 * 78, 3 * 52. So there are just 2 pairs. Hence, ans is 2. |

Shobs said: (Aug 5, 2016) | |

Only 2 pairs is correct. Know the meaning of co-prime first "two numbers are said to be coprime if their H.C.F is 1 "which suits only the (1, 12) and (3, 4) pairs. |

Sai Ravi Teja Vallabhuni said: (Sep 25, 2016) | |

Guys, 2 and 6 are not co-primes, that means 2 is divided only by 1 and 6 is divided by both 2 and 3. Co primes are the numbers where they should be divided only by one, in this case, they are 3 and 4. |

Rbjii said: (Dec 4, 2016) | |

Well, many of you got confused with Prime, co-Prime, why not (2,6) and all.....!!! Lets we will Understood what's the question is ...!!! The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is: step 1 : A * B = 2028 (Let A & B makes 2028 on multiply) step 2 : (x*13) * (y*13) = 2028 (as mentioned HCF=13 for both A & B, so I took it as A=x*13, B=y*13) ******Note : HCF in the sense, only one number should be common in both A & B, hence it is 13 none other should be...!! ( are you cleared ??)***** step 3 : x*y=2028/13*13 step 4 : x * y = 12 ( Product of X and Y makes 12, so we will separate 12 for and x & y) *** The Sub-multiples of 12 are (1,12) (2,6) (3,4) .....still now ok...!! (Once again remember the NOTE..plzz)**** case 1 : (1,12) A = 1 * 13 B = 12 * 13 (Product of A & B makes 2028, fine) also only 13 is common in both A & B. Hence it is satisfied case 2 : (3,4) A = 13 * 3 B = 13 * 4 ( Here also only 13 is common in both A & B. Hence this also satisfied ) case 3 : (2,6) A = 13 * 2 B = 13 * 6(also 39 * 2 ) See clearly in both A and B the common numbers are 13 as well 2. It shouldn't be according to our NOTE Hence this is not a correct pair with HCF as 13 alone. So it is said to not co-prime JUST FOR UNDERSTANDING BIG DESCRIPTION..!! Paper Pen practice works faster ..!! ATB |

Yashraj said: (Feb 11, 2017) | |

I can't understand this. Please help me to get it. |

Jatin007 said: (Feb 14, 2017) | |

Another way I'm telling you without co-primes. 2028/13=156, again divide 156/13 = 12. Now factorize 12 = 1, 12, 3, 4. There are 4 numbers so the pair is 2. |

Klark said: (Feb 28, 2017) | |

Why we can't use the pair (4, 3) & (12, 1). We are asked to find the total number of such pairs. So (1, 12), (3, 4), (4, 3), (12, 1)? |

Garima said: (Apr 28, 2017) | |

Can we take a pair of 13, 1? |

Anik said: (May 12, 2017) | |

Why not 2, 6? Please explain in clear. |

Sudip Choudhury said: (Jun 25, 2017) | |

LCM=2028/13=156 which means 13x12 or 12*13 so there are 2 ways. |

Disha said: (Jul 22, 2017) | |

Not getting this. Can anyone explain it to me? |

Balachandra Padiyar said: (Jul 28, 2017) | |

The co prime numbers those numbers which are having only one common factor that is 1 in (2&6) the factors are, 2 = 1,2. 6 = 1,2,3. Here in 2&6 there are two factors ie, 1&2 hence they are not co primes hence it can't be used here. Let us take 4&3, 4=1,2,4, 3=1,3, Here only one common factor ie,1 and same for 12&1. |

Shivangi said: (Aug 22, 2017) | |

But Question is how many no of the pair. Not co factors. Please clear my doubt is no of pair means co-factor? |

Kishan said: (Sep 16, 2017) | |

The product of two numbers is 2028 and their HCF is 13 If the numbers of two digits find the numbers. Can anyone answer me? |

Nagendra said: (Oct 16, 2017) | |

In the question, they doesn't mention about the co-prime numbers, right? |

Vishal said: (Dec 27, 2017) | |

4 is not a co-prime because it has a factor 1*4 and 2*2 also. |

Shubham said: (Jan 21, 2018) | |

ab=12, now possible solutions can be 1*12, 3*4, 2*6 but we won't consider 2*6 because it will lead to HCF 13*2, as 2 is a factor of both 2 & 6. Hope you get it. |

Mahesh said: (Jun 18, 2018) | |

If we take 2 and 6 then we get (13*2)*(13*6)=26*78. HCF of 26 and 78 will be 26 it's against the given question. When we took 3 and 4 we get HCF as 13 only. That's the reason why we shouldn't take 2 and 6. |

Gokul said: (Aug 3, 2018) | |

xy=2028/13=156. Then how to solve it? |

Abhishek S G said: (Aug 14, 2018) | |

We should not take 2 and 6 because if we take 2 and 6, the product will be 2028 but HCF is 26 not 13. |

Jagrati said: (Aug 21, 2018) | |

Product of 2 numbers = HCF * LCM. 2028= 13 * LCM. 2028/13=LCM. Since, LCM=156. Factors of 156 are 2*2*39. That is the pair is 2. |

Zoha said: (Oct 9, 2018) | |

The product of two numbers is 2028 and their HCF is 13 find their LCM. Ans: so, there LCM is 2 is that correct. |

Shahrukh said: (Feb 18, 2019) | |

If 13x * 13y = 2028 then; It should have been 13(x*y)= 2028 And x*y= 156 But your solution is telling that 13*13(x*y)= 2028. x*y = 2028/ 169 = 12. So , x*y = 12. I mean why divided by 169 why not by 13 after taking common 13? Please tell me. |

Julie said: (Mar 11, 2019) | |

Thank you for giving the clear answer @Snehesh. |

Randy Ranjith said: (May 19, 2019) | |

Then why we take (4, 3) 4 is not co-prime? |

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