LCM & HCF together

I am struck by the large number of questions on Quora that demonstrate an inability to cope with this minor topic. Though it may be that what I see is merely a demonstration of the will to seek answers rather than battle with a problem.

Here, LCM(x,y) is labeled LCM,  HCF(x,y) is labeled HCF,  Diff=|x-y|  Product is x . y or xy



1. Decide whether Diff( x,y)  is itself a factor or multiple  of HCF (x,y).

2. Find the pair of  numbers with LCM = 56 and HCF=4 

3.  Find a pair of numbers both bigger than five such that LCM =7HCF 

4. Find a pair of  numbers both bigger than five such that ( LCM, diff) = (24,5)

5. Find two pairs of relatively prime numbers such that  LCM= 240

6. Find at least two pairs of relatively prime numbers such that  LCM= 420

7. Find at least three pairs of  numbers (x,y)  such that LCM - HCF = x+y

8. Find at least three pairs of  numbers (x,y)  such that LCM - HCF = Diff (x,y)

9.  The difference between LCM and HCF of two natural numbers x,y is 57. What is the minimum value of x+y?

10. Can 6 and 15 be the LCM and HCF of two numbers?

11. Can two numbers have HCF=12 and LCM = 350?

12. The LCM and the HCF of two numbers are 45 and 3 respectively, and their sum is 24. What is their difference?

13. The LCM and the HCF of two numbers are 495 and 5 respectively, and their sum is 100. What is their difference?

14. The sum of the LCM and HCF of two numbers is 2196 and their difference is 2172. If the difference between the numbers is 12, what is the sum of the numbers?


1. HCF (2,8)=2, Diff(2,8) = 6  So HCF is a factor of the difference; the difference is always a multiple of the HCF.

2. HCF.LCM = xy => 56 . 4=224 = 28 . 8   Answer  (28,8)

3. LCM.HCF=7HCF², any pair (n, 7n) will succeed for {n,7} relatively prime, e.g. (6,42), (8,56)

4. (3,8) works.

5. Relatively prime means HCF=1 so because 240 = 2⁴.3.5 and so (3,80) and (5,72) succeed 

6.  420 = 2².3.5.7  so valid pairs are (3,140), (5,84), (7,60)

7.  Multiples of (2,3) succeed.

8. Multiples of (1,N) succeed. So do (n, Nn). Extra mark for either generalisation and another for finding any of the n>1 set.

9. X+Y = 27  Let h be HCF(X,Y), so X=hx, Y=hy, LCM(X,Y)= hxy 
Given that 57=LCM-HCF = hxy-h = h(xy-1)    h=1, => xy=58 => (2,29) =>sum =31
 h=3=> xy=20 =>(X,Y)=(4h,5h)=>(12,15)=>sum=27

10. XY=6 . 15 = 90 = 6x.6y = 36xy =>xy not integer.  LCM/HCF should be integer.

11. XY=12 . 350 = 4200 = 12x.12y fails. LCM and HCF must have common factor of HCF

12. XY = 45.3 = 135 = 3x.3y =>x.y=15 Suggests (X,Y)=(15,9) X+Y=24, X-Y=6. 
Or set X+Y=24 XY=135 and solve by substitution.

13. XY=5x . 5y = 25 . 99 = X(100-X)  suggests (X,Y) = (45,55)  Difference 10.

14.  By elimination, HCF=12, LCM=2184 = 12 . 182 = 12 . (2.7.13)  So XY=12².2.7.13 = 12x.12y and  X-Y=12 so x-y=1 =>(x,y)=(13,14), so X+Y = 12(13+14) = 12.27 = 324

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