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Quantz- Plz Solve.
by PRASHANT AGARWAL - Sunday, 13 July 2008, 08:11 PM
 

If the sum of two natural numbers and their LCM is 89, then how many such pairs of numbers are possible?

a> 7

b> 6

c> 5

d> 4

e> less than 4

 

Please let me know the approach also , that you would follow to crack this one.

Re: Quantz- Plz Solve.
by CATendra 2008 - Sunday, 13 July 2008, 08:27 PM
  take p to be the hcf of both the natural numbers.
the numbers thus are px,py whr x and y are coprime

lcm = px * py/HCF  = px * py /p = pxy

px + py + pxy = 89
p(x + y + xy) = 89

both the factors on the LHS are integers
take  p= 1 and x + y + xy = 89
(x+1)(y+1) = 90

Factorise RHS and get the possible values of x and y
e.g (2,44),(5,14)....etc
Re: Quantz- Plz Solve.
by TG Team - Sunday, 13 July 2008, 08:38 PM
  Let the two numbers be x and y
such that HCF(x,y) = h
=> x = ha, y = hb and LCM(x,y) = hab
=> h(a + b + ab) = 89
=> h = 1
and a + b + ab = 89
=> a = (89 - b)/(1 + b) = 90/(1 + b) - 1.
Number of positive divisors of 90 (i.e. 2*5*32) = 2*2*3 = 12.
So, 1 + b = 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90
=> (a,b) = (89,0), (44,1), (29,2), (17,4), (14,5), (9,8), (8,9), (5,14), (4,17), (2,29), (1,44) or (0,89)
Now a and b are non zero b'coz they are natural numbers.
Also HCF(a,b) = 1.
So final values for (a,b) = (1,44) , (2,29) , (4,17) , (5,14) , (8,9) i.e. 5 pairs. smile
Re: Quantz- Plz Solve.
by PRASHANT AGARWAL - Sunday, 13 July 2008, 08:47 PM
 

Yeah , thats how they quoted the solution in answer key also, but then i dont understand the basic fact that would 89 be the LCM of any of the above pairs ?????  Thts wat question demands. right ?

Or am i contemplating the question wrong ??

Re: Quantz- Plz Solve.
by ATOM ANT - Sunday, 13 July 2008, 11:11 PM
  Prashant,

The lcm is not given as 89.Only the sum of the 2 nos and their lcm is equal to 89.

eg: 44 + 1 + (LCM) 44 = 89
      8 + 9 + (LCM) 72 = 89

Got it?
Re: Quantz- Plz Solve.
by PRASHANT AGARWAL - Sunday, 13 July 2008, 11:38 PM
  Ohh yeah ok ! Thx dude.
Re: Quantz- Plz Solve.
by Vaibhav Panchal - Saturday, 18 June 2016, 01:03 AM
  Case 1 : a,b are co prime HCF = 1
a + b + ab = 89
a + b + ab + 1 = 90
(a + 1)(b + 1) = 90 = 2 * 3^2 * 5
No of factors = 12
Solutions 12/2 - 1 = 5; a cannot be 0
Case 2 : HCF>1 Let it be h
h(a + b + ab/h^2) = 89
h cannot be 1 and 89 is prime.
No solutions