The product of two numbers is '2028' and their H.C.F.is '13'. The number of such pairs is:

A.1
B.2
C.3
D.4

• ans=2
  let the numbers be 13x and 13y
  13x*13y=2028
  xy=12
  so the co-primes with their product 12 are(1,12)and(3,4)
  so the required numbers are(13*1,13*12)and(13*3,13*4)
  so, there are 2 such pairs.
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• ans: C
  78*26
  39*52
  13*156
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• l.c.m*h.c.f=product
  l.c.m=2028/13=156
  2028=156*13&52*39,so only two pairs are possible.ans=2(b)
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• 13x*13y=2028=>xy=12
  3 possibilities:
  1. 1,12
  2. 3,4
  3. 2,6
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