Question

p,q and r are natural numbers such that p:q:r=2:3:6 sum of their squares is 3a6bc where a,b,c are unknown digits now consider following 1)number of factors of 3a6bc is 9 2)the three digit number 6bc divides the five digit number 3a6bc is 1&2 are true or false or 1 true

Answer 1

p, q, r are positive integer numbers.  a, b, c are single digit integers from 0 to 9.

  p : q : r = 2 : 3 : 6     Let the constant of proportionality be x (rational number).
  p = 2x    q = 3x     r = 6x
  p² + q² + r² = 3a6bc 
  4x²+9x²+36x² = 30,000+a*1,000+600+b*10+c
  49 x² = 30,600+1000*a+10*b+c    or    3a6bc

  Then factors of LHS are : 
         1, 7, 49, x, 7x, 49x, x², 7 x², 49x²
   there are 9 factors.
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   6bc  divides  3a6bc ??
   3a6bc is in the range:    39,699 ---  30,600

   3a6bc  =  3a * 1000 + 6bc
   3a6bc / 6bc =  3a*1000/6bc  + 1
  
   49 * 25 * 25  = 30625    that is :  a = 0, b = 2  and c = 5.
         so 6bc divides  3a6bc  for  the above numbers.
         perhaps more numbers too..
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The solution:
   the range of  3a6bc  is 39699   to   30600.
   so range of x²  is    from  39699/49  = 810.1  to   624.48
   so range of  x  is  from:    28 to 25.

   so  49 * 625 = 30625        =======>  this is right.
         49 * 26²  = 33124
         49 * 27²  = 35721
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Deductive way of solving this problem.

N =  49 * x²  = 3a6bc
   if 6bc is a factor of  3a6bc, then 6bc is a multiple of 7, 7², x².
      625 = 25²  is a perfect square.   trying it works.