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Perhaps there are multiple solutions.  The solution i have is 720, 880 and 801.
       720 + 880 + 801 = 2401,      720 + 880 = 40²,   880 + 801 = 41²
          and  720 + 801 = 39²

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Let the three natural numbers be  M , N , P.  Let there be some natural numbers x, y, and z such that the following is true.

          M + N + P = 2401 = 49²      -- equation 1
          M + N = x²                    -- eq 2
          N + P = y²                            -- eq 3
          P + M = z²                              -- eq 4

     Since the above sums are less than 2401, x, y, and z are less than 50 and around 40.

     Sum these above three equations and/ or subtract to get the following :        

          x² + y² + z² = 2 * (M+N+P) = 4802    --  eq 5
            x² + y² - z²  =  2 N          --- equations  6
           y² + z² - x² = 2 P
           z² + x² - y² = 2 M                -- eq 8

            x² + y² = 2401 + N = 49² + N    -- eq  9
            y² + z² = 2401 + P =  49² + P
           z² + x²  = 2401 + M = 49² + M  --  eq 11

Hence, we get
             N = 49² - z² = (49 - z) (49 + z)    -- equation 12
              P = 49² - x² = (49 - x)(49 + x)    -- eq 13
              M = 49² - y²  = (49 - y) (49 + y)    -- eq 14

From the above, we know that ,  M, P and N are not prime. They have factors.

Further,  we get  the following equations  15 to eq 17:
              (x y z)² = (49² - M) (49² M + NP) = (49² - N)(49² N + MP)
                          = (49² - P) (49² P + MN)

We get equations  18 to 20 as
           y² z² = 49² P + M N ,       x² y² = 49² N + M P,
           x² z² = 49² M + P N

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 Now we solve the given problem as below.   From equation 5, we get that x, y and z are all even or two of them odd and one of them is even.  Then only the sum on RHS will be even.  On RHS, we have 4802, which is not multiple of 4.  Hence, x, y, z are not all even.  Two of them are odd. One is even.

   M, N, P are not equal among themselves, as per equations 2, 3, and 4, we get that if they are equal  2M = x^2.  it is not possible.   =========================================

Let us take      x² + y² + z² =  4802    -- equation 5

Let us look at the units digits of x, y  and z that can make units digit 2 on RHS. 

A square number has units digit as 0, 1, 4, 5, 6, 9 only.   Three possible units digits of x², y², and z² which will give 2 as units digit in the sum are :
          ( 4, 9, 9 ),  (0, 1, 1), (1, 5, 6) and (0, 6, 6)

  Here (0, 6, 6) have all even. So we dont consider that.

  The possible units digits of x, y, and z are the following.  As the problem is symmetric about, M, N and P and also about x, y, and z,  we only look at the combinations of values.   

    Possible units digits a, b, c  of  x, y, z respectively are listed as:

       (2, 3, 3),  (2, 7, 3), (2, 7, 7), (8, 3, 3), (8, 7, 3), (8, 7, 7), (0, 1, 1),
       (0, 1, 9), (0, 9, 9),(1, 5, 6), (1, 5, 4), (9, 5, 4), (9, 5, 6)

   The equation  5, can be rewritten as

      (10 x' + a)² + (10 y' + b)² + (10 z' + c)² = 4802

  Here x', y', and z' are between 0 and 4 only. as x, y, z are < 50.

     100 (x'² +y'² + z'²) + 20 ( a x' + b y' + c z') = 4802 - (a²+b²+c²)
 
        5 (x'² + y'² + z'²) + (a x' + b y' + c z') = 240 + [ 2 - a² - b² - c² ] / 20  --  equation 23

   Hence  a² + b² + c² - 2 must be a multiple of 20. 

We check the above combinations and see what matches this criteria.

I find that  (0, 1, 9) matches.    a =0, b = 1, c = 9.

    then  5 (x'² + y'² + z'² ) + ( y' + 9 z' ) = 244
                       so  y' + 9 z' = 4 (mod 5)    ie., 4, 9, 14, 19, 24, 29 etc.

   I choose,    x' = 4    y' = 4,    z = 3    , that works.

          So x = 10 x' + a =  40,
                  y = 10 y' + b = 41    and
                             z = 10 z' + c = 39

Hence,  M = 720,  N = 880,    P = 801    using equations  12, 13 and 14.

2 5 2
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