If (1+3+5+…+p)+(1+3+5+…+q)=(1+3+5+…+r)where each set of parentheses contains the sum of consecutive odd integers as shown, the smallest possible value of p+q+r, (where p>6)is

If (1+3+5++p)+(1+3+5++q)=(1+3+5++r)

where each set of parentheses contains the sum of 

consecutive odd integers as shown, the smallest possible value of p+q+r, (where p>6)

is

  1. A

    12

  2. B

    21

  3. C

    45

  4. D

    54

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    Solution:

    We know that 1+3+5++(2k1)=k2

    Thus, the given equation can be written as

        p+122+q+122=r+122(p+1)2+(q+1)2=(r+1)2

    Therefore, (p+1, q+1, r+1) forms a Pythagorean triplet. As p>6,p+1>7.

    The first Pythagorean triplet containing a number > 7 is (6, 8, 10).

     We may take

          p+q+r=21.

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