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(w h e r ep >6)` is `12` b. `21` c. `45` d. `54`

Options

A. 12
B. 21
C. 45
D. 54

Answer

B

Explanation

We know that 1+3+5+….+(2k-1)=`k^(2)`. Thus, the given equation can be written as
`((p+1)/2)^(2)+((q+1)/2)^(2)=((r+1)/2)^(2)`
`rArr(p+1)^(2)+(q+1)^(2)=(r+1)^(2)`
As `pgt6,p+1gt7`, we may take p+1=8,q+1=6,
`r+1=10`. Hence,
p+q+r=21