If `(1+3+5++p)+(1+3+5++q)=(1+3+5++r)`where each set of parentheses con – InterviewSolution

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1.	If `(1+3+5++p)+(1+3+5++q)=(1+3+5++r)`where each set of parentheses contains the sum of consecutive oddintegers as shown, the smallest possible value of `p+q+r(w h e r ep >6)`is`12`b. `21`c. `45`d. `54`A. 12B. 21C. 45D. 54
Answer» Correct Answer - B 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

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