1WT ABT Q AND R?????
If (1+3+5+...+p)+(1+3+5+..+q)=(1+3+5+..+r) where each set of parantheses contains the sum os consecutive odd integers as shown ,the smallest possible value of p+q+r,(where p>6) is ???
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4 Answers
62(1+3+5+...+p)+(1+3+5+..+q)=(1+3+5+..+r)
This is a good question
(p+1)2/2+ (q+1)2/2 = (r+1)2/2
This is same as
(p+1)2+ (q+1)2 = (r+1)2
so this is for the smallest odd numbers p,q,r such that they form a pythagorean triplet!
That will be 6,8, 10
p=7, q=5, r=9
Hence p+q+r = 21
