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Challenge Question # 21

5 ,12,13 and 11,60,61 are triplets

Now in both triplets one of the element is divisible by 3 i.e. 12 and 60

In first case 5^2 + 12^2 = 13^2

The element which is divisible by nine always comes out to be 9 after digit sum so we can cast out 9.

L.H.S 25 + 3^2 =7+9 =16 =7 ( by digit sum)

In L.H.S. if cast out 9 then also the digit sum comes out to be 7

R.H.S. 13^2= 4^2 = 16 = 7

In triplet 11,60,61

60 is divisible by 3 , the square of 60 is comes out to be 9 after digit sum ( 60^2= 6^2= 36= 9)

So we can only take the digit sum of 11 in L.H.S.

11^2 =1+1 ^2= 2^2=4

R.H.S. 61^2 = 7^2=49=4+9=13=1+3=4

L.H.S. =R.H.S.

So in triplet we can skip the element which is divisible by 3 and only take the digit sum of other element on L.H.S.