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Question is \((10^x)^y - 64\) . Let say \((10^x)^y\) as Number1

Say Number1 - 64 = Number2 ==>

100 - 64 = 36 [ Number1: No. of zeroes = 2 , Number2: No. of 9's = zero ] and Sum of digits of Number 2 : 9*0 + (3+6) = 1*9 = 9

1000 - 64 = 936 [ Number1: No. of zeroes = 3 , Number2: No. of 9's = 1] and Sum of digits of Number 2 : 9*1 + (3+6) = 9 + 9 = 2*9 = 18

10000 - 64 = 9936 [ Number1: No. of zeroes = 4 , Number2: No. of 9's = 2] and Sum of digits of Number 2 : 9*2 + (3+6) = 18 + 9 = 3*9= 27

100000 - 64 = 99936 [ Number1: No. of zeroes = 5 , Number2: No. of 9's = 3] and Sum of digits of Number 2 : 9*3 + (3+6) = 27 + 9 =4*9= 36

so lets go from right to left for the sum of digits of number2 i.e given as 279

so 279 = 31*9 = 9*30 + (3+6) => Number2: Number of 9's = 30 ==> Number1: Number of zeros = 32

So the Number1 i.e. \((10^x)^y = 10000.....(32 zeroes)\)

Now, as we now, \(10^1\) = 10 (1 zero)

\(10^2\) = 100 (2 zeroes)

\(10^3\) = 1000 (3 zeroes)

same way, 10000.....(32 zeroes) = \(10^32\)

\((10^x)^y = 10^(xy) = 10^32\)

==> xy = 32

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