Digital mathematics
Let a, b and c be the 356th , 357th and 358th digits respectively after decimal places, of the decimal expansion of the fraction ( 98235672325937/99999999999999). Let z be the the biggest perfect square such that z<(100a+10b+c) and y be the smallest perfect square such that y>(100a+10b+c). Let the positive square root of z be k and that of y be l. Find k+l.
The answer is 51.
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1 solution
The decimal expansion of 98235672325937/99999999999999 is 0.98235672325937........... Here, the seventh digit is 7. Thus the next 7 in the seventh place of the second recurring decimal expansion will be at 21st place. Thus, this AP will end in 357. Thus there will be 25 E W terms. Thus, the 357th term will be 7. Thus, b=7. Thus the preceding term a and succeeding term c will be 6 and 2 respectively. Thus 100a+10b+c =672. Thus, z(maximum possible perfect square) < 672 is 625 whose positive square-root, k is 25. And y(the smallest possible perfect square number)>672 is 676, whose positive square-root,l is 26. Thus, k+l=25+26=51.