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✏️ perfect squares always end with digits 1,4,5,6,9 6 is succeeded by an odd number 5 is succeeded by an even number. And always found two zeroes together at the end ( if we have zero at the end) ✏️ If NET14 and NET15 are 5d numbers st their sum is 175229 then N+E+T = 7+8+6 = 21 ✏️If N,E,T are distinct positive integers such that N*E*T = 2013 Then maximum possible sum N+E+T = 671+3+1 = 675..
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In option (B)-2 in units digit in perfect squares 2 won't come in units digit (C)-units digit 6 before number must odd D-8 won't come as unit digit in perfect squares so answer is (A) Question (8):if we add the sum 3(100A+10B+C)=100C+10C+C (D) only satisfy this equation So answer (D) Thank you very much for your explanation sir iam following your videos regularly Thank you sir 👌🙏
1(a) , Reason : as last digit of any perfect square number is 0,1,4,9,6,5 only so (b, d) are neglected and as you told any perfect square having last digit 6 must have previous odd number so (c) discarded and (a) is correct. 2(d) by checking all options.