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Sum of the digits of the number (10 4n2+8 +1) 2 , where n is a positive integer is ? Sum of the digits of the number (104n2+8+1)2 , where n is a positive integer is ?
HiiBe it any number just calculate the total number of zeros that will precede the unit digit in the numbereg 101*101 = 10201 1001 *1001 =1002001 10001*10001 = 100020001Hence in any case the sum of digigts is always going to be 4
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