Grade 10Differential Calculus
Sum of the digits of the number (104n2+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 ?
1 Answer
Sourabh Singh
Hii
Be it any number just calculate the total number of zeros that will precede the unit digit in the number
eg 101*101 = 10201
1001 *1001 =1002001
10001*10001 = 100020001
Hence in any case the sum of digigts is always going to be 4
Be it any number just calculate the total number of zeros that will precede the unit digit in the number
eg 101*101 = 10201
1001 *1001 =1002001
10001*10001 = 100020001
Hence in any case the sum of digigts is always going to be 4
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