Sum of the digits of the number (10 4n2+8 +1) 2 , where n is a positive integer is ?

Question

Sourabh Singh · Answer

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

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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 ?}

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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 ?

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Sum of the digits of the number (10⁴ⁿ²⁺⁸+1)^{2 , where n is a positive integer is ?}