12345678987654321

Number Theory Level 3

N = 11111 ⋯ 11111 ⏟ Number of 1s = 2018 \large N=\underbrace{11111 \cdots 11111}_{\text{Number of 1s }= 2018}

What is the digit sum of N 2 N^2 ?


The answer is 18148.

X X by X X , 17, Taiwan

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1 solution

Alex Greist
Aug 6, 2018

Lets first try to develop a pattern:

  • 1 2 1^{2} = 1 1 | 1 1 = 1 2 1^{2}

  • 1 1 2 11^{2} = 121 121 | 1 1 + 2 2 + 1 1 = 2 2 2^{2}

  • 11 1 2 111^{2} = 12321 12321 | 1 1 + 2 2 + 3 3 + 2 2 + 1 1 = 3 2 3^{2}

  • 111 1 2 1111^{2} = 1234321 1234321 | 1 1 + 2 2 + 3 3 + 4 4 + 3 3 + 2 2 + 1 1 = 4 2 4^{2}

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  • 111111111 1 2 1111111111^{2} = 12345678987654321 12345678987654321 | 1 1 + 2 2 + 3 3 + 4 4 + 5 5 + 6 6 + 7 7 + 8 8 + 9 9 + 8 8 + 7 7 + 6 6 + 5 5 + 4 4 + 3 3 + 2 2 + 1 1 = 9 2 9^{2}

  • 1111111111 1 2 11111111111^{2} = 12345678900987654321 12345678900987654321 | 1 1 + 2 2 + 3 3 + 4 4 + 5 5 + 6 6 + 7 7 + 8 8 + 9 9 + 0 0 + 0 0 + 9 9 + 8 8 + 7 7 + 6 6 + 5 5 + 4 4 + 3 3 + 2 2 + 1 1 = 1 2 1^{2} + 9 2 9^{2}

The pattern that appears is every time 9 9 ones are added, the sum of its digits increase by 9 2 9^{2} (after the number is squared). While every individual 1 1 added, increases the value being squared by 1 1 (after the number is squared). Using this, we can divide the number of ones N N has, by 9 9 . This leaves us with: 2018 2018 = ( 224 224 x 9 9 ) + 2 2 . So the final equation will have to include the 224 224 sets of 9 2 9^{2} and also 2 2 2^{2} for the remainder of 2 2 .

  • This will be the final equation: ( 224 224 x 9 2 9^{2} ) + 2 2 2^{2} = 18148 \boxed{18148}
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