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Number Theory Level 3

111 … 111 ⏟ Number of 1s = 4034 − 222 … 222 ⏟ Number of 2s = 2017 \sqrt{\underbrace{111\dots111}_{\text{Number of 1s }= \ 4034}-\underbrace{222\dots222}_{\text{Number of 2s }= \ 2017 }} Find the sum of the digits of the number above.


The answer is 6051.

Áron Bán-Szabó by Áron Bán-Szabó , 17, Hungary

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4 solutions

Uros Stojkovic
Jul 5, 2017

A = 111...111 ⏟ 4034 − 222...222 ⏟ 2017 = 111...111 ⏟ 4034 − 111...111 ⏟ 2017 − 111...111 ⏟ 2017 = 111...111 ⏟ 2017 000...000 ⏟ 2017 − 111...111 ⏟ 2017 = 111...111 ⏟ 2017 × ( 1 000...000 ⏟ 2017 − 1 ) = 111...111 ⏟ 2017 × 999...999 ⏟ 2017 = 111...111 ⏟ 2017 × 111...111 ⏟ 2017 × 9 = 111...111 ⏟ 2017 × 3 = 333...333 ⏟ 2017 \begin{aligned} A & = \sqrt{\underbrace{111...111}_{4034}-\underbrace{222...222}_{2017}}=\sqrt{\underbrace{111...111}_{4034}-\underbrace{111...111}_{2017}-\underbrace{111...111}_{2017}} \\ &= \sqrt{\underbrace{111...111}_{2017}\underbrace{000...000}_{2017}-\underbrace{111...111}_{2017}} \\ & =\sqrt{\underbrace{111...111}_{2017}\times (1\underbrace{000...000}_{2017}-1)} \\ & =\sqrt{\underbrace{111...111}_{2017}\times \underbrace{999...999}_{2017}} \\ & = \sqrt{\underbrace{111...111}_{2017}\times \underbrace{111...111}_{2017}\times 9} \\ & = \underbrace{111...111}_{2017}\times 3 \\ & =\underbrace{333...333}_{2017} \end{aligned}

Hence, the sum of digits is 3 × 2017 = 6051 3\times 2017=\boxed{6051} .

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N = 111...111 ⏟ # of 1s = 4034 − 222...222 ⏟ # of 2s = 2017 = 1 0 4034 − 1 9 − 2 ( 1 0 2017 − 1 ) 9 = ( 1 0 2017 − 1 ) ( 1 0 2017 + 1 ) − 2 ( 1 0 2017 − 1 ) 9 = ( 1 0 2017 − 1 ) ( 1 0 2017 + 1 − 2 ) 9 = ( 1 0 2017 − 1 ) 2 9 = 1 0 2017 − 1 3 = 999...999 ⏞ # of 9s = 2017 3 = 333...333 ⏟ # of 3s = 2017 \begin{aligned} N & = \sqrt{\underbrace{111...111}_{\text{\# of 1s = 4034}} - \underbrace{222...222}_{\text{\# of 2s = 2017}}} \\ & = \sqrt{\frac {10^{4034}-1}9 - \frac {2(10^{2017}-1)}9} \\ & = \sqrt{\frac {(10^{2017}-1)(10^{2017}+1)- 2(10^{2017}-1)}9} \\ & = \sqrt{\frac {(10^{2017}-1)(10^{2017}+1- 2)}9} \\ & = \sqrt{\frac {(10^{2017}-1)^2}9} \\ & = \frac {10^{2017}-1}3 = \frac {\overbrace{999...999}^{\text{\# of 9s = 2017}}}3 = \underbrace{333...333}_{\text{\# of 3s = 2017}} \end{aligned}

Therefore, the sum of digits of N N is 3 × 2017 = 6051 3 \times 2017 = \boxed{6051} .

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A = 111...111 ⏟ 4034 − 222...222 ⏟ 2017 A = 1 9 ( 10 4034 − 1 ) − 2 9 ( 10 2017 − 1 ) A = 1 9 ( 10 2017 + 1 ) ( 10 2017 − 1 ) − 2 9 ( 10 2017 − 1 ) A = 1 9 ( 10 2017 − 1 ) { 10 2017 + 1 − 2 } A = 1 9 ( 10 2017 − 1 ) 2 A = 1 3 ⋅ ( 10 2017 − 1 ) A = 1 3 ⋅ 999...999 ⏟ 2017 A = 333...333 ⏟ 2017 A=\underbrace { 111...111 }_{ 4034 } -\underbrace { 222...222 }_{ 2017 } \\ A=\frac { 1 }{ 9 } \left( { 10 }^{ 4034 }-1 \right) -\frac { 2 }{ 9 } \left( { 10 }^{ 2017 }-1 \right) \\ A=\frac { 1 }{ 9 } \left( { 10 }^{ 2017 }+1 \right) \left( { 10 }^{ 2017 }-1 \right) -\frac { 2 }{ 9 } \left( { 10 }^{ 2017 }-1 \right) \\ A=\frac { 1 }{ 9 } \left( { 10 }^{ 2017 }-1 \right) \left\{ { 10 }^{ 2017 }+1-2 \right\} \\ A=\frac { 1 }{ 9 } { \left( { 10 }^{ 2017 }-1 \right) }^{ 2 }\\ \sqrt { A } =\frac { 1 }{ 3 } \cdot \left( { 10 }^{ 2017 }-1 \right) \\ \sqrt { A } =\frac { 1 }{ 3 } \cdot \underbrace { 999...999 }_{ 2017 } \\ \sqrt { A } =\underbrace { 333...333 }_{ 2017 }

Thus, the sum of the digit digits of A \sqrt { A } is 2017 ⋅ 3 = 6051 2017 \cdot 3 = 6051 .

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We will show that: 111 … 111 ⏟ 2 k = 222 … 222 ⏟ k + 333 … 33 3 2 ⏟ k \underbrace{111\dots111}_{2k}=\underbrace{222\dots222}_{k}+\underbrace{333\dots333^2}_{k}

222 … 222 ⏟ k + 333 … 33 3 2 ⏟ k = 2 ∗ 111 … 111 ⏟ k + ( 3 ∗ 111 … 111 ⏟ k ) 2 = 111 … 111 ⏟ k ( 2 + 999 … 999 ⏟ k ) = 111 … 111 ⏟ k ∗ ( 1 000 … 000 ⏟ k + 1 ) = 111 … 111 ⏟ k 000 … 000 ⏟ k + 111 … 111 ⏟ k = 111 … 111 ⏟ 2 k \underbrace{222\dots222}_{k}+\underbrace{333\dots333^2}_{k}=2*\underbrace{111\dots111}_{k}+(3*\underbrace{111\dots111}_{k})^2=\underbrace{111\dots111}_{k}(2+\underbrace{999\dots999}_{k})=\underbrace{111\dots111}_{k}*(1\underbrace{000\dots000}_{k}+1)=\underbrace{111\dots111}_{k}\underbrace{000\dots000}_{k}+\underbrace{111\dots111}_{k}=\underbrace{111\dots111}_{2k}

The answer is: 2017 ∗ 3 = 6051 2017*3=\boxed{6051}

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