Number base and binomial coefficient

C 0 0 x + C 0 1 C 1 1 x + C 0 2 C 1 2 C 2 2 x + + C 0 2016 C 2016 2016 x = 111 111 ( x + 1 ) 2017 1 s \overline{C_{0}^{0} } _{x} + \overline{C_{0}^{1} C_{1}^{1}} _{x} + \overline{C_{0}^{2} C_{1}^{2} C_{2}^{2}} _{x} + \cdots + \overline{C_{0}^{2016} \dots C_{2016}^{2016}}_{x} = \underbrace{\overline{111\dots111}_{(x+1)}}_{2017 \ 1's}

Find the minimum possible value of x x which satisfies the equation above. (Note: The number base must be positive integer.)

Notation : C k n = ( n k ) = n ! k ! ( n k ) ! \large C_{k}^{n}= { n \choose k } = \frac{n!}{k!(n-k)!} denotes of binomial coefficient .

C 1008 2017 C_{1008}^{2017} 2016! C 1008 2016 + 1 C_{1008}^{2016}+1 2016 C 1008 2016 C_{1008}^{2016} 2017!

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1 solution

Tommy Li
Aug 21, 2016

C 0 0 x + C 0 1 C 1 1 x + C 0 2 C 1 2 C 2 2 x + + C 0 2016 C 2016 2016 x \overline{C_{0}^{0} } _{x} + \overline{C_{0}^{1} C_{1}^{1}} _{x} + \overline{C_{0}^{2} C_{1}^{2} C_{2}^{2}} _{x} + \dots + \overline{C_{0}^{2016} \dots C_{2016}^{2016}}_{x}

= 1 x + 11 x + 121 x + 1331 x + + C 0 2016 C 2016 2016 x = \overline{1} _{x} + \overline{11} _{x} + \overline{121} _{x} + \overline{1331} _{x} + \dots + \overline{C_{0}^{2016} \dots C_{2016}^{2016}}_{x}

= 1 + ( x + 1 ) + ( x 2 + 2 x + 1 ) + ( x 3 + 3 x 2 + 3 x 2 + 1 ) + + ( x 2016 + 2016 x 2015 + + 2016 x + 1 ) = 1 +(x+1)+(x^2+2x+1)+(x^3+3x^2+3x^2+1)+ \dots +(x^{2016}+2016x^{2015}+\dots+2016x+1)

= 1 + ( x + 1 ) + ( x + 1 ) 2 + ( x + 1 ) 3 + + ( x + 1 ) 2016 = 1+(x+1)+(x+1)^2+(x+1)^3+ \dots+ (x+1)^{2016}

= 111 111 ( x + 1 ) 2017 1 s =\underbrace{\overline{111\dots111}_{(x+1)}}_{2017 \ 1's}

The calculation above shows that we can put any value to satisfy the equation but it does not .

Considering the boundary of the number base , the value of each single digit must be smaller than the value of number base and number base must be positive integer .

e.g. 2016 6 \large \overline{2016}_{6} is an incorrect expression but 2016 7 \large \overline{2016}_{7} is a correct expression .

Therefore , we need to find the maximum value of the digit which exists in the equation .

Finding patterns , the maximum value of C k n \large C_{k}^{n} is C n 2 n \large C_{\frac{n}{2}}^{n} ( for even numbers) . C 1008 2016 \large C_{1008}^{2016} is biggest and exists in the equation . C 1008 2016 \large C_{1008}^{2016} is the binomial coefficient in the expansion of ( x + 1 ) 2016 (x+1)^{2016} .

The value of x x cannot be smaller than and equal to C 1008 2016 \large C_{1008}^{2016} .Then , C 1008 2016 + 1 \large C_{1008}^{2016}+1 is the minimum possible value of x x which satisfies the equation above .

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