C 0 0 x + C 0 1 C 1 1 x + C 0 2 C 1 2 C 2 2 x + ⋯ + C 0 2 0 1 6 … C 2 0 1 6 2 0 1 6 x = 2 0 1 7 1 ′ s 1 1 1 … 1 1 1 ( x + 1 )
Find the minimum possible value of x which satisfies the equation above. (Note: The number base must be positive integer.)
Notation : C k n = ( k n ) = k ! ( n − k ) ! n ! denotes of binomial coefficient .
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C 0 0 x + C 0 1 C 1 1 x + C 0 2 C 1 2 C 2 2 x + ⋯ + C 0 2 0 1 6 … C 2 0 1 6 2 0 1 6 x
= 1 x + 1 1 x + 1 2 1 x + 1 3 3 1 x + ⋯ + C 0 2 0 1 6 … C 2 0 1 6 2 0 1 6 x
= 1 + ( x + 1 ) + ( x 2 + 2 x + 1 ) + ( x 3 + 3 x 2 + 3 x 2 + 1 ) + ⋯ + ( x 2 0 1 6 + 2 0 1 6 x 2 0 1 5 + ⋯ + 2 0 1 6 x + 1 )
= 1 + ( x + 1 ) + ( x + 1 ) 2 + ( x + 1 ) 3 + ⋯ + ( x + 1 ) 2 0 1 6
= 2 0 1 7 1 ′ s 1 1 1 … 1 1 1 ( x + 1 )
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. 2 0 1 6 6 is an incorrect expression but 2 0 1 6 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 is C 2 n n ( for even numbers) . C 1 0 0 8 2 0 1 6 is biggest and exists in the equation . C 1 0 0 8 2 0 1 6 is the binomial coefficient in the expansion of ( x + 1 ) 2 0 1 6 .
The value of x cannot be smaller than and equal to C 1 0 0 8 2 0 1 6 .Then , C 1 0 0 8 2 0 1 6 + 1 is the minimum possible value of x which satisfies the equation above .