2 0 1 6 % × 2 0 1 5 % × 2 0 1 4 % × ⋯ × 3 % × 2 % × 1 % = 1 0 x % 2 0 1 6 !
Find x which satisfies the equation above .
Notation : ! denotes the factorial notation. For example, 8 ! = 1 × 2 × 3 × ⋯ × 8 .
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Great solution. I'd have my method posted up soon.
Relevant wiki: Rules of Exponents - Algebraic
We know,
x % = 1 0 0 x
∵ 2 0 1 6 % × 2 0 1 5 % × 2 0 1 4 % × ⋯ × 3 % × 2 % × 1 % ⟹ 2 0 1 6 % × 2 0 1 5 % × 2 0 1 4 % × ⋯ × 3 % × 2 % × 1 % ⟹ 2 0 1 6 % × 2 0 1 5 % × 2 0 1 4 % × ⋯ × 3 % × 2 % × 1 % = 1 0 0 2 0 1 6 × 1 0 0 2 0 1 5 × 1 0 0 2 0 1 4 × ⋯ × 1 0 0 3 × 1 0 0 2 × 1 0 0 1 = 1 0 0 × 1 0 0 × 1 0 0 × ⋯ × 1 0 0 × 1 0 0 2 0 1 6 × 2 0 1 5 × 2 0 1 4 × ⋯ × 3 × 2 × 1 2 0 1 6 t i m e s = 1 0 0 2 0 1 6 2 0 1 6 !
Substitute 2 0 1 6 × 2 0 1 5 % × 2 0 1 4 % × ⋯ × 3 % × 2 % × 1 % = 1 0 0 2 0 1 6 2 0 1 6 ! into
2 0 1 6 % × 2 0 1 5 % × 2 0 1 4 % × ⋯ × 3 % × 2 % × 1 % = 1 0 x % 2 0 1 6 ! we have:
1 0 0 2 0 1 6 2 0 1 6 ! = 1 0 x % 2 0 1 6 !
Cancel out numerators, we have:
1 0 0 2 0 1 6 ( 1 0 2 ) 2 0 1 6 1 0 0 × ( 1 0 2 ) 2 0 1 6 1 0 2 × ( 1 0 2 ) 2 0 1 6 1 0 2 × 1 0 2 × 2 0 1 6 1 0 2 + 4 0 3 2 1 0 4 0 3 4 x = 1 0 x % = 1 0 0 1 0 x = 1 0 x = 1 0 x = 1 0 x = 1 0 x = 1 0 x = 4 0 3 4 .
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2 0 1 6 % × 2 0 1 5 % × 2 0 1 4 % × ⋯ × 3 % × 2 % × 1 % = 1 0 x % 2 0 1 6 !
( 2 0 1 6 × 2 0 1 5 × 2 0 1 4 × ⋯ × 3 × 2 × 1 ) × ( 2 0 1 6 times 1 0 − 2 × 1 0 − 2 × 1 0 − 2 × ⋯ × 1 0 − 2 ) = 1 0 x − 2 2 0 1 6 !
2 0 1 6 ! × 1 0 − 4 0 3 2 = 1 0 x − 2 2 0 1 6 !
1 0 4 0 3 2 2 0 1 6 ! = 1 0 x − 2 2 0 1 6 !
x = 4 0 3 2 + 2 = 4 0 3 4
Note : % = 1 0 − 2