We know that
( 1 − 2 2 1 ) ∗ ( 1 − 3 2 1 ) ∗ ( 1 − 4 2 1 ) ∗ ⋯ ∗ ( 1 − n 2 1 ) = k
where n is a positive integer.
If k can't be 2 1 and can't be 2 0 1 7 1 0 0 8 , then choose 0.
If k can't be 2 1 , but it can be 2 0 1 7 1 0 0 8 , then choose 1.
If k can't be 2 0 1 7 1 0 0 8 , but it can be 2 1 , then choose 2.
If k can be 2 1 and 2 0 1 7 1 0 0 8 too, then choose 3.
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We will show that ( 1 − 2 2 1 ) ∗ ( 1 − 3 2 1 ) ∗ ( 1 − 4 2 1 ) ∗ ⋯ ∗ ( 1 − n 2 1 ) > 2 1 .
We know that x 2 = ( x − 1 ) ( x + 1 ) + 1 , so 1 − y 2 1 = y 2 ( y − 1 ) ( y + 1 ) . From that we can see that
( 1 − 2 2 1 ) ∗ ( 1 − 3 2 1 ) ∗ ( 1 − 4 2 1 ) ∗ ⋯ ∗ ( 1 − n 2 1 ) = 2 ∗ 2 1 ∗ 3 ∗ 3 ∗ 3 2 ∗ 4 ∗ ⋯ ∗ n 2 ( n − 1 ) ( n + 1 ) = 2 1 ∗ 2 3 ∗ 3 2 ∗ 3 4 ∗ 4 3 ∗ 3 4 ∗ ⋯ ∗ n − 1 n ∗ n n − 1 ∗ n n + 1 .
Since b a ∗ a b = 1 , ( 1 − 2 2 1 ) ∗ ( 1 − 3 2 1 ) ∗ ( 1 − 4 2 1 ) ∗ ⋯ ∗ ( 1 − n 2 1 ) = 2 1 ∗ 1 ∗ 1 ∗ 1 ∗ ⋯ ∗ 1 ∗ n n + 1 . Now we can see that 2 1 ∗ n n + 1 > 2 1 (because n n + 1 > 1 ), so ( 1 − 2 2 1 ) ∗ ( 1 − 3 2 1 ) ∗ ( 1 − 4 2 1 ) ∗ ⋯ ∗ ( 1 − n 2 1 ) > 2 1 .
The answer is 0, because 2 1 = 2 1 > 2 0 1 7 1 0 0 8 .
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Plugging in some numbers, we get:
For n = 2 : k = 4 3
For n = 3 : k = 3 2
For n = 4 : k = 8 5
For n = 5 : k = 5 3
Expanding some fractions, we get the pattern 4 3 , 6 4 , 8 5 , 1 0 6 . . . 2 n n + 1 for n = 2 , 3 , 4 , . . . , n .
To prove that the series above is equal to 2 n n + 1 , we test it for n = 2, which is 4 2 + 1 , therefore true.
Assuming that ( 1 − 2 2 1 ) ( 1 − 3 2 1 ) ( 1 − 3 2 1 ) . . . ( 1 − n 2 1 ) = 2 n n + 1 , we try to prove that the same is true for n + 1 :
( 1 − 2 2 1 ) ( 1 − 3 2 1 ) ( 1 − 3 2 1 ) . . . ( 1 − ( n + 1 ) 2 1 ) = 2 n + 2 n + 2
Notice that on the left side, since we assume that ( 1 − 2 2 1 ) ( 1 − 3 2 1 ) ( 1 − 3 2 1 ) . . . ( 1 − n 2 1 ) = 2 n n + 1 , we can exchange all the factors infront of ( n + 1 ) 2 1 with 2 n n + 1 :
2 n n + 1 ⋅ ( 1 − ( n + 1 ) 2 1 = 2 n + 2 n + 2
Simplyfing the second factor on the left hand side yields:
2 n n + 1 ⋅ ( n + 1 ) 2 n 2 + 2 n = 2 n + 2 n + 2
⟺ 2 n 1 ⋅ n + 1 n 2 + 2 n = 2 n + 2 n + 2
⟺ 2 n 1 ⋅ n + 1 n = 2 n + 2 1
⟺ 2 1 ⋅ n + 1 1 = 2 n + 2 1
⟺ 2 n + 2 1 = 2 n + 2 1
, which is obviously true. Therefore, the assumption ( 1 − 2 2 1 ) ( 1 − 3 2 1 ) ( 1 − 3 2 1 ) . . . ( 1 − n 2 1 ) = 2 n n + 1 is true.
Testing for 2 n n + 1 = 2 1 solves to 0 = 1 , which is false. Similarily, solving 2 n n + 1 = 2 0 1 7 1 0 0 8 yields n = − 2 0 1 7 , but n ≥ 2 , therefore this is also no possible solution. Therefore the correct answer is 0