Christmas Streak 39/88: Broken Regularity
A polynomial f ( x ) satisfies:
f ( 1 ) = 1 , f ( 2 ) = 2 , f ( 3 ) = 3 , f ( 4 ) = 4 , f ( 5 ) = 5 , f ( 6 ) = 6 , f ( 7 ) = 8 .
Which of the followings are necessarily true?
A. If f ( 9 ) is an integer, f ( 8 ) is rational.
B. If f ( x ) has a degree of 6 , then for every integer k , f ( k ) is an integer.
C. If f ( x ) has a degree of 7 and f ( 9 ) = 9 , then for every integer k , f ( k ) is an integer.
This problem is a part of <Christmas Streak 2017> series .
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2 solutions
Your example for A doesn't match f ( 7 ) = 8 . It should be easy to fix it though.
A.
This sentence didn't give the degree of f , and therefore f ( 8 ) can be any real numbers.
∴ FALSE
B.
Use the difference chart.
1 1 2 0 1 0 3 0 0 1 0 0 4 0 0 1 1 0 1 5 0 1 1 1 6 1 2 8
And therefore the last row has a degree of 0 (constant) and is an integer.
Therefore, we can infer that for all integers k , f ( k ) must be an integer, since integer + integer = integer.
∴ TRUE
C.
Using the difference chart again,
1 1 2 0 1 0 3 0 0 1 0 0 4 0 0 1 1 0 1 ( x − 1 5 ) 5 0 1 ( x − 1 4 ) 1 1 ( x − 1 3 ) ( 7 7 − 7 x ) 6 1 ( x − 1 2 ) ( 6 3 − 6 x ) 2 ( x − 1 1 ) ( 5 0 − 5 x ) 8 ( x − 1 0 ) ( 3 8 − 4 x ) ( x − 8 ) ( 2 7 − 3 x ) x ( 1 7 − 2 x ) ( 9 − x ) 9
And since the last row must be a constant, x − 1 5 = 7 7 − 7 x ⇔ 8 x = 9 2 ⇔ x = 2 2 3 .
Therefore f ( 8 ) = 2 2 3 and this directly contradicts the statement.
∴ FALSE
From above, only B is correct.
What a great problem! I didn't think of that solution even though I knew the technique.
What you mean by "difference chart?
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Imagine taking the difference quotient (what's used in the definition of a derivative), but with h = 1 . It behaves just like taking the derivative by bringing the degree down by 1. He is setting up the table with the y-values shown, but the x-values they correspond to are spaced 1 apart. They are 1, 2, 3, 4, 5, 6, 7, 8, 9. Here's an example. Let y = x 3 . Then do 1 ( x + 1 ) 3 − x 3 . You get a quadratic polynomial with leading coefficient 3. If you choose a bunch of x-values spaced 1 apart for this function, and do the differences, the fourth row would end out being constant.
https://brilliant.org/wiki/method-of-differences/
Take a look at this wiki!
A is false: try e.g. f ( x ) = x + 6 ! 1 ( x − 1 ) ( x − 2 ) ( x − 3 ) ( x − 4 ) ( x − 5 ) ( x − 6 ) + π ( x − 1 ) ( x − 2 ) ( x − 3 ) ( x − 4 ) ( x − 5 ) ( x − 6 ) ( x − 7 ) ( x − 9 ) .
B is true: the conditions imply that f ( x ) = x + 6 ! 1 ( x − 1 ) ( x − 2 ) ( x − 3 ) ( x − 4 ) ( x − 5 ) ( x − 6 ) , so f ( k ) = k + ( 6 k − 1 ) is always an integer for positive values of k , and it's an integer for non-positive values as well, since in that case it equals k + ( 6 6 − k ) .
C is false: the conditions imply that f ( x ) = x − 6 ! ⋅ 2 1 ( x − 1 ) ( x − 2 ) ( x − 3 ) ( x − 4 ) ( x − 5 ) ( x − 6 ) ( x − 9 ) , and it's easy to check that f ( 8 ) = 2 3 / 2 .