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Problem Of The Week #457 March 1st 2021

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
Here is this week's POTW:

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Consider functions $f$ defined for all real numbers and taking real numbers as values such that

$f(x+14)-14\le f(x) \le f(x+20)-20$, for all real numbers $x$.

Determine all possible values of $f(8765)-f(4321)$.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
No one answered last week's POTW. (Sadface) However, you can find the suggested solution below:

From $140=7 \times 20 =10 \times 14$, we have the following chains of inequalities:

$f(x)\le f(x+20)-20 \le f(x+40)-40 \le \cdots \le f(x+140)-140\\ f(x)\ge f(x+14)-14 \ge f(x+28)-28 \ge \cdots \ge f(x+140)-140$

So by the squeeze principle, equality holds throughout and we have $f(x)=f(x+20)-20$ and $f(x)=f(x+14)-14$ for all real numbers $x$.

Since $8765-4405=4360$ is a multiple of 20 and $4405-4321=84$ is a multiple of 14, it follows that

$f(8765)-f(4405)=8765-4405$ and $f(4405)-f(4321)=4405-4321$

Adding these equations yields $f(8765)-f(4321)=8765-4321=4444$.

Since $f(x)=x$ satisfies the conditions of the problem, the only possible value of the expression $f(8765)-f(4321)$ is 4444.
 
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