edited by
1,615 views
3 votes
3 votes

Let $f$ be a function with both input and output in the set $\{0,1,2, \dots ,9\}$, and let the function $g$ be defined as $g(x) = f(9-x)$. The function $f$ is non-decreasing, so that $f(x) \geq f(y)$ for $x \geq y$. Consider the following statements:

  1. There exists $x \in \{0, \dots ,9\}$ so that $x = f(x)$
  2. There exists $x \in \{0, \dots ,9\}$ so that $x = g(x)$
  3. There exists $x \in \{0, \dots ,9\}$ so that $x = (f(x) +g(x)) \text{mod }10$

Which of the above statements must be TRUE for ALL such functions $f$ and $g$ ?

  1. Only $\text{(i)}$ 
  2. Only $\text{(i)}$  and $\text{(ii)}$ 
  3. Only $\text{(iii)}$ 
  4. None of them
  5. All of them
edited by

1 Answer

11 votes
11 votes
Option A.

We have $10$ elements for the domain set to map to. If we want to ensure $x \neq f(x)$ lets do as follows:

$f(0) = 1, f(1) = 2, f(2) = 3, \ldots, f(8) = 9.$ Now the only option is for $f(9) = 9$ or else we will break the non-decreasing condition.

Like this for any mapping we will get at least one $x$ such that $x = f(x).$

Now consider $f(0) = 0, f(1) = 0, f(2) = 0, \ldots  ,f(8) = 0, f(9) = 1.$ This implies, $g(9) = g(8) = g(7) = \ldots =g(1) = 0, g(0) = 1.$ That is we have a non-decreasing $f$ and for no $x,$ $x = g(x).$ So, (ii) is FALSE.

Let $f(0) = f(1) = f(2) = \ldots =f(8) = 0,f(9) = 1.$ Now, for no $x,$ $x = f(x) + g(x) \mod 10.$ So, (iii) is FALSE.
edited by
Answer:

Related questions

8 votes
8 votes
6 answers
1
7 votes
7 votes
1 answer
3