let f: A `to` A, where A is a set of non-negative integers and satisfying the condition.
(i) `x-f(x)=19[x//19]-90[f(x)//90],AAxinA`
(ii) `1900ltf(1990)lt2000`
then f(1990) equals, (where [.] denote greatest integer function)

Options

A. 1994
B. 1900
C. 1904
D. 1890

Answer

To solve the problem, we need to find the value of \( f(1990) \) given the conditions. Let's break it down step by step.

### Step 1: Understand the given conditions

We have two conditions:
1. \( x - f(x) = 19 \left\lfloor \frac{x}{19} \right\rfloor - 90 \left\lfloor \frac{f(x)}{90} \right\rfloor \) for all \( x \in A \)
2. \( 1900 < f(1990) < 2000 \)

### Step 2: Substitute \( x = 1990 \) into the first condition

Using the first condition, we substitute \( x = 1990 \):

\[ 1990 - f(1990) = 19 \left\lfloor \frac{1990}{19} \right\rfloor - 90 \left\lfloor \frac{f(1990)}{90} \right\rfloor \]

### Step 3: Calculate \( \left\lfloor \frac{1990}{19} \right\rfloor \)

Calculating \( \frac{1990}{19} \):

\[ \frac{1990}{19} \approx 104.7368 \implies \left\lfloor \frac{1990}{19} \right\rfloor = 104 \]

### Step 4: Substitute back into the equation

Now substituting back into the equation:

\[ 1990 - f(1990) = 19 \times 104 - 90 \left\lfloor \frac{f(1990)}{90} \right\rfloor \]

Calculating \( 19 \times 104 \):

\[ 19 \times 104 = 1976 \]

So we have:

\[ 1990 - f(1990) = 1976 - 90 \left\lfloor \frac{f(1990)}{90} \right\rfloor \]

### Step 5: Rearranging the equation

Rearranging gives:

\[ f(1990) = 1990 - 1976 + 90 \left\lfloor \frac{f(1990)}{90} \right\rfloor \]

This simplifies to:

\[ f(1990) = 14 + 90 \left\lfloor \frac{f(1990)}{90} \right\rfloor \]

### Step 6: Define \( f(1990) \)

Let \( f(1990) = 14 + 90k \) where \( k = \left\lfloor \frac{f(1990)}{90} \right\rfloor \).

### Step 7: Apply the second condition

From the second condition, we know:

\[ 1900 < f(1990) < 2000 \]

Substituting \( f(1990) \):

\[ 1900 < 14 + 90k < 2000 \]

### Step 8: Solve the inequalities

1. For the left side:

\[ 1900 < 14 + 90k \implies 1886 < 90k \implies k > \frac{1886}{90} \approx 20.96 \implies k \geq 21 \]

2. For the right side:

\[ 14 + 90k < 2000 \implies 90k < 1986 \implies k < \frac{1986}{90} \approx 22.07 \implies k \leq 22 \]

### Step 9: Determine possible values for \( k \)

From the inequalities, \( k \) can only be \( 21 \) or \( 22 \).

### Step 10: Calculate \( f(1990) \)

1. If \( k = 21 \):

\[ f(1990) = 14 + 90 \times 21 = 14 + 1890 = 1904 \]

2. If \( k = 22 \):

\[ f(1990) = 14 + 90 \times 22 = 14 + 1980 = 1994 \]

### Step 11: Verify the values

Both values \( 1904 \) and \( 1994 \) satisfy \( 1900 < f(1990) < 2000 \).

### Conclusion

The possible values for \( f(1990) \) are \( 1904 \) and \( 1994 \).

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