Number of 4-digit numbers that are less than or equal to 2800 and either divisible by 3 or by 11 , is equal to
Step 1: Find the number of 4-digit numbers divisible by 3. The range is from 1000 to 2799. We use the formula for the number of terms in an arithmetic sequence: \[ 1002 + (n - 1) \times 3 = 2799 \] Solving for \(n\): \[ n = 600 \] So, there are 600 numbers divisible by 3 between 1000 and 2799.
Step 2: Find the number of 4-digit numbers divisible by 11. We use the floor function to find the total number of multiples of 11: \[ \left\lfloor \frac{2799}{11} \right\rfloor = 254 \] \[ \left\lfloor \frac{999}{11} \right\rfloor = 90 \] Therefore, the number of numbers divisible by 11 between 1000 and 2799 is: \[ 254 - 90 = 164. \]
Step 3: Find the number of 4-digit numbers divisible by both 3 and 11 (i.e., divisible by 33). We again use the floor function: \[ \left\lfloor \frac{2799}{33} \right\rfloor = 84 \] \[ \left\lfloor \frac{999}{33} \right\rfloor = 30 \] So, the number of numbers divisible by 33 between 1000 and 2799 is: \[ 84 - 30 = 54. \]
Step 4: Apply the inclusion-exclusion principle to find the total number of 4-digit numbers divisible by 3 or 11. The formula is: \[ n(3) + n(11) - n(33). \] Substituting the values: \[ 600 + 164 - 54 = 710. \] Thus, the total number of 4-digit numbers divisible by 3 or 11 is 710.
Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.
Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.