Question

Determine the number of positive integers x, where x <100 and x is divisible by 2,3 or 5

Answer 1

Answer on Question #62173 – Math – Discrete Mathematics

Question

Determine the number of positive integers $x$, where $x < 100$ and $x$ is divisible by 2, 3 or 5.

Solution

Method 1

The first positive integer divisible by 2 is $2 = 2 \cdot 1$. Solving inequality $2k \leq 99$, where $k$ is integer, we get the number of positive integers less than 100 and divisible by 2.

The $[y]$ means the integer part of $y$ (the largest integer less than or equal to $y$).

Let us find the number of positive integers $x_1$, where $x_1$ is divisible by 2:

$$N(A) = \left[ \frac{99}{2} \right] = [49.5] = 49.$$

Further we shall find the number of positive integers $x_2$, where $x_2$ is divisible by 3:

$$N(B) = \left[ \frac{99}{3} \right] = 33.$$

In the next step we shall find the number of positive integers $x_3$, where $x_3$ is divisible by 5:

$$N(C) = \left[ \frac{99}{5} \right] = [19.8] = 19.$$

Then we shall determine the number of positive integers $x_4$, where $x_4$ is divisible by $6 = 3 \cdot 2$:

$$N(A \cap B) = \left[ \frac{99}{6} \right] = [16.5] = 16.$$

Then we shall determine the number of positive integers $x_5$, where $x_5$ is divisible by $10 = 2 \cdot 5$:

$$N(A \cap C) = \left[ \frac{99}{10} \right] = [9.9] = 9.$$

Then we shall determine the number of positive integers $x_6$, where $x_6$ is divisible by $15 = 3 \cdot 5$:

$$N(B \cap C) = \left[ \frac{99}{15} \right] = [6.6] = 6.$$

Now we shall determine the number of positive integers $x_7$, where $x_7$ is divisible by

$$30 = 2 \cdot 3 \cdot 5:$$ $$N(A \cap B \cap C) = \left[ \frac{99}{30} \right] = [3.3] = 3.$$

Method 2

We shall use the general formula for the $l$-th term of the arithmetic sequence:

$$a_l = a_1 + (l - 1)d,$$

hence

$$l = \frac{a_l - a_1}{d} + 1.$$

The first positive integer divisible by 2 is $a_1 = 2$, the last positive integer less than 100 and divisible by 2 is $a_l = 98$, hence the number of positive integers $x_1$, where $x_1$ is divisible by 2:

$$N(A) = \frac{98 - 2}{2} + 1 = \frac{96}{2} + 1 = 49.$$

The first positive integer divisible by 3 is $b_1 = 2$, the last positive integer less than 100 and divisible by 3 is $b_l = 99$, hence the number of positive integers $x_2$, where $x_2$ is divisible by 3:

$$N(B) = \frac{99 - 3}{3} + 1 = \frac{96}{3} + 1 = 33.$$

The first positive integer divisible by 5 is $c_1 = 5$, the last positive integer less than 100 and divisible by 5 is $c_l = 95$, hence the number of positive integers $x_3$, where $x_3$ is divisible by 5:

$$N(C) = \frac{95 - 5}{5} + 1 = \frac{90}{5} + 1 = 19.$$

The first positive integer divisible by $6 = 2 \cdot 3$ is $d_1 = 6$, the last positive integer less than 100 and divisible by 6 is $d_l = 96$, hence the number of positive integers $x_4$, where $x_4$ is divisible by 6:

$$N(A \cap B) = \frac{96 - 6}{6} + 1 = \frac{90}{6} + 1 = 16.$$

The first positive integer divisible by $10 = 2 \cdot 5$ is $e_1 = 10$, the last positive integer less than 100 and divisible by 10 is $e_l = 90$, hence the number of positive integers $x_5$, where $x_5$ is divisible by 10:

$$N(A \cap C) = \frac{90 - 10}{10} + 1 = \frac{80}{10} + 1 = 9.$$

The first positive integer divisible by $15 = 3 \cdot 5$ is $f_1 = 15$, the last positive integer less than 100 and divisible by 15 is $f_l = 90$, hence the number of positive integers $x_6$, where $x_6$ is divisible by 15:

$$N(B \cap C) = \frac{90 - 15}{15} + 1 = \frac{75}{15} + 1 = 6.$$

The first positive integer divisible by $30 = 2 \cdot 3 \cdot 5$ is $g_{1} = 30$, the last positive integer less than 100 and divisible by 30 is $g_{l} = 90$, hence the number of positive integers $x_{7}$, where $x_{7}$ is divisible by 30:

$$N (A \cap B \cap C) = \frac {9 0 - 3 0}{3 0} + 1 = \frac {6 0}{3 0} + 1 = 3.$$

Finally using the inclusion-exclusion principle we determine the required value:

$$\begin{array}{l} N (A \cup B \cup C) = N (A) + N (B) + N (C) - N (A \cap B) - N (A \cap C) - N (B \cap C) + N (A \cap B \cap C) \\ = 4 9 + 3 3 + 1 9 - 1 6 - 9 - 6 + 3 = 7 3. \end{array}$$

Answer: 73.

www.AssignmentExpert.com