Question

how many integers between 1 and 99 inclusive are divisible by either 2 or 3 or both

Answer 1

Answer on Question #60957 – Math – Discrete Mathematics

Question

How many integers between 1 and 99 inclusive are divisible by either 2 or 3 or both?

Solution

Let's determine sets A and B in the following way:

A = 'integers between 1 and 99 inclusive, which are divisible by 2'; $A = \{a_1, a_2, \ldots, a_n\}$;

B = 'integers between 1 and 99 inclusive, which are divisible by 3'; $B = \{b_1, b_2, \ldots, b_m\}$.

Thus,

A ∪ B = 'integers between 1 and 99 inclusive, which are divisible by either 2 or 3 or both';

A ∩ B = 'integers between 1 and 99, which are divisible by both 2 and 3, thus, they are divisible by 6'; A ∩ B = $\{c_1, c_2, \ldots, c_k\}$.

Let's find the number $|\mathbf{A}|$ of elements in the set A:

If $a_1 = 2$, $d = 2$, $a_n = 98$, then $a_n = a_1 + (n - 1)d = 2 + 2(n - 1) = 98$, hence

$n - 1 = \frac{98 - 2}{2}$, that is, $n = 1 + \frac{96}{2}$, $|\mathbf{A}| = n = 49$.

Let's find the number $|\mathbf{B}|$ of elements in the set B:

If $b_1 = 3$, $e = 3$, $b_m = 99$, then $b_m = b_1 + (m - 1)e = 3 + 3(m - 1) = 99$, hence

$m - 1 = \frac{99 - 3}{3}$, that is, $m = 1 + \frac{96}{3}$, $|\mathbf{B}| = m = 33$.

Let's find the number $|\mathbf{A} \cap \mathbf{B}|$ of elements in the set $\mathbf{A} \cap \mathbf{B}$:

If $c_1 = 6$, $f = 6$, $c_k = 96$, then $c_k = c_1 + (k - 1)f = 6 + 6(k - 1) = 96$, hence

$k - 1 = \frac{96 - 6}{6}$, that is, $k = 1 + \frac{90}{6}$, $|\mathbf{A} \cap \mathbf{B}| = k = 16$.

Thus,

$$|\mathbf{A}| = 49; |\mathbf{B}| = 33; |\mathbf{A} \cap \mathbf{B}| = 16.$$

Let's find the number $|\mathbf{A} \cup \mathbf{B}|$ of elements in the set $\mathbf{A} \cup \mathbf{B}$:

$$|\mathbf{A} \cup \mathbf{B}| = |\mathbf{A}| + |\mathbf{B}| - |\mathbf{A} \cap \mathbf{B}|;$$ $$|\mathbf{A} \cup \mathbf{B}| = 49 + 33 - 16 = 66.$$

Answer: 66.

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