Counting Integers Divisible by 4 or 10

The question asks us to find the number of integers between 299 and 501 that are divisible by 4 or 10. This means we need to consider the integers strictly greater than 299 and strictly less than 501. The range of integers we are considering is thus from 300 to 500, inclusive.

To find the number of integers divisible by 4 or 10 in this range, we can use the Principle of Inclusion-Exclusion. The formula for two sets A and B is:

$\text{Number of elements in (A or B)} = |\text{A} \cup \text{B}| = |\text{A}| + |\text{B}| - |\text{A} \cap \text{B}|$

In this case:

• Set A is the set of integers between 299 and 501 that are divisible by 4.
• Set B is the set of integers between 299 and 501 that are divisible by 10.
• Set A $\cap$ B is the set of integers between 299 and 501 that are divisible by both 4 and 10.

Step 1: Find the number of integers divisible by 4 (Set A)

We need to count integers $n$ such that $300 \le n \le 500$ and $n$ is divisible by 4.

The first integer in this range divisible by 4 is 300 ($300 = 4 \times 75$).

The last integer in this range divisible by 4 is 500 ($500 = 4 \times 125$).

To find the count, we can think of this as an arithmetic progression: 300, 304, ..., 500. The terms are $4 \times 75, 4 \times 76, ..., 4 \times 125$. The number of terms is the number of possible multiples of 4, which is from 75 to 125 inclusive.

Number of terms = (Last multiple index - First multiple index) + 1

$|\text{A}| = 125 - 75 + 1 = 50 + 1 = 51$

Alternatively, using floor division: $\lfloor \text{Upper Bound} / \text{Divisor} \rfloor - \lfloor (\text{Lower Bound} - 1) / \text{Divisor} \rfloor$

$|\text{A}| = \lfloor 500/4 \rfloor - \lfloor 299/4 \rfloor = 125 - 74 = 51$

Step 2: Find the number of integers divisible by 10 (Set B)

We need to count integers $n$ such that $300 \le n \le 500$ and $n$ is divisible by 10.

The first integer in this range divisible by 10 is 300 ($300 = 10 \times 30$).

The last integer in this range divisible by 10 is 500 ($500 = 10 \times 50$).

Using the arithmetic progression approach, the terms are $10 \times 30, 10 \times 31, ..., 10 \times 50$. The number of terms is from 30 to 50 inclusive.

$|\text{B}| = 50 - 30 + 1 = 20 + 1 = 21$

Using floor division: $\lfloor 500/10 \rfloor - \lfloor 299/10 \rfloor = 50 - 29 = 21$

Step 3: Find the number of integers divisible by both 4 and 10 (Set A $\cap$ B)

An integer is divisible by both 4 and 10 if it is divisible by their least common multiple (LCM). The prime factorization of 4 is $2^2$, and the prime factorization of 10 is $2 \times 5$.

$\text{LCM}(4, 10) = 2^{\max(2,1)} \times 5^{\max(0,1)} = 2^2 \times 5^1 = 4 \times 5 = 20$.

So, we need to count integers $n$ such that $300 \le n \le 500$ and $n$ is divisible by 20.

The first integer in this range divisible by 20 is 300 ($300 = 20 \times 15$).

The last integer in this range divisible by 20 is 500 ($500 = 20 \times 25$).

Using the arithmetic progression approach, the terms are $20 \times 15, 20 \times 16, ..., 20 \times 25$. The number of terms is from 15 to 25 inclusive.

$|\text{A} \cap \text{B}| = 25 - 15 + 1 = 10 + 1 = 11$

Using floor division: $\lfloor 500/20 \rfloor - \lfloor 299/20 \rfloor = 25 - 14 = 11$

Step 4: Apply the Principle of Inclusion-Exclusion

Now we use the formula: $|\text{A} \cup \text{B}| = |\text{A}| + |\text{B}| - |\text{A} \cap \text{B}|$

$|\text{A} \cup \text{B}| = 51 + 21 - 11$

$|\text{A} \cup \text{B}| = 72 - 11$

$|\text{A} \cup \text{B}| = 61$

Thus, there are 61 integers between 299 and 501 that are divisible by 4 or 10.

Revision Table: Counting Divisible Integers

Understanding how to count integers divisible by a number within a specific range is a common type of problem. Here’s a quick review of the method used:

• Identify the range of integers: Between 299 and 501 means 300 to 500 inclusive.
• To count integers divisible by 'd' in the range [a, b]: Find the first multiple of 'd' $\ge$ a and the last multiple of 'd' $\le$ b. If the first multiple is $d \times m_1$ and the last is $d \times m_2$, the count is $m_2 - m_1 + 1$. Alternatively, $\lfloor b/d \rfloor - \lfloor (a-1)/d \rfloor$.
• For "divisible by X or Y", use the Principle of Inclusion-Exclusion: Count(X or Y) = Count(X) + Count(Y) - Count(X and Y).
• "Divisible by X and Y" means divisible by LCM(X, Y).

Additional Information: Principle of Inclusion-Exclusion

The Principle of Inclusion-Exclusion is a counting technique used to find the number of elements in the union of multiple sets. It is particularly useful when the sets overlap (have common elements).

For two sets A and B, the principle is as used above: $|\text{A} \cup \text{B}| = |\text{A}| + |\text{B}| - |\text{A} \cap \text{B}|$. We add the sizes of the individual sets, but because the intersection (elements common to both) is counted twice, we subtract it once.

For three sets A, B, and C, the principle extends to:

$|\text{A} \cup \text{B} \cup \text{C}| = |\text{A}| + |\text{B}| + |\text{C}| - (|\text{A} \cap \text{B}| + |\text{A} \cap \text{C}| + |\text{B} \cap \text{C}|) + |\text{A} \cap \text{B} \cap \text{C}|$

This principle is fundamental in combinatorics and number theory problems involving counting elements that satisfy one of several conditions.