Finding Natural Numbers Not Divisible by 10, 15, or 25

The problem asks us to find the count of natural numbers from 1 to 1000 that are not divisible by 10, nor by 15, nor by 25. We can solve this using the principle of inclusion-exclusion.

Let $S$ be the set of natural numbers from 1 to 1000. The total number of elements in $S$ is $|S| = 1000$.

Let $A$ be the set of numbers in $S$ that are divisible by 10.

Let $B$ be the set of numbers in $S$ that are divisible by 15.

Let $C$ be the set of numbers in $S$ that are divisible by 25.

We are looking for the number of elements that are not in $A$, not in $B$, and not in $C$. This is given by $|S| - |A \cup B \cup C|$.

According to the principle of inclusion-exclusion for three sets:

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

Calculating the Sizes of Individual Sets

• The number of elements in $A$ (divisible by 10) is the number of multiples of 10 less than or equal to 1000. $$|A| = \left\lfloor \frac{1000}{10} \right\rfloor = 100$$
• The number of elements in $B$ (divisible by 15) is the number of multiples of 15 less than or equal to 1000. $$|B| = \left\lfloor \frac{1000}{15} \right\rfloor = 66$$
• The number of elements in $C$ (divisible by 25) is the number of multiples of 25 less than or equal to 1000. $$|C| = \left\lfloor \frac{1000}{25} \right\rfloor = 40$$

Calculating the Sizes of Pairwise Intersections

The intersection of two sets $X$ and $Y$ ($X \cap Y$) contains elements divisible by both the number defining $X$ and the number defining $Y$. This means the elements are divisible by the Least Common Multiple (LCM) of the two numbers.

• $A \cap B$ (divisible by 10 and 15): Numbers divisible by LCM(10, 15).

  LCM(10, 15) = 30.

  $$|A \cap B| = \left\lfloor \frac{1000}{30} \right\rfloor = 33$$
• $A \cap C$ (divisible by 10 and 25): Numbers divisible by LCM(10, 25).

  LCM(10, 25) = 50.

  $$|A \cap C| = \left\lfloor \frac{1000}{50} \right\rfloor = 20$$
• $B \cap C$ (divisible by 15 and 25): Numbers divisible by LCM(15, 25).

  LCM(15, 25) = 75.

  $$|B \cap C| = \left\lfloor \frac{1000}{75} \right\rfloor = 13$$

Calculating the Size of the Triple Intersection

The intersection of three sets $A$, $B$, and $C$ ($A \cap B \cap C$) contains elements divisible by 10, 15, and 25. This means the elements are divisible by LCM(10, 15, 25).

• LCM(10, 15, 25) = LCM(LCM(10, 15), 25) = LCM(30, 25) = 150.
• $A \cap B \cap C$ (divisible by 10, 15, and 25): Numbers divisible by 150. $$|A \cap B \cap C| = \left\lfloor \frac{1000}{150} \right\rfloor = 6$$

Applying the Inclusion-Exclusion Principle

Now, we calculate the total number of elements divisible by 10 or 15 or 25 ($|A \cup B \cup C|$):

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

Substitute the values we calculated:

$$|A \cup B \cup C| = 100 + 66 + 40 - (33 + 20 + 13) + 6$$

$$|A \cup B \cup C| = 206 - (66) + 6$$

$$|A \cup B \cup C| = 140 + 6$$

$$|A \cup B \cup C| = 146$$

This means there are 146 natural numbers less than or equal to 1000 that are divisible by at least one of 10, 15, or 25.

Finding the Numbers Not Divisible by 10, 15, or 25

The number of natural numbers less than or equal to 1000 which are neither divisible by 10 nor 15 nor 25 is the total number of elements in $S$ minus the number of elements divisible by at least one of the numbers.

Required Count = $|S| - |A \cup B \cup C|$

Required Count = $1000 - 146$

Required Count = $854$

So, there are 854 natural numbers less than or equal to 1000 that are not divisible by 10, 15, or 25.

Revision Table: Key Calculations

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 finite sets. When calculating the size of the union of sets, we initially sum the sizes of the individual sets. However, this overcounts elements that belong to more than one set. We then subtract the sizes of all pairwise intersections to correct for the overcounting. If there are three or more sets, elements belonging to three or more sets might have been subtracted too many times, so we add back the size of the triple intersection, and so on. The general formula for the union of $n$ sets involves alternating sums and differences of the sizes of intersections of various combinations of the sets.

For two sets A and B:

$$|A \cup B| = |A| + |B| - |A \cap B|$$

For three sets A, B, and C:

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

This principle is fundamental in combinatorics and probability theory for solving counting problems involving overlapping categories.