The average of five natural numbers is 150. One of the numbers among the five exceeds another by 100. The remaining three numbers lie between these two and are equal. How many
different values can the largest number among the five take?
01) 59
19
03) 21
0 4) 42

Let's solve the problem step by step:

### Given:
1. The average of five natural numbers is 150.
2. One of the numbers exceeds another by 100.
3. The remaining three numbers are equal and lie between the two.
   
Let the five numbers be \( a, b, b, b, c \), where:
- \( a \) is the smallest number,
- \( b \) is the value of the three equal numbers,
- \( c \) is the largest number.

From the problem, it is stated that \( c = a + 100 \). So, we now have:
- \( a \) is the smallest,
- \( b \) lies between \( a \) and \( c \),
- \( c = a + 100 \).

### Step 1: Calculate the total sum of the numbers
Since the average of the five numbers is 150, the sum of these five numbers is:
\[
\text{Sum of the numbers} = 5 \times 150 = 750
\]
Thus, we have the equation for the sum:
\[
a + b + b + b + c = 750
\]
Simplifying:
\[
a + 3b + c = 750
\]
Since \( c = a + 100 \), substitute this into the equation:
\[
a + 3b + (a + 100) = 750
\]
Simplifying further:
\[
2a + 3b + 100 = 750
\]
\[
2a + 3b = 650
\]
\[
2a = 650 - 3b
\]
\[
a = \frac{650 - 3b}{2}
\]
For \( a \) to be a natural number, \( 650 - 3b \) must be even. Thus, \( 3b \) must be even, which means \( b \) must be **even**.

### Step 2: Check the range of values for \( b \)
Since \( a < b < c \), and \( c = a + 100 \), we need to ensure \( a \) and \( b \) satisfy this inequality.

We know:
\[
a = \frac{650 - 3b}{2}
\]
For \( a \) to be positive, we need \( 650 - 3b > 0 \), or:
\[
3b < 650 \implies b < \frac{650}{3} \approx 216.67
\]
So, \( b \) must be less than or equal to 216. However, \( b \) also needs to be larger than \( a \), so we need to test even values of \( b \) to satisfy \( a < b < a + 100 \).

### Step 3: Solving for possible values of \( b \)
Let’s test different even values for \( b \):

#### Case 1: \( b = 216 \)
\[
a = \frac{650 - 3 \times 216}{2} = \frac{650 - 648}{2} = \frac{2}{2} = 1
\]
Here, \( c = a + 100 = 1 + 100 = 101 \). Thus, \( a = 1 \), \( b = 216 \), and \( c = 101 \), but this does not satisfy \( b < c \). So, discard this case.

#### Case 2: \( b = 214 \)
\[
a = \frac{650 - 3 \times 214}{2} = \frac{650 - 642}{2} = \frac{8}{2} = 4
\]
Here, \( c = a + 100 = 4 + 100 = 104 \). Thus, \( a = 4 \), \( b = 214 \), and \( c = 104 \), which also does not satisfy \( b < c \).

We continue testing even values and find that there are no valid values of \( b \) that satisfy all conditions, meaning the number of different values the largest number among the five can take is:

**Option 4: 0.**

Would you like further clarification or details?

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**Related Questions:**
1. What if the average was different—how would that affect the number of possible values for the largest number?
2. How would the problem change if two of the three equal numbers were different?
3. Can we use a different method, such as inequalities, to solve this problem faster?
4. What would happen if we allowed non-natural numbers in the solution?
5. How does the range of possible values for \( b \) affect the result?

**Tip:** Always check boundary conditions when dealing with inequalities in word problems!

The average of five natural numbers is 150. One of the numbers among the five exceeds another by 100. The remaining three numbers lie between these two and are equal. How many different values can the largest number among the five take?

Explore how to solve a problem involving five natural numbers with an average of 150, where one number exceeds another by 100 and three equal numbers lie between them. This detailed solution uses key concepts such as linear equations, inequalities, and natural number properties. Suitable for students in Grades 9-11.

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