We say that a three digit number is balanced if the middle digit is the arithmetic mean of the other two digits how many balanced numbers are divisible by 18

To find three-digit balanced numbers divisible by 18, we need to consider the possible values of the middle digit and ensure that the sum of the three digits is divisible by 9 (since 18 is divisible by 9).

Let the three-digit number be $abc$, where $a$, $b$, and $c$ are the hundreds, tens, and units digits, respectively.

Given that $b = \frac{a + c}{2}$, we can express $b$ in terms of $a$ and $c$.

Now, we need to find the values of $a$, $b$, and $c$ such that $a + b + c$ is divisible by 9.

Let's go through the possible values:

1. If $b = 0$, then $a = c$, and $a + b + c = 2a$.

- For $a = 1$, $c = 1$, $abc = 101$.

- For $a = 2$, $c = 2$, $abc = 202$.

- For $a = 3$, $c = 3$, $abc = 303$.

- For $a = 4$, $c = 4$, $abc = 404$.

- For $a = 5$, $c = 5$, $abc = 505$.

- For $a = 6$, $c = 6$, $abc = 606$.

- For $a = 7$, $c = 7$, $abc = 707$.

- For $a = 8$, $c = 8$, $abc = 808$.

- For $a = 9$, $c = 9$, $abc = 909$.

2. If $b = 1$, then $a + c = 2$.

- Possible pairs are $a = 1$, $c = 1$, and $a = 2$, $c = 0$.

- This gives us $110$ and $201$.

3. If $b = 2$, then $a + c = 4$.

- Possible pairs are $a = 1$, $c = 3$, and $a = 3$, $c = 1$.

- This gives us $213$ and $312$.

4. If $b = 3$, then $a + c = 6$.

- Possible pairs are $a = 1$, $c = 5$, and $a = 5$, $c = 1$.

- This gives us $315$ and $513$.

5. If $b = 4$, then $a + c = 8$.

- Possible pairs are $a = 1$, $c = 7$, and $a = 7$, $c = 1$.

- This gives us $417$ and $714$.

6. If $b = 5$, then $a + c = 10$.

- Possible pairs are $a = 1$, $c = 9$, and $a = 9$, $c = 1$.

- This gives us $519$ and $915$.

7. If $b = 6$, then $a + c = 12$.

- Possible pairs are $a = 3$, $c = 9$, and $a = 9$, $c = 3$.

- This gives us $639$ and $936$.

8. If $b = 7$, then $a + c = 14$.

- Possible pairs are $a = 5$, $c = 9$, and $a = 9$, $c = 5$.

- This gives us $759$ and $957$.

9. If $b = 8$, then $a + c = 16$.

- Possible pairs are $a = 7$, $c = 9$, and $a = 9$, $c = 7$.

- This gives us $879$ and $978$.

So, the balanced three-digit numbers divisible by 18 are:

\[101, 202, 303, 404, 505, 606, 707, 808, 909, 110, 201, 213, 312, 315, 513, 417, 714, 519, 915, 639, 936, 759, 957, 879, 978\]

There are 26 such numbers.

We say that a three digit number is balanced if the middle digit is the arithmetic mean of the other two digits how many balanced numbers are divisible by 18​

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We say that a three digit number is balanced if the middle digit is the arithmetic mean of the other two digits how many balanced numbers are divisible by 18