Simple equations word problem

Question

In a three-digit number, the difference between its hundreds digit and its tens digit is equal to the difference between its tens digit and its units digit. Also the sum of the digits is $9$. How many numbers satisfy the given condition?

I know that, since $a-b = b-c$ and $a+b+c=9$, then $b=3$ and so $a+c=6$ and hence there are at least $6$ numberss. But the answer is $9$ and when I searched the question online, the answer includes another condition which is $a-b = c-b$, I am not able to understand the logic as to how this equation got formed. Any ideas??

Answer 1

I guess your first condition translates to $|a-b|=|b-c|$. This gives either $a-b=b-c$ (as in $531$) or $a-b=c-b$ (as in $171$).

Answer 2

I think the question probably said this $ |(a-b)|=|(b-c)|$ which includes your two of the conditions

$$a-b=b-c$$

and

$$a-b=c-b$$

Answer 3

if $b=3$ and $a+c=6$ The only possible answers would be ${135,234,333,423,531}$ notice that this will all satisfy $|(a-b)|=|(b-c)|$ and $ a+b+c=9$

but if the first conditons transalates like this $(a-b)=(b-c)$ then the only answers that would satisfy are ${135,234,333,531}$