Question

The reverse of the number 129 is 921, and these add to 1050, which is divisible by 30.
How many three-digit numbers have the property that, when added to their reverse,
the sum is divisible by 30?

Answer 1

Answer:

Step-by-step explanation:

Write the number as $abc$ and use the divisibility rules for 10 and 3. From the rule for 10, we know that the last digit of $abc+cba$ is $o$, hence $a+c=10$. Thus the options are $a=1$ & $c=9$, $a=2$, & $c=8$, etc.

From the rule for 3, namely that the digit sum must be divisible by 3, we find that $2a+2b+2c=20+2b$  is divisible by $3$. By hand it's easy to check that $b=2,5,8$ are the only options that work.

So there are 9 choices for $a$ and $c$ and $3$ for $b$ , giving 27 total.