What's the digit "a" in this number?

Question

a and b are digits in a four-digit natural number 7a5b. If 7a5b is divisible by 18, how many different possible values can "a" have?

Answer 1

A number is divisible by $18$ iff it's divisible by $2$ and $9$. So, we must have $b \in \{0,2,4,6,8\}$ and $7+a+5+b$ divisible by $9$, since a number is divisible by $9$ iff the sum of its digits is divisible by $9$. I think you can solve it by now.

Answer 2

One way:

$$7a5b=7050+100a+b\equiv 12+10a+b\pmod{18}$$

Clearly, $b$ must be even

and $9$ must divide $12+10a+b=12+10a+b\iff 3+a+b$ must be divisible by $9$

For example,

if $b=0,3+a$ must be divisible by $9\implies a=6$ as $0\le a\le9\iff3\le a+3\le12$

if $b=2,5+a$ must be divisible by $9\implies a=4$

and so on

Answer 3

Since it is divisible by $18$, $7a5b$ must be divisible by $9$ and $2$. Using divisibility rules, $$b \text{ must be even}$$ $$7+a+5+b \text{ must be divisible by 9}$$ So if $b$ is even it is of the form $2k$ for some integer $k$. So then $7+a+5+2k=9m$. Now use cases for $b$. For example, when $b=2$, $$7+a+5+2=14+a$$ So $a$ must be 4 since $14+4=18=9\cdot2$. Now, what about $a=4,6,8,10$? You can try them on your own.