Checking whether $4a2b$ is can be divided by $36$

Question

Imagine we have a number, which is given as seen below:

$$4a2b $$

The first thing I thought is $b$ should be an even number. If a number can be divided by $36$, then it must be able to be divided by $2$, $3$ and $6$. Let's give numbers now.

$b \implies 0,2,4,\boxed {6},8$

Sum of the numerals will be multiples of $6$ and $3$, which means that we can only take $6$ among them.

What about $a$? or is it enough to just find $b$?

Regards!

Answer 1

$4000+100a+20+b \equiv 0 \pmod{36}$

$24+28a+b \equiv 0\pmod{36}$

$28a+b \equiv 12 \pmod{36}$

Solution set: $(a,b) \in \{(3,0),(4,8),(8,4)\}$

Answer 2

Note that this notation is generally understood to mean multiplication, rather than concatenation, so you should make that clearer. I'll use brackets to denote concatenation.

[4a2b] = [4a]*100+[2b] = ([4a]*25)*4+[2b]. Thus, [4a2b] is divisible by 4 iff [2b] is. So b must be 0, 4, or 8. For [4a2b] to be divisible by 9, the sum of the digits must be divisible by 9, so for each value of b, there is a different possible value of a:

b = 0, a = 3
b = 4, a = 8
b = 8, a = 4

So there are three solutions: 4320, 4824, or 4428.

Answer 3

In order for a number to be divisible by $36$, it needs to be divisible by both $4$ and $9$.

The test for divisibility by $9$ is that the digits add up to some multiple of $9$. Here the known digits add up to $6$ so we need $a+b$ to be either $3$ or $12$ (for a digit sum of $9$ or $18$)

The test for divisibility by $4$ is that the last two digits (tens and units) form a number divisible by four. This gives a set of options for $b \in \{0,4,8\}$ (note that $b=6$ is not possible) which then determine the corresponding possibilities for $a$. There are thus three possible $(a,b)$ solutions.

Answer 4

The greatest possible number is $4999$ and this is $$138\cdot 36=4968$$ the smalles one is $4000$ and this is $$112\cdot 36$$ and now by trial and error we get $$118\cdot 36=4248$$ Can you find the others?