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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!

Busi
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4 Answers4

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

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

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

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

SlipEternal
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  • I don't know what $\text {mod}$ means. Can you keep it on? – Busi May 14 '18 at 18:37
  • $a \equiv b \pmod{c}$ means there exists an integer $k$ such that $a=ck+b$. It can be thought of as the remainder after division plus some multiple of the dividend. – SlipEternal May 14 '18 at 18:39
  • I've been told that the only value that satisfies with the condition is $3$, is this wrong? – Busi May 14 '18 at 18:45
  • There are three solutions. That is correct. You have: $4320 = 120\cdot 36$, $4428 = 123\cdot 36$, $4824 = 134\cdot 36$.

    Those are the only solutions. Check out this set of tables from Wolframalpha. You are looking for pairs where the final element is 12 (make sure to copy the full url and paste it... it is not formatting correctly): http://www.wolframalpha.com/input/?i=TABLE%5BTABLE%5B(a,b,28*a%2Bb+mod+36),%7Bb,0,9%7D%5D,%7Ba,0,9%7D%5D

     – SlipEternal May 14 '18 at 19:03
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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.

Joffan
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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.

Acccumulation
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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?

Dr. Sonnhard Graubner
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