Find the number of integers $n$ of the form $30a0b03$ that are divisible by $13$.

Question

Find the number of integers $n$ of the form $30a0b03$ that are divisible by $13$.

What I Tried: This problem looks simple but I cannot find it's solution. I recalled all the divisible rules of $13$ and tried to put them here, but none of them really worked. One thing which I got was :- $$\rightarrow 13 | bo3 - 0a0 + 3$$ I only got this by applying one divisibility rule of $13$. I might try and test for each values of $a$ and $b$ but there will be $100$ configurations, and I don't want to do that.

Can anyone help me?

Edit: After the hints @Daniel gave me, I figured out the solution and found $7$ solutions, can anyone tell me if this is correct or not?

Answer 1

For four or more digit numbers, we do have a divisibility test by $13$ owing to $1001$ being divisible by $13$ ie, $1000$ leaving remainder $-1$. One checks the alternating sum of blocks of three consecutive digits. Eg, $\overline{abcdefgh}$ is divisible by $13$ if $\overline{ab}-\overline{cde}+\overline{fgh}$ is.

So we check how many $n$ there for which following is multiple of $13$ $$\overline{3}-\overline{0a0}+\overline{b03}=\overline{(b-1)(10-a)6}$$

We have $130+26=156$ one such number $\Rightarrow b=2$, $a=5$.

Others can be found by adding/subtracting $130$.

Answer 2

You can crunch it out like this:

$$3\cdot 10^6+a\cdot 10^4 + b\cdot 100 + 3 \equiv 3\cdot(-3)^6+a\cdot (-3)^4 + b\cdot 9 + 3 \pmod{13}.$$

Note that $3^3 = 27 \equiv 1 \pmod{13}$ so the above is $\equiv 3+3a+9b+3 \pmod{13},$ which you want to be congruent to $0$. That reduced to

$$3a+9b \equiv 7 \equiv 33 \pmod{13}$$

or

$$a+3b \equiv 11 \pmod{13}.$$

You can solve this congruence for $b$ for each value of $a = 0,1,\ldots 9.$ Three of the solutions for $b$ are $10$, $11$, and $12$. The other seven solutions work.

Answer 3

This can alternately be written as 30,000,003+10,000a+100b

1.30,000,003≡6(mod 13)

Thus, 10,000a+100b≡7(mod 13)

• 3a+9b≡7(mod 13)
• 3a+9b=7: No possibilities
• 3a+9b=20:No possibilities
• Now you have to realize that the RHS has to divide 3 by Bezouts lemma. So, 13x+7=3y
• 3y-13x=7
• 2x(mod 3)≡1(mod 3). Thus, x≡2(mod 3) and can be represented as 3a+2.
• 3y-13(3a+2)=7 so 3y-39a=33 and y-13a=11. y=13a+11. y also has to be less than or equal to 9. However, no solutions for y satisfy this.