$N$ is an $80-$digit positive integer (in the decimal scale). All digits except the $44th$ digit (from the left) are $2$. If N is divisible by $13,$ find the $44{th}$ digit ?
P.S: This isn't a homework question. Found it on a website and the explanation for the answer was very vague.
Link: http://iim-cat-questions-answers.2iim.com/quant/number-system/other/number-system_8.shtml
My interpretation of the given answer:
We see that $222222$ is divisible by $13$. There's two tools in use here:
$$222222[\text{some digits}] \equiv [\text{some digits}] \pmod {13}$$
and
$$[\text{some digits}]222222$$ is divisible by $13$ if and only if $$[\text{some digits}]$$ is divisible by $13$.
This means, in order to determine divisibility by $13$, we can eliminate $2$'s in groups of six from the start and the end of the number.
We can eliminate $42=6 \times 7$ most significant $2$'s and $36=6 \times 6$ least significant $2$'s, which leaves the original number being divisible by $13$ if and only if the number $2a$ is divisible by $13$ (here $2a$ is a $2$-digit number written in decimals, i.e., $=20+a$).
Then we just check which number of the form $2a$ is divisible by $13$, and it happens to be $26$. So $a=6$.
222222 is divisible by 13.
Therefore 222...(42 digits) is divisible by 13. Since (7 * 6 = 42)
And 222...(36 digits) is divisible by 13. Since (6 * 6 = 36)
Therefore 42 + 36 = 78 digits
We have the 43rd and 44th digit in the middle, which have to be divisible by 13.
We know 43rd digit is also 2.
26 is the only possibility which is divisible by 13.
Therefore 44th digit is 6.
A number ${d_{1}}\dots{d_{80}}$ is divisible by $13$ if and only if:
${d_{1}}{d_{2}}-{d_{3}}{d_{4}}{d_{5}}+{d_{6}}{d_{7}}{d_{8}}-\dots+{d_{78}}{d_{79}}{d_{80}}$ is divisible by $13$.
We know that $-{d_{45}}{d_{46}}{d_{47}}+{d_{48}}{d_{49}}{d_{50}}-\dots+{d_{78}}{d_{79}}{d_{80}}=0$.
We also know that $-{d_{3}}{d_{4}}{d_{5}}+{d_{6}}{d_{7}}{d_{8}}-\dots+{d_{36}}{d_{37}}{d_{38}}=0$.
Hence ${d_{1}}{d_{2}}-{d_{39}}{d_{40}}{d_{41}}+{d_{42}}{d_{43}}{d_{44}}$ has to be divisible by $13$.
${d_{1}}{d_{2}}-{d_{39}}{d_{40}}{d_{41}}+{d_{42}}{d_{43}}{d_{44}}=22-222+220+{d_{44}}=20+{d_{44}}$.
Therefore $20+{d_{44}}$ has to be divisible by $13$, and therefore ${d_{44}}$ has to be $6$.
