Simplify $\frac{m}{n}$ if:$m=2244851485148514627\\n=8118811881188118000$

Question

I found this task in Selected Problems in Elementary Mathematics by Neven Elezović a few years ago and tried to find a better way of solving this today than I did back in high-school. Unfortunately, I haven't done it, so I was wondering if there is another, faster approach. When I searched for the task, there was only one comment saying it's impossible without a calculator, but that was false.

Simplify $\frac{m}{n}$ if:

$$m=2244851485148514627\\n=8118811881188118000$$

I noticed the patterns: $p_m=8514\\p_n=8118.$

$m=2244\cdot10^{15}+8514\cdot(10^{11}+10^7+10^3)+627\\\;\;\;=(8514-6270)\cdot10^{15}+8514\cdot(10^{11}+10^7+10^3)+627\\\;\;\;=8514\cdot(10^{15}+10^{11}+10^7+10^3)-627\cdot(10^{16}-1)\\\;\;\;=8514\cdot(10^{15}+10^{11}+10^7+10^3)-627\cdot(10^4-1)\cdot(10^{12}+10^8+10^4+1)\\\;\;\;=8514\cdot10^3(10^{12}+10^8+10^4+1)-627\cdot(10^4-1)\cdot(10^{12}+10^8+10^4+1)\\\;\;\;=(10^{12}+10^8+10^4+1)(8514\cdot10^3-6270\cdot10^3+627)\\\;\;\;=2244627\cdot(10^{12}+10^8+10^4+1)$

$n=8118\cdot(10^{15}+10^{11}+10^7+10^3)=8118\cdot10^3\cdot(10^{12}+10^8+10^4+1)$ $$\frac{m}{n}=\frac{2244627\cdot(10^{12}+10^8+10^4+1)}{8118\cdot10^3\cdot(10^{12}+10^8+10^4+1)}=\frac{2244627}{4059\cdot2\cdot10^3}=\frac{503}{2000}$$ /typo corrected from $503$ to $553$/ Thanks for the remark in comments: what I meant to ask initially was if there was some other way of solving this to reduce calculations to the minimum.

Answer 1

The first step in a simple solution would be to look for common factors to reduce the numbers $m,n$ to something more manageable. The repeated $8118$ is a good place to start, as that number is readily recognized as divisible by $2$ and $9$ (since its digital root is $9$). Quickly, you find that $8118=2\cdot 9\cdot 11\cdot 41$. $m$ is odd, hence not divisible by $2$, but its digital root is also $9$, so it is divisible by $9$. I was able to do all next steps with pencil and paper, no calculator needed.

$\frac{m}{9}=249427942794279403$. Now try division of that result by $11$, and it works, yielding $22675267526752673$. So try division by $41$, and it works too, yielding $553055305530553$. This can be represented as $553(1000100010001)$

$\frac{n}{9}=902090209020902000$; division of that by $11$ yields $82008200820082000$; and division of that by $41$ yields $2000200020002000$. This can be represented as $2000(1000100010001)$

Final step: $\frac{553(1000100010001)}{2000(1000100010001)}=\frac{553}{2000}$

I'm not sure this is entirely easier than the approach you took, which seems to me to try to discover the large factor of $1000100010001$ first, but it requires no arithmetic other than simple division.

Answer 2

If I had to do it by hand and didn't notice the pattern, I would just use the Euclidean algorithm to find $\gcd(n,m)$, so $$ 8118811881188118000 \bmod 2244851485148514627=1384257425742574119\\ 2244851485148514627\bmod 1384257425742574119=860594059405940508\\ 1384257425742574119\bmod 860594059405940508 =523663366336633611\\ 860594059405940508 \bmod 523663366336633611 =336930693069306897\\ 523663366336633611 \bmod 336930693069306897= 186732673267326714\\ 336930693069306897 \bmod 186732673267326714 =150198019801980183\\ 186732673267326714 \bmod 150198019801980183= 36534653465346531\\ 150198019801980183 \bmod 36534653465346531= 4059405940594059\\ 36534653465346531 \bmod 4059405940594059=0$$ and we have the factor we can pull out. Clearly noticing the digit repetition reduced the work considerably. The problem was designed to make that work. Most problems in books with numbers this large are designed that way.