A problem in fractions from a very old arithmetic textbook

Similar in vein to a problem I posted before here, I would be interested if anyone can give me any pointers as to how one might solve this question from the same arithmetic textbook:

"Simplify

$$\frac{23401369863013698630136986301369629}{34500729927007299270072992700729582}"$$

It is important to note that the book was originally published in 1947, well before any electronic calculation aids were available in schools, so there must be a method which can be carried out on pencil and paper without a ridiculous amount of effort (and paper).

I have tried looking for ways of spotting some pretty large common factors of the numerator and denominator which could be easily cancelled, but without success. There ought to be some cunning way to do this which makes use of the "almost periodic" nature of the digits in both numbers, but I have failed to see it.

Note: the simplified answer has just 7 digits in the numerator and denominator.

Solution 1:

The repeating block in the numerator is $01369863$; the last block, $01369629$, is defective by exactly $234$, the first three digits. The same thing happens in the denominator: the repeating block is $00729927$, and the last block, $00729582$, is defective by $345$, the leading three digits. Let $a=01369863$, $b=00729927$, and $x=10^8$; then the fraction is

$$\begin{align*} \frac{234x^4+a(x^3+x^2+x+1)-234}{345x^4+b(x^3+x^2+x+1)-345}&=\frac{234+\frac{a}{x-1}}{345+\frac{b}{x-1}}\cdot\frac{x^4-1}{x^4-1}\\\\ &=\frac{234(x-1)+a}{345(x-1)+b}\\\\ &=\frac{234\cdot99999999+1369863}{345\cdot99999999+729927}\\\\ &=\frac{23400000000-234+1369863}{34500000000-345+729927}\\\\ &=\frac{23401369629}{34500729582}\\\\ &=\frac{2600152181}{3833414398}\\\\ &=\frac{236377471}{348492218}\;, \end{align*}$$

Still not quite down to six digits, but better; the last two steps took out factors of $9$ and $11$, which are recognizable by elementary tests. Knowing that it can be further simplified, I might think of testing for divisibility by $101$: $71-74+37-36+2=0$ and similarly in the denominator. That reduces it finally to

$$\frac{2,340,371}{3,450,418}\;.$$

(In fact I got that ex post facto from Old John’s comment.)

Solution 2:

Evidently the gcd is a long calculation by hand, and prone to error in any case. So, trickery it is. Does it say anything relevant in the text?

Tue Nov 26 13:37:03 PST 2013
GCD ( 34500729927007299270072992700729582, 23401369863013698630136986301369629 ) 
 quotient 1
GCD ( 23401369863013698630136986301369629, 11099360063993600639936006399359953 ) 

 quotient 2
GCD ( 11099360063993600639936006399359953, 1202649735026497350264973502649723 ) 

 quotient 9
GCD ( 1202649735026497350264973502649723, 275512448755124487551244875512446 ) 

 quotient 4
GCD ( 275512448755124487551244875512446, 100599940005999400059994000599939 ) 

 quotient 2
GCD ( 100599940005999400059994000599939, 74312568743125687431256874312568 ) 

 quotient 1
GCD ( 74312568743125687431256874312568, 26287371262873712628737126287371 ) 

 quotient 2
GCD ( 26287371262873712628737126287371, 21737826217378262173782621737826 ) 

 quotient 1
GCD ( 21737826217378262173782621737826, 4549545045495450454954504549545 ) 

 quotient 4
GCD ( 4549545045495450454954504549545, 3539646035396460353964603539646 ) 

 quotient 1
GCD ( 3539646035396460353964603539646, 1009899010098990100989901009899 ) 

 quotient 3
GCD ( 1009899010098990100989901009899, 509949005099490050994900509949 ) 

 quotient 1
GCD ( 509949005099490050994900509949, 499950004999500049995000499950 ) 

 quotient 1
GCD ( 499950004999500049995000499950, 9999000099990000999900009999 ) 

 quotient 50
GCD ( 9999000099990000999900009999, 0 ) 

Tue Nov 26 13:37:03 PST 2013