Talk:Vedic Mathematics/Sutras/Ekadhikena Purvena

"There will be (divisor-1) terms in the answer." No, there won't. At least, not in every case. It's probably a coincidence that it works for 1/19 and 1/29, because e.g. for 1/49 the answer has 42 repeating digits, not 48. The book of Vedic Mathematics doesn't say that there should be (divisor-1) digits in the answer, but it says that when you come to the (divisor-1) number, you have half the answer. And they have in mind the current remainder, not the number of digits. If you look for 48 as a remainder in the answer, it'll show up at step 21, which is exactly half the correct 42-digit answer.

But even still this technique doesn't work as a whole for 1/39, and I don't know why. I mean, the division/multiplication by (x-1) still works, but the stop condition doesn't. You never achieve the number 38 (divisor-1) as a remainder in the answer, and it doesn't have 38 digits either - it has only 6 repeating digits: 0.(025641)... . The 38 isn't even divisible by 6, so it can't match the repeating pattern with more cycles at 38. And even the rule that half the answer is 9-complement the first half of the answer - it doesn't work too.

What is so special about 1/39 that this method doesn't work for it? It looks it doesn't matter that it isn't prime, because 49 isn't prime also, but the technique works for 1/49. -- SasQ