184 7. ANCIENT NUMBER THEORY7.10. Take as a unit of time Ô = ^ of a year, about 37 hours, 18 minutes, say a
day and a half in close approximation. Then one average lunar month is Ì = 19T,
and one average solar year is Õ — 235T. Given that the Moon was full on June 1,
1996, what is the next year in which it will be full on June 4? Observe that June
4 in whatever year that is will be 3 days (2Ã) plus an integer number of years. We
are seeking integer numbers of months (x) and years (y), counting from June 1,
1996, such that Mx = Yy + 2T, that is (canceling T), 19x = 235j/ + 2. Use the
kuttaka to solve this problem and check your answer against an almanac. If you
use this technique to answer this kind of question, you will get the correct answer
most of the time. When the answer is wrong, it will be found that the full moon
in the predicted year is a day earlier or a day later than the prescribed date. The
occasional discrepancies occur because (1) the relation Ì = 19T is not precise, (2)
full moons occur at different times of day, and (3) the greatest-integer function is
not continuous.
7.11. Use Bhaskara's method to find two integers such that the square of their sum
plus the cube of their sum equals twice the sum of their cubes. (This is a problem
from Chapter 7 of the Vija Ganita.)
7.12. The Chinese mutual-subtraction algorithm (the Euclidean algorithm) can
be used to convert a decimal expansion to a common fraction and to provide ap-
proximations to it with small denominators. Consider, for example, the number
e « 2.71828. By the Euclidean algorithm, wc get
271,828 = 2-100,000 + 71,828
100,000 1 · 71,828+28,17271,828 = 2 · 28,172+ 15,484
28,172 1 · 15,484+ 12,68815,484 = 1 · 12,688+2,796
12,688 = 4-2,796+ 1,504
2,796 = 1 • 1,504+ 1,292
1,504 1 • 1,292 + 212
1,292 â-212 + 20(^212) = ßï •20+12
(^20) = 1 • 12 + 8
(^12) = 2 • 8 + 4
(^8) = 2-^4
Thus the greatest common divisor of 271,828 and 100,000 is 4, and if it is divided
out of all of these equations, the quotients remain the same. We can thus write