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XV.—Essay towards a General Solution of Numerical Equations of all Degrees 
having Integer Roots. By H. F. Tarzot, F.R.S. 
(Read 17th May 1875.) 
The method of solution which I have the honour to lay before the Society 
is founded upon an old and well-known arithmetical process called “ casting out 
the nines.” It consists in substituting for any number the remainder which 
that number leaves when divided by nine. As this remainder contains only a 
single figure, whereas the original number may contain any number of figures, 
it is obvious that if the remainder can be employed instead of the number itself, 
a great saving of time and trouble must follow. 
_ Thus, for example, let 274511358777 be the proposed number. In order to 
find the remainder, it is not necessary to divide it by 9, which might occasionally 
be troublesome. It is sufficient to add the figures together by 2 or 3 at a time, 
till the sum equals (or surpasses) zine; then erase those figures, and employing 
the surplus only, continue the process. Thus, in the given example, 2 + 7 
making 9, are erased or cast aside. Then come 4 + 5 = 9, which are similarly 
dropped ; then come 1 + 1 + 3 +5 = 10, which is to be accounted as 1 only 
(9 being omitted.) This 1, with 8 following, make 9, and are therefore cast 
aside. Finally, 7 + 7 + 7 = 21, which exceeds a multiple of 9 by 3, and there- 
fore 3 is written for it. The remainder of the given number is accordingly 
three. 
In future I shall call the remainder of a number, when divided by 9, its 
reduced form, or more simply, its reduct. 
Another mode of obtaining the reduct, but less simple, is to add together a// 
the figures of the given number (which in this example would make 57) : then 
to treat this result as a new number and find its reduct, thus: 5 + 7 = 12. 
Again, treating 12 as a new number, we have 1 + 2 = 3, therefore the final 
reduct is 3, as we found before 
Theorem 1.—If numbers are added together, the reduct of their sum equals 
the sum of their (partial) reducts. 
Theorem I1.—If numbers are multiplied echee the reduct of their product 
equals the product of the (partial) reducts. 
| This is only an extension of Theorem I. 
Examples: Add together 84147, whose reduct is 6, 
| and 13201 . reduct 7 
97348 . reduct 4. 
VOL. XXVII. PART III. aK 
