New PDF release: A Key to Chinese Speech and Writing: Vol. I

By Joel Bellassen, Zhang Pengpeng, Joel Bellasen

This introductory path not just serves as a textbook but additionally as a consultant which, via quite a few routes (etymology, semantics, linguistics, pics and culture), might be useful imagine in chinese language. the purpose of this direction is to give the chinese because it is: a global of indicators with their foundation and the good judgment in their composition, represented via the main often used ones. The four hundred characters featured in quantity 1 will enable you realize no less than 70% of the characters utilized in books and newspapers.

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Extra resources for A Key to Chinese Speech and Writing: Vol. I

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9 ); 100 = 1 (mod. 9 ); 1000 = 1 (mod. 9 ); ___ Hence, the given number is congruent to a + b + c + d + . . (mod. 9). Thus, 275 = 5 + 7 + 2 = 14 == 5 (m od. 9). It follows from these considerations that an integer, written in the decimal system, yields the same remainder when divided b y 9 as does the sum of its digits. In particular, a number is divisible b y 9 if, and only if, the sum of its digits is also divisible b y 9. Hence, if we m ultiply two large numbers and obtain a product P whose correct­ ness is in doubt, we may em ploy the following check: Instead of mul­ tiplying the original numbers, we m ultiply the sums of their digits; the product of these two sums must be congruent (mod.

If the sums in question exceed 9, we m ay in their stead take the sums of their digits, that is, the sums of the digits of the sums of the digits. Exam ple: 7744-88 = 681472. Check: The sum of the digits of 7744 is 22; the sum of the digits of 22 is 4. The sum of the digits of 88 is 16, whose digits in turn add up to 7. The sum of the digits of 681472 is 28. 19 W hat we have here said regarding 9 is true of 3 as well, for 10 is also congruent to 1 modulo 3. This check, however, is not as powerful since, even if our calculations were false, there would still be a prob­ ability of 1 /3 that the wrong result is congruent to the correct one modulo 3, while in the case of 9 the probability is only 1/9.

The first problem is fairly simple. It is evident that we can decom ­ pose any natural number n into prime factors. For, b y a lim ited num­ ber of trials, we m ay ascertain whether among the natural numbers smaller than n there are such as divide n\ if there is no such number, n is prime, and the desired factorization is sim ply n = n. If n has at least one proper divisor ni such that n = n\-n2 (« i and n% differing from 1, hence also from n ) 2 we m ay repeat the attem pt at factorizing with respect to ni and n2.

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