"Remainder multiplication" divides one by the other
Proof of Sun Tzu's Theorem: An Imperfect and Unstable Manifestation
By Jinggang Zhang zhangyi003@yahoo.cn
www.sxcopyright.com 2008.08.08
Sun Tzu's Theorem:
Example Solve the congruence group
Solution Since 3, 5, 7 are mutually prime, the solution can be given by Sun Tzu's Theorem, = 3 5 7 = 105,
Therefore, by Sun Tzu's Theorem, the solution to the congruence equation is given as follows: ≡ 2 35 2 + 1 21 3 + 1 15 2 (mod 105) that is
≡ 23 (mod 105).
The solution to Sun Tzu's theorem above is to calculate the multiplicity × derivative × remainder of the sum of the items, minus the two products and get a number, it is not perfect and the solution is more complex, popularization of the application of a certain degree of difficulty, and also unstable.
With the "residual multiplication method" to "Sun Tzu's theorem" is simplified into a general solution, so that the remaining problem is solved, that is, there are positive bases, there are negative bases, there are positive residuals, there are also negative residuals. Mutually divisible by 1 can be solved, mutually divisible by less than 1 can also be solved (without limiting the problem of large residuals), the solution is transformed into a general algorithm, making it perfect, stable and universal application.
With Pan Chengdong, Pan Chengbiao 2005, Peking University Press, page 157, a simple number theory:
Example X ≡ 3 (mod 8)
X ≡ 1 (mod 5)
X ≡ 1 (mod 3) Answer X ≡ -29 (mod 120)
Using the "residual times" simplified comparison. The answer □ = 91 using the "residual multiplication" simplified comparison.
3......1
□ ÷ 5......1
8......3
< p> According to the counterfactual: the minority of the remainder of the lower equation is the "complementary number" of the upper equation (e.g., 4 ÷ 3 = quotient 1 remainder 1, if = quotient 2 is less than 2), which is called the negative remainder.3......1 is less than 2
□ ÷ 5......1 is less than 4
8......3 is less than 5
To find the positive and negative numbers by doubling the number of points:
Positive base 40 + 96 + 105 = 241
Divisor 3 × 5 × 8 = 120
Negative base 80 + 24 + 15 = 119
Solve by formula method 1: add the residue × base items and divide by the product residue.
① positive base, positive remainder
(1×40 + 1×96 + 3×105) ÷ (3×5×8)
=451÷120......91
② positive base, negative remainder
(2×40 + 4×96 + 5×105) ÷ (3×5×8)
=451 ÷ 120
② positive base, negative remainder
(2×40 + 4×96 + 5×105) ÷ (3×5×105)
= 989÷120......29
③ Negative base, negative remainder
(2×80 + 4×24 + 5×15) ÷ (3×5×8)
= 331÷120... ...91
④ Negative base, positive remainder
(1 × 80 + 1 × 24 + 3 × 15) ÷ (3 × 5 × 8)
= 149 ÷ 120 ......29
Obviously, reducing with 29 Adding a remainder, reducing a number, doesn't fit the the question, reduction with negative -29 meets the question minus the remainder, plus a few, but the -29 comes from an implicitly obvious and unconvincing source. (Low-level schools can not accept)
With 91 reduction minus the remainder, plus a few, in line with the meaning of the question, 91 is the correct answer.
The solution is basically the same as Sun Tzu's Law, but with two answers.
If you use the "residual division method" to divide the remainder by one and the minority by one, there are no two answers.
Solve by method 2:
① Positive base, positive remainder
(3×□+1)÷5=□......1
｛6(5+1-1)+1｝÷(5×3)
=31÷15 ......1
(15×□+1)÷8=□......3
｛105(8+3-1)+1｝÷(8×15)
=1051÷120... ...91
Method 2 solution:
③ Use a positive base and a negative remainder
(3 × □ - 2) ÷ 5 = □ ...-4
｛6 (5 - 4 + 2) -2｝÷(3×5)
=16÷15......1
(15×□ + 1)÷8=□...-5
｛105(8 -5-1) + 1｝÷(8×15)
= 211÷120......91
Method 3 solution:
② Negative base, negative remainder
(3×□ -2)÷5=□...-4
｛9(5+4-2)-2｝÷(3×5)
= 61÷15... ...1
(15 × □ + 1) ÷ 8 = □ ...-5
｛15 (8 + 5 + 1) + 1｝｝÷ (8 × 15)
=211÷120......91
Method 3 solution:
④ Negative base, positive remainder
(3 x □ + 1) ÷ 5 = □......1
｛9(5-1 + 1) + 1｝÷(5 x 3)
=46÷15 ......1
(15×□+1)÷8=□......3
｛15(8-3+1)+1｝÷(8×15)
=91÷120 ......91
Answer □ = 91
Prove it again, using the "residual multiplication method" to solve: "Things don't know how many"
3......2
□ ÷ 5...3
□ ÷ 5... ...3
7......2
According to the counterfactual method: the minority of the remainder of the lower equation is the "complement" of the upper equation (e.g., 5 ÷ 3 = quotient 1 is 2, if = quotient 2 is 1 less), which is called the negative remainder.
3 ......2 is 1 less
□ ÷ 5 ......3 is 2 less
7 ......2 is 5 less
To calculate the positive and negative by doubling the number:
The base number:
The negative and positive numbers are calculated by doubling the number. Base:
Positive base 70 + 21 + 15 = 106
Divide 3 × 5 × 7 = 105
Negative base 35 + 84 + 90 = 209
Using the formula residue multiplication method, method 1 to solve: residue × base of the sum of the items, at the product of the residue is both.
① with a positive base, positive residual solution:
(2 × 70 + 3 × 21 + 2 × 15) ÷ (3 × 5 × 7)
= 233 ÷ 105 ......23
② with a positive base, negative residual solution:
(1 × 70 + 2 × 21 + 5 × 15) ÷ (3 × 5 × 15) ÷ (3 × 5 × 15) ÷ (3 × 5 × 15) ÷ (3 × 5 × 15) ÷ (3 × 5 × 5 × 5) ÷ (3 × 5 × 5 × 5) ÷ (3 × 5 × 5 × 5) 15) ÷ (3×5×7)
=187÷105......82
③ Solve with a negative base, negative remainder
(1×35 + 2×84 + 5×90) ÷ (3×5×7)
=653 ÷105... ...23
④ Negative base, positive remainder Solution:
(1 x 35 + 2 x 84 + 5 x 90) ÷ (3 x 5 x 7)
=502 ÷ 105......82
Reduce the remainder by 23, plus a few.
Reduce the plus remainder with 82, and the minus number. Reduce the minus remainder with -82, plus a few. (Unacceptable in lower schools)
The above solution is essentially the same as Sun Tzu's Law, but with two answers.
If you use the "residue multiplication method" to divide the remainder by one and the minority by one, there are no two answers.
Method 2:
① Positive base, positive remainder
(3 × □ + 2) ÷ 5 = □ ......3
｛6 (5 + 3-2) + 2｝ ÷ (5 × 3)
= 38 ÷ 15 ......8
(15×□+8)÷7=□......2
｛15(7+2-8)+8｝÷(7×15)
=23÷105... ...23
Method 2 Solution
② Use a positive base and a negative remainder
(3 × □-1) ÷ 5 = □...-2
｛6 (5 - 2 + 1) -1｝÷(3×5)
=23÷15......8
(15×□ + 8)÷7=□...-5
｛15 (7 -5-8) + 8｝÷(7×15) (according to the note: 7 can be expanded by a factor of 2)
=23÷105......23
Method 3 solution:
③ Negative base, negative remainder
( 3×□-1)÷5=□...-2
｛9(5 + 2-1)-1｝÷(3×5)
=53÷15... ...8
(15 × □ + 8) ÷ 7 = □ ...-5
｛90 (7 + 5 + 8) + 8｝｝÷ (7 × 15)
=1808÷105...... 23
Method 3 Solution
④ Negative Base, Positive Remainder
(3 x □ + 2) ÷ 5 = □......3
｛9 (5 - 3 + 2) + 2｝÷ (5 x 3)
=38÷15 ......8
(15×□+8)÷7=□......2
｛90(7-2+8)+8｝÷(7×15)
=1178÷105 ......23
Answer □ = 23
From the above comparison that "Sun Tzu's Theorem", the solution is complex, and sometimes unstable, "residual multiplication method" is stable no matter what the situation, and the solution is simple, easy to popularize, more suitable for solving application problems.
Example: A student who lives in a school, his family gives him 36 dollars a week for living expenses. The actual daily living expenses of the student only 5 yuan, one day his sister-in-law to the school to see him and gave 50 yuan, he used this money to buy two favorite extracurricular books to spend 10 yuan, to buy learning tools to spend 2 yuan, after returning home after the holidays to explain the situation and return 55 yuan to the parents.
Q: How many weeks of living expenses did the student bring? How many days actually live in school? A **** how much money? How much money was spent?
Solve by method 2:
Equation (36 × □ + 50-10-2) ÷ 5 = □......55 yuan
｛36 × (5 + 55-50 + 10 + 2) + 50 -10-2｝÷(5×36)
= (36×22 + 50-10-2)÷180
= 830÷180...... 110
Answer; 1, (110 - 50 + 10 + 2) ÷ 36 = 2, (the smallest number in □ inside the parentheses)
2, (110 - 55) ÷ 5 = 11, (the smallest number in □ outside the parentheses)
3 36 × 2 + 50 = 122,
4. 122 - 55 = 67.
A. The student brought 2 weeks of living expenses, actually lived in school for 11 days, and had $122 in one **** and spent $67.