BÀI BÁO CÁO KHOA HỌC TẠI BẮC KINH ATCM 2009 THE SCIENTIFIC REPORT IN BEIJING - CHINA

The 14 th ASIAN TECHNOLOGY CONFERENCE IN MATHEMATICS, BEIJING, CHINA 2009 BEIJING NORMAL UNIVERSITY December 17-21, 2009 THE MINH TRAN HCMC UNIVERSITY OF TECHNOLOGY VIETNAM TITLE : USING SCIENTIFIC CALCULATORS TO SOLVE MATHEMATICAL PROBLEMS Authors : THE MINH TRAN DEPARTMENT OF APPLIED MATHEMATICS HCMC UNIVERSITY OF TECHNOLOGY (HCMUT) VIETNAM Introduction • Using a calculator is an important skill for students. As they grow and progress through their education. • They will learn advanced math skills that certainly require them to use a calculator . • A few quick steps can help anyone use the device and solve mathematical problems . • I have organized exams : “Excellent students use scientific calculators to solve mathematical problems ” Therefore, I have built math-calculator subject and technology in mathematics WE CAN USE SCIENTIFIC CALCULATORS : 1.VIETNAM CALCULATOR VN- 570RS 3.CASIO FX-570MS 4.CASIO FX-570ES2.TEXAS INSTRUMENTS 83 Plus A. ARITHMETIC • We calculate : 3281978 × 8791803 • We input 3281978 × 8791803 on the screen • Press the = key. Answer : 2.885450403 • We know that answer have 14 digits but the digit 3 can have rounded • We delete the digit 3 at the first number and the digit 8 at the second number then we multiply two numbers together • Press the = key. Answer : 2.232710263 We continue to delete the digit 2 at the first number and the digit 7 at the second number then we multiply 81978 and 91803 to get 7525826334 • Result : 3281978 × 8791803 = 28854504026334 Note : when you multiply two numbers, you need to thoroughly check digits 13 10 × 1.Multiply two numbers 11 10× 2. Find x, know that : • We input 381978 ÷ 382007 on the screen • Press the = key. Answer : 0.999924085 • We continue to press : × 3 - 8 then we press 9 times the = key. The final result have saved to the Ans key • Now,we have then we press : Ans - 1 = Result : x = - 1.11963298 3 8  3 8  3 8  3 8  3 8  3 8  3 8  3 8  3 8  1 1  x  381978 382007 1 − x x Ans + = 1 1 1− x 3. Find a natural number n , know that the number contains the first four digits and the end four digits are digit 1 Step 1 : Find the number contains the end four digits are digit 1. Solution : The number contains the end digit is digit 1 The number contains the end two digits are digit 1 The number contains the end three digits are digit 1 The number contains the end four digits are digit 1 Answer : the number contains the end four digits are digit 1 n1 = 8471 3 n 3 1 3 71 3 471 3 8471 3 8471 Step 2 : Find the number contains the first four digits are digit 1. Solution : We have ( we get the number 223 ) We try ( answer contains two digit are digit 1 ) . we do not choose the number 223 The number contains the first two digit are digit 1 We have ( we get the number 480 ) We try ( answer contains four digits are digit 1 ) We choose the number 480 . Answer : n2 = 480 If we do not find n2, we can calculate to find n2 Result : n = 4808471 and 193 10 .1216.12238471 ×= 7.480111101111 3 = 203 10 .11178.14808471 = .11111000111;1111001111 33 14.22311111111 3 = 3 22308471 3 4808471 1111. . .1111 = 4. Find natural number x,y, know that : ( and y include a digit) Solution : we guess : x is digit 2 or 3 ● When x = 2 : (we do not choose x = 2 ) ● We choose the digit 3 , we have : ( ) 2 2 2 5 541 1 ( )4481448( ) x y y y x x y= 2 x 4 4 (5 ) 59 12117361y ≤ = 9 9 9 5410109448144809 5 541 1 94481448 9 5419199448144899y y y y ≤ = ≤ 55,99303467 5 56, 00348007 6y y ⇔ ≤ ≤ ⇒ = Result : x = 3 , y = 6 and 5416169448144896 Furthermore , we can find y by the SOLVE key.We input ( ) 9 13 12 11 10 9 2 50 541 10 10 1 10 10 9 10 4481448 10 90y y y y + = × + × + × + × + × × × + + on the screen. We press SHIFT SOLVE key and press 1 key 6y ⇒ = 9 56 = 5. Find the fourteenth digit after of the decimal point of We have : the digit 5 can have rounded We try the digit 5 can not have rounded We set : The second power of two sides of equation : We have : (1)  We use function of calculator to solve equation : We have roots : We only get root 12 12 3.464101615 . = Solution : 10 12 3.464101615 1.4 10 0 − − ≈ × > 12 3,464101615 ( 0)x x = + > 2 2 2 3.464101615 3.464101615 12 0 (1)x x + × + − = 2 10 3.464101615 12 9,54391775 10 − − = − × . 2 10 2 3.464101615 9,54391775 10 0x x − + × − × = 10 1 1,37754 10x − = × 2 6.92820323x = − 1 x 12 3, 464101615137754 . ⇒ = Result : 5 [...]... Str2 : Input “H(X, Y, Z, T) = 0 =”, Str3 : Input “K(X, Y, Z, T) = 0 =”, Str4 : Lbl 1 : Input « Xo = », A : Input « Yo = », B : Input « Zo = », C : Input « To = », D :1 →N : String  Equ(Str1, Y1) : String  Equ(Str2, Y2) : String  Equ(Str3, Y3) : String  Equ(Str4, Y4) : While abs(X-A) > 1E-12 or abs(Y-B) > 1E – 12 or abs(Z-C) > 1E – 12 or abs(T-D) > 1E-12 : Radian : N+1 → N : if (N The end five numerals of P are : P = 1019739 - 82689 = 37050 Result : P = 526837050 (we are sure that the digit 8 is true because the after digit of the digit 8 is the digit 3) 7 solve linear equations with four unknowns :  x + y + z + t = 100  x, y , z , t ∈ Ζ + 0,5 x + 6 y + 4 z + 7t = 488... Y2][O, P, Q, Y3][S, U, V, Y4]])/J → A : End : End : End : Disp « X= »,X : Disp “Y =”, Y : Disp “Z =”,Z : Disp « PRESS QUIT TO EXIT » (1) : Goto 1 Ex : Solve the following unlinear equations : Solution : The TI calculator will have result in 2 minutes x=1.073805344; y=1.896586209; z=13.99719752; t=3.999983857 x=0.5624970866; y=1.514473157; z=17.34656344; t=3.995849184 x=0.2226617235 ; y=1.081330382;... nDeriv(Y3, Z, Z) → Q : nDeriv(Y3, T, T) → R : nDeriv(Y4, X, X) → S : nDeriv(Y4, Y, Y) → U : nDeriv(Y4, Z, Z) → V : nDeriv(Y4, T, T) → W : det([[E, F, G, H][I, K, L, M][O, P, Q, R][S, U, V, W]]) → J : if(J=0) : Then : Disp “NO ROOT, MULTIPLICITY ROOT OR.XO, YO” : Goto 1 : Else : X – det ([[Y1, F, G, H][Y2, K, L, M][Y3, P, Q, R][Y4, U, V, W]])/J → A : Y – det ([[E, Y1, G, H][I, Y2, L, M][O, Y3, Q, R][S, Y4, V,... String  Equ(Str3, Y3) : String  Equ(Str4, Y4) : While abs(X-A) > 1E-12 or abs(Y-B) > 1E – 12 or abs(Z-C) > 1E – 12 or abs(T-D) > 1E-12 : Radian : N+1 → N : if (N . The 14 th ASIAN TECHNOLOGY CONFERENCE IN MATHEMATICS, BEIJING, CHINA 2009 BEIJING NORMAL UNIVERSITY December 1 7-2 1, 2009 THE MINH TRAN HCMC. : The number contains the end digit is digit 1 The number contains the end two digits are digit 1 The number contains the end three digits are digit 1 The