標題:
Maths olympiad
發問:
1. 6x=8^x find all real x2. 54321^43210=? (mod100000) i.e. last five digits =?3. a,x are both real,so that 17x^4+34ax^3+(25a^2+10)x^2+(8a^3+10a)x+(a^4+25)=0. find the product of all possible values of a 4.How many four-digit positive integers have their product of digits being a positive square... 顯示更多 1. 6x=8^x find all real x 2. 54321^43210=? (mod100000) i.e. last five digits =? 3. a,x are both real,so that 17x^4+34ax^3+(25a^2+10)x^2+(8a^3+10a)x+(a^4+25)=0. find the product of all possible values of a 4.How many four-digit positive integers have their product of digits being a positive square number? 更新: 回答者： ?雨後陽光? ( 知識長 ) Thanks for answing. Q1~3 are correct. Q4 The answer should be 809 but I can get it. Here are some problem of your solution: 4) Case 2 : OOXX type , for example : 2288 , 6116 , 9595. 9C2 * 4!/(2! 2!) = 216 ways (How about 6169 as 6*1*6*9=18*18)
最佳解答:
1)6x = 8? 2(3x) = 2^(3x) 3x = 2^(3x - 1)Let 3x - 1 be y , then y+1 = 2?For y > 1 , y+1 2? , no solutions.For y = 0 , it is a solution. i.e. 3x - 1 = 0 , x = 1/3So x = 1/3 or x = 2/3 . 2)54321 ^ 43210 = (54321)^(149 x 290)54321 ^ 149 =...+ 149C4 * 54320?+ 149C3 * 543203 + 149C2 * 543202 + 149 * 54320 + 1149C4 * 54320?(mod 100000) = 149C4 * 5432?(mod 10) * 10000 = 19720001 * 2?(mod 10) * 10000 = 1 * 16 (mod 10) * 10000 = 60000149C3 * 543203 (mod 100000) = 540274 * 54323 (mod 100) * 1000 = 74 * 323 (mod 100) * 1000 = 2424832 (mod 100) * 1000 = 32000149C2 * 543202 (mod 100000) = 11026 * 54322 (mod 1000) * 100 = 26 * 4322 (mod 1000) * 100 = 4852224 (mod 1000) * 100 = 22400149 * 54320 (mod 100000) = 93680So 54321^149 = 60000 + 32000 + 22400 + 93680 + 1 = 208081 = 8081 (mod 100000) So 54321 ^ 43210 (mod 100000) = 8081 ^ 290 (mod 100000) = ... + 290C3 * 80803 + 290C2 * 80802 + 290 * 8080 + 1 (mod 100000)290C3 * 80803 (mod 100000) = 4022880 * 8083 (mod 100) * 1000 = 80 * 83 (mod 100) * 1000 = 40960 (mod 100) * 1000 = 60000 290C2 * 80802 (mod 100000) = 41905 * 8082 (mod 1000) * 100 = 905 * 652864 (mod 1000) * 100 = 905 * 864 (mod 1000) * 100 = 781920 (mod 1000) * 100 = 92000290 * 8080 (mod 100000) = 2343200 (mod 100000) = 43200Hence 54321 ^ 43210 = 60000 + 92000 + 43200 + 1 = 195201 = 95201 (mod 100000) 3)17x? + 34ax3 + (25a2+10)x2 + (8a3 +10a)x + (a?+25) = 0 (16x? + 32ax3 + 24a2x2 + 8a3x + a?) + (x?+ 2ax3 + (a2+10)x2 + 10ax + 25) = 0 (2x + a)?+ (x? + 2ax3 + 10x2 + a2x2 + 10ax + 25) = 0 (2x + a)?+ (x? + 2x2 (ax + 5) + (ax + 5)2) = 0 (2x + a)?+ (x2 + ax + 5)2 = 0 ==> 2x + a = 0 and x2 + ax + 5 = 0x = - a/2 , sub. into x2 + ax + 5 = 0 :a2/4 - a2/2 + 5 = 0 a2 = 20∴ The product of all possible values of a = - 20 4)Let ABCD be the four-digit positive integer. (A , B , C ,D = 1 to 9)Case 1 : OOOO type , for example : 8888 9 ways Case 2 : OOXX type , for example : 2288 , 6116 , 9595. 9C2 * 4!/(2! 2!) = 216 ways 2012-05-20 19:30:25 補充： Case 3 : OOOX type If O is not a perfect square , then O , X = 2 , 8 2P2 * 4C1 = 8 ways If O is a perfect square , then X must be another perfect square , for example , 4449 , 9994 , 1999. 3P2 * 4 = 24 ways Total = 8 + 24 = 32 ways 2012-05-20 19:30:38 補充： Case 4 : OOXY type Then X , Y = 1 , 4 , 9 or X , Y = 2 , 8 7 * 3C2 * 4! / 2! + 7 * 2C2 * 4! / 2! = 252 + 84 = 336 ways 2012-05-20 19:30:51 補充： Case 5 : OXYZ type Type 1 : For example , when O*X , Y*Z are both perfect square , then O*X , Y*Z = 1*4 , 2*8 or 4*9 , 2*8 or 1*9 , 2*8 i.e. O , X , Y , Z = 1 , 4 , 2 , 8 or 4 , 9 , 2 , 8 or 1 , 9 , 2 , 8 4P4 * 3 = 72 ways 2012-05-20 19:31:02 補充： Type 2 : For example , when O*X*Y , Z are both perfect square , then O*X*Y = 2*3*6 , Z = 1 , 4 , 9 or O*X*Y = 2*4*8 , Z = 1 , 9 3C1 * 4P4 + 2C1 * 4P4 = 120 ways 2012-05-20 19:31:08 補充： Type 3 : For example , when O*X*Y*Z is a perfect square , then O*X*Y*Z = 1 * 2 * 3 * 6 4P4 = 24 ways Total 72 + 120 + 24 = 216 ways ∴ There are 9 + 216 + 32 + 336 + 216 = 809 four-digit positive integers have their product of digits being a positive square number. 2012-05-20 19:33:31 補充： 自由自在( 知識長 ) : You help me a lot ! Thankyou very much! 2012-05-20 19:41:24 補充： 起初我以為第4題最易 , 怎料原來它才是最難的! 2012-05-21 16:44:46 補充： 6169 is included in case4 OOXY ==> 6619.
其他解答:
第２題，由 Euler-Carmichael 定理，對任何 (a, 10^5) = 1, a^5000 ≡ 1 (mod 10^5)。 輕易可得： 54321^1250 ≡ 1 (mod 10^5) 4321^625 ≡ 1 (mod 10^5) 因此， 54321^43210 ≡ 4321^85 ≡ (4320+1)^85 ≡ 1 + (85)(4320) + (85)(42)(4320)^2 + (85)(83)(14)(4320)^3 ≡ 1 + 67200 + 68000 + 60000 ≡ 95201 (mod 10^5) 2012-05-21 17:52:15 補充： 也可用 321^625 ≡ 1 (mod 10^5)： 54321^43210 ≡ (321^85) + 40000 ≡ (320 + 1)^85 + 40000 ≡ 40001 + (85)(320) + (85)(42)(320)^2 + (85)(83)(14)(320)^3 ≡ 40001 + 27200 + 68000 + 60000 ≡ 95201 (mod 10^5)。|||||Alternate method for question 2: rewrite 54321 ^ 43210 = (50000 + 4000 + 300 + 20 + 1) ^ (43210) = (1 ^ 43210) + (20 ^ 1) x (1 ^ 43209 ) x 43210 C 1 + (20 ^ 2) x (1 ^ 43208 ) x 43210 C 2 + (20 ^ 3) x (1 ^ 43207 ) x 43210 C 3 + (20 ^ 4) x (1 ^ 43206 ) x 43210 C 4 + (300 ^ 1) x (1 ^ 43209 ) x 43210 C 1 + (300 ^ 2) x (1 ^ 43208 ) x 43210 C 2 + (4000 ^ 1) x (1 ^ 43209 ) x 43210 C 1 + (20 ^ 1) x (300 ^ 1) x (1 ^ 43208 ) x 43210 C 1 x 43209 C 1 + (20 ^ 2) x (300 ^ 1) x (1 ^ 43207 ) x 43210 C 2 x 43208 C 1 + terms divisible by 100000 = 1 + 20 x 43210 + 400 x (43210 x 43209 / 2) + 8000 x (43210 x 43209 x 43208 / 6) + 160000 x (43210 x 43209 x 43208 x 43207/ 24) + 300 x 43210 + 90000 x (43210 x 43209 / 2) + 4000 x 43210 + 6000 x (43210 x 43209) + 120000 x (43210 x 43209 x 43208 / 2) + terms divisible by 100000 = 1 + 64200 + 78000 + 60000 + 63000 + 50000 + 40000 + 40000 + 10A = 95201 + 10A (where A is an integer)|||||培正決賽題目 ？|||||Only 809 cases, please check.
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