plot summary:
[Machine Xiao Wei] in the [engineer Ah Wei] escorted into the [nine turn Elixir] seventh turn of the cultivation.
This is to be studied [elementary number theory Preliminary].
Drama Start:
Star Calendar May 08, 2016 10:05:37, the Milky Way Galaxy Earles the Chinese Empire Jiangnan Line province.
[Engineer Ah Wei] is working with [machine Xiao Wei] to study [elementary Number theory Preliminary].
<span style= "FONT-SIZE:18PX;" >710316 can be divisible by 3 710316 can be divisible by 9 710316 cannot be divisible by 11 710316 cannot be divisible by 7 # According to the Law Division of the Division of Sex # is actually not practical, just explain the Rules def TMP (): number = 710316; Array = numberarray (number); size = len (array); #被11整除 array_odd = Array[0:size:2]; Array_even = Array[1:size:2]; #被7整除 array_last3 = number%1000; Array_front = (number-array_last3)//1000; #判断被3整除 if sum (array)%3==0:print (' {0} can be divisible by 3 '. Format (number)); Else:print (' {0} cannot be divisible by 3 '. Format (number)); #判断被9整除 if sum (array)%9==0:print (' {0} can be divisible by 9 '. Format (number)); Else:print (' {0} cannot be divisible by 9 '. Format (number)); #判断被11整除 if (sum (array_odd)-sum (array_even))%11 = = 0:print (' {0} can be divisible by 11 '. Format (number)); Else:print (' {0} cannot be divisible by 11 '. Format (number)); #判断被7整除 if (array_last3-array_front)%7==0:print (' {0} can be divisible by 7 '. Format (number)); Else:print (' {0} cannot be divisible by 7 '. Format (number)); #把一个整数的各位数字放入数组 # the Def numberarray (number) is arranged by a single digit: array = []; bit = Len (stR (number)); For I in range (bit): Array.append (NUMBER%10); number//=10; #array. reverse (); Return array;</span>
<span style= "FONT-SIZE:18PX;" > #判断一个数是否质数 def Prime (num): if (num < 2): return False; SQR = Int (math.sqrt (num)) +1; For I in range (2, SQR): if (num%i==0): return False; return True; </span>
<span style= "FONT-SIZE:18PX;" > #最大公约数def gcd (M, N): m, n = max (M, N), Min (m, n); i = 1; While N: print (' Step ', I, ': ', M, n); i + = 1; m, n = n, m% n return m</span>
<span style= "FONT-SIZE:18PX;" > #最小公倍数def LCM (M, N): return M*N/GCD (M, N);</span>
<span style= "FONT-SIZE:18PX;" >>>> 2625.0print (LCM (375,;</span>))
<span style= "FONT-SIZE:18PX;" > #分解质因数 def primefactor (num, lists): if (num < 2): lists.append (num); return lists; Elif (prime (num) = = True): lists.append (num); return lists; else: sqr = int (math.sqrt (num)) +1; i = 2; While I <= SQR: if (num% i = = 0 and prime (i) = = True): lists.append (i); num = num//i; break; I+=1; return Primefactor (num, lists); </span>
<span style= "FONT-SIZE:18PX;" >>>> 720 = 2 * 2 * 2 * 2 * 3 * 3 * 5152 = 2 * 2 * 2 * 19813680.0def TMP2 (): Number = [720]; For j in Range (len (number)): factors = primefactor (Number[j], []); Count = len (factors); s = str (number[j]) + ' = '; For I in range (count): s + = str (factors[i]); if (I < count-1): s + = ' * '; print (s); Print (GCD (720,152)); Print (LCM (720));</span>
<span style= "FONT-SIZE:18PX;" >>>> 21 Euler function value φ (+) = 12.0def Tmp3 (number): print (' {0} ' Euler function value φ ({0}) = {1} '. Format (numbers, round ( Eulerphi (number), 3))) #欧拉函数def Eulerphi (number): factors = set (Primefactor (number, [])); factors = sorted (factors); #print (factors); phi = number; For I in range (len (factors)): Phi *= (1-1/factors[i]); Return phi;</span>
<span style= "FONT-SIZE:18PX;" >>>> 25 Euler function value φ = 20.0def tmp4 (): For I in range (1): if (18**i)%25==1: print (i); Break;</span>
<span style= "FONT-SIZE:18PX;" >>>> [4.0, 9.0, 14.0]def Tmp5 (): result = []; MoD =; remain = 6; quotient = 9; For I in range (1): x = (mod*i + remain)/quotient; If ABS ((int (x))-X) < 1e-6: result.append (x); If x > mod: Break ; Print (Result);</span>
<span style= "FONT-SIZE:18PX;" >>>> [9.0] #例6def TMP5 (): result = []; MoD =; remain = 1; quotient =; For I in range (1): x = (mod*i + remain)/quotient; If ABS ((int (x))-X) < 1e-6: result.append (x); If x > mod: Break ; Print (Result);</span>
<span style= "FONT-SIZE:18PX;" >var s = [' F (a) =e, F (b) =f, F (c) =g ', ' a=-1, e=2; ', ' b=0, f=3; ', ' c=1, g=6; ', ' f (x) = 2 (x-0) (x-1)/(-1)/(-2) ' , ' +3 (x+1) (x-1)/(1)/( -1) ', ' +6 (x+1) (x-0)/(2)/(1) ',];</span>
<span style= "FONT-SIZE:18PX;" >[58, 238, 418, 598, 778, 958, 1138, 1318, 1498] #孙子定理的变通def tmp6 (): remain = [2, 3, 4]; MoD = [4, 5, 9]; Count = Len (mod); Multi = 1; For I in range (count): multi *= mod[i]; #直接遍历这个范围 bound = SUM (remain) *multi+1; result = []; For I in range (1, bound): for J in Range (count): if I%MOD[J]!=REMAIN[J]: Break ; If J >= count-1: result.append (i); Print (Result);</span>
<span style= "FONT-SIZE:18PX;" >[282, 975, 1668, 2361, 3054, 3747, 4440, 5133, 5826, 6519, 7212, 7905] #孙子定理的变通def tmp6 (): #模数 mod = [7, 9, 11] ; #余数 remain = [2, 3, 7]; Count = Len (mod); Multi = 1; For I in range (count): multi *= mod[i]; #直接遍历这个范围 bound = SUM (remain) *multi+1; result = []; For I in range (1, bound): for J in Range (count): if I%MOD[J]!=REMAIN[J]: Break ; If J >= count-1: result.append (i); Print (Result);</span>
<span style= "FONT-SIZE:18PX;" >[2111, 4421, 6731, 9041, 11351, 13661, 15971, 18281, 20591, 22901, 25211, 27521, 29831, 32141, 34451, 36761, 39071, 41 381, 43691, 46001] #韩信点兵 # The Flexible def TMP6 () of Sun Tzu's theorem: #模数 mod = [5, 6, 7, one]; #余数 remain = [1, 5, 4, ten]; Count = Len (mod); Multi = 1; For I in range (count): multi *= mod[i]; #直接遍历这个范围 bound = SUM (remain) *multi+1; result = []; For I in range (1, bound): for J in Range (count): if I%MOD[J]!=REMAIN[J]: Break ; If J >= count-1: result.append (i); Print (Result);</span>
<span style= "FONT-SIZE:18PX;" >2 30 685 25 708 20 7211 15 7414 10 7617 5 7820 0 80# Hundred-watt def TMP8 (): a =; b =; c = n/A; result = []; For I in Range (a+1): for J in range ( b+1): for K in range (c+1): if (ABS (((i*3 + j*2 + k*0.5)) < 1e-6) and i+j+k = =: print (I, J, K);</span>
<span style= "FONT-SIZE:18PX;" > #不定方程def TMP7 (): #ax + by = C #b为大于1的整数 A, B, C = 3, 2, 5; U, V, I, j = A, B, 0, 1; R = u%v; K = J; While r!=1: u, v = V, R; Q = u//v; R = u%v; j = i-q*j; i = k; K = J; x = c*k; y = c* (1-a*k)/b; Print (x, y);</span>
>>>-34 6.0# indefinite Equation def tmp7 (): #ax + by = C #b为大于1的整数 A, b, C = 2; U, V, I, j = A, B, 0, 1; R = u%v; K = J; While r!=1: u, v = V, R; Q = u//v; R = u%v; j = i-q*j; i = k; K = J; x = c*k; y = c* (1-a*k)/b; Print (x, y);
The end of this section, to know how to funeral, please see tell.
[ab initio mathematics] The NO. 205 Section Elementary Number theory preliminary