[ab initio mathematics] The NO. 205 Section Elementary Number theory preliminary

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