Polynomial Remainder Theorem and multiple of  2,3,4,5,6,7,8,9,10,11,13

 

        To determine if a number is a multiple of 2, 5, or 10 is trivial.  Let’s check the cases for others and formulate this kind of problems by using polynomials.

 

Question:  (1) Check if 245327 is a multiple of  3.

                   (2) Check if 245327 is a multiple of 9.

 

         Sol:

                   Let  f(x) = 2x⁵ + 4x⁴ + 5x³ + 3x² + 2x + 7  .

                  Then

                           f(1) = 2+4+5+3+2+7=23

                           f(10)=245327

 

                   With Polynomial Remainder Theorem, we have

                        f(x) = (x-1)q(x) + f(1) ,  and q(x)  Z[x]

                     (q(x) is with coefficients in integers because all the

                       coefficients of f(x) and (x-1) are integers and the

                       highest-order coefficient of (x-1) is 1 )

 

                     And

                           24327=f(10)=(10-1)q(10) + f(1)

                           24327 = 9q(10) + f(1)

 

               

(1)        To check if 245327 can be divided exactly by 3,

          we only need to check f(1).

                          f(1) is not a multiple of 3  245327 is not a multiple of 3.

 

(2)        f(1) is not a multiple of 9  245327 is not a multiple of 9.

 

 

Question: Check if 245327 is a multiple of 11.

                

         Sol:

                 Let  f(x) = 2x⁵ + 4x⁴ + 5x³ + 3x² + 2x + 7 

                          f(-1) = (-2)+4+(-5)+3+(-2)+7 = 5

                          f(10) = 245327

 

                 With polynomial remainder theorem,

                          f(x) = (x+1)q(x) + f(-1)  ,  q(x)  Z[x]

 

                 Then

                            245327=f(10)=(10+1)q(10) + f(-1)

                                        = 11q(10) + f(-1)

 

                 Thus,  it only needs to check if f(-1) is a multiple of 11.

                  f(-1) is not a multiple of 11   245327 is not a multiple of 11.

 

 

Question: (1) Check if 189245327 is a multiple of 7 .

(2)   Check if 189245327 is a multiple of 13.

 

      Sol:

               

Let f(x)=189x² + 245x + 327 .

Then

           f(-1) = 189-245+327 = 271

           f(1000)=189245327

 

With polynomial remainder theorem,

          f(x) = (x+1)q(x) + f(-1),  q(x)  Z[x]

 

 189245327=f(1000)=(1000+1)q(1000) + f(-1)

                   = 1001q(1000) + f(-1)

 

           1001 is a multiple of 7 and 13

      f(-1) can determine if 189245327 is a multiple of 7 or 13.

 

(1)     f(-1) = 271 is not a multiple of 7. Neither is 189245327.

(2)     f(-1) is not a multiple of 13. Neither is 189245327.

 

 

Question:  (1)  Check if  86433252 is a multiple of 4.

                  (2)  Check if 86433252 is a multiple of 8.

 

       Sol:

                 The idea to tackle this problem is:

                           100 is a multiple of 4;

                           1000 is a multiple of 8.

 

                (1)

                         Let f(x) = 86x³ + 43x² + 32x + 52 .

                         Then

                                f(0)=52

                                 86433252=f(100)

 

                       And from polynomial remainder theorem:

                            f(x)=xq(x) + f(0) , q(x)  Z[x]

 

                            86433252 = f(100) = 100q(100) + f(0)

 

                            f(0) is not a multiple of 4. Neither is 86433252 .

 

  

 

              (2)

                         Let g(x)= 86x² + 433x + 252 .

                         Then

                                   g(0) = 252

                                   g(1000) = 86433252

 

                         And we have

                               g(x) = xq(x) + g(0) ,  q(x)  Z[x]

 

                               86433252=g(1000) = 1000q(1000) + g(0)

 

                           g(0) is not a multiple of 8. Neither is 86433252.