Respuesta :

There are  [tex]55[/tex] ways of these integers be selected to give a sum.

Given that:

It has [tex]12[/tex] consecutive integers .

Now,

Definition of consecutive integers :

" Consecutive integers are the integers that follow in the fixed sequence .

Consecutive integers is represented by  [tex]n,n+1,n+2,n+3,....[/tex] where [tex]n[/tex] is an integer."

By given :

we have [tex]12[/tex] consecutive integers .

Thus,[tex]n=1[/tex] and substitute the equation is,

[tex]1,(n+1),(1+2),(1+3),(1+4),(1+5),(1+6),(1+7),(1+8),(1+9),(1+10),(1+11)\\\\\implies 1,2,3,4,5,6,7,8,9,10,11,12[/tex]

Now,

Split all the integers into 4 equal parts,

Part 1:  Those integers are divisible by [tex]4[/tex] and the remainder be 0.

Then,  

   [tex]a=0(mod 4)[/tex]

Part 2:  Those integer producing the remainder [tex]1[/tex] when it is divisible by [tex]4[/tex].

Then,

   [tex]a=1(mod 4)[/tex]

Part 3:  Those integer producing the remainder [tex]2[/tex]  when it is divisible by[tex]4[/tex].

Then,

    [tex]a=2(mod 4)[/tex]

Part 4: Those integer producing the remainder [tex]3[/tex]  when it is divisible by[tex]4[/tex].

Then,

   [tex]a=3(mod4)[/tex]

Since, three of these integers be selected to give a sum which divides by [tex]4[/tex] is,

In any 12 consecutive integers there are [tex]12\div4[/tex]

i.e. exactly 3 numbers of each category of mod4 namely ,

[tex]0(mod4), 1(mod4), 2(mod4), 3(mod4)[/tex]

Thus, the  total combinations of above [tex]5[/tex] categories of sets of [tex]3[/tex] integers are

  • All the 3 numbers are  [tex]0(mod4)[/tex]

                 [tex]3C_1=3(1)=3[/tex]

  • One number be   [tex]0(mod4)[/tex]  and other two numbers are  [tex]2(mod4)[/tex]

            [tex]3C_1*3C_2=3(1)*3=9[/tex]

  • One number  [tex]0(mod4)[/tex]  and other numbers  [tex]1(mod4)[/tex] & [tex]3(mod4)[/tex]

           [tex]3C_1*3C_1*3C_1=3*3*3=27[/tex]

  • Two numbers be [tex]1(mod4)[/tex] and one  number be [tex]2(mod4)[/tex]  

           [tex]3C2 * 3C1 = 3*3 = 9[/tex]

  • One number  be  [tex]2(mod4)[/tex] and two numbers be [tex]3(mod4)[/tex]

           [tex]3C1 * 3C2 = 3*3 = 9[/tex]

Thus, sum of the total ways be [tex]12[/tex] consecutive integers of three integers is divisible by 4 is,

       [tex]3+9+27+9+9=55[/tex]

Hence, it has [tex]55[/tex] ways.

For more information,

https://brainly.com/question/24144187

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