Question

Problem 12
Given 12 consecutive integers, in how many ways can three of these integers
be selected to give a sum which divides by 4?
A step by step explanation.

Answer 1

9514 1404 393

Answer:

  55

Step-by-step explanation:

Here's one way to count them.

12 consecutive integers will have 3 each of the integers whose value is 0, 1, 2, or 3 mod 4. To make a total that is 0 mod 4, any of these combinations of mod 4 values may be used:

  000 -- 1 combination

  013 -- 27 combinations, 3 each of mod 0, mod 1, and mod 3

  022 -- 9 combinations, 3 of mod 0, and 3 ways to choose 2 of mod 2

  112 -- 9 combinations, 3 of mod 2, and 3 ways to choose 2 of mod 1

  233 -- 9 combinations, 3 of mod 2, and 3 ways to choose 2 of mod 3

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Consider the set of 12 consecutive integers 1 .. 12. Then here are the 9 "233" combinations. Mod 2 integers are 2, 6, 10, and mod 3 integers are 3, 7, 11.

  (2, 3, 7), (2, 3, 11), (2, 7, 11),

  (3, 6, 7), (3, 6, 11), (6, 7, 11),

  (3, 7, 10), (3, 10, 11), (7, 10, 11)

The total number of combinations that will be divisible by 4 is ...

  1 + 27 + 9 + 9 + 9 = 55 . . . . combinations with a sum divisible by 4

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Additional comment

The number of ways that 3 can be chosen from 12 is 220, so this is 1/4 of that total. It is perhaps not surprising that the remainders from division by 4 are uniformly distributed among the possibilities.

A computer program to list the 220 subsets and the mod 4 remainders of their totals confirms this is the correct number.

Answer 2

There are  $55$  ways of these integers be selected to give a sum.

Given that:

It has  consecutive integers .

Now,

Address the formula for  $C(n,r)$,

$C(n,r)=\frac{n!}{(n-r)!n!}$

Here,

$n=12,r=3$

      $C(12,3)=\frac{12!}{(12-3)!3!} \\\\C(12,3)=\frac{12!}{9!3!} \\\\C(12,3)=\frac{12*11*10*9*8*7*6*5*4*3*2*1}{9*8*6*7*5*4*3*2*1*3*2*1} \\\\C(12,3)=\frac{12*11*10}{3*2*1} \\\\C(12,3)=4*5*11\\\\C(12,3)=220$

The sum $mod 4$  of any given subset increases by $1$ when the set is “slid” to the left.

Every subset gives rise to all four values $mod 4$  by sliding each of four places.

Since, The sum of the integer is divides by $4$.

Thus,

     $220\div 4=55$

Hence, it has $55$ ways .

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