The numbers 1,6,8,13,15,20 can be placed in the circle below, each exactly once, so that the sum of each pair of numbers adjacent in the circle is a multiple of seven in fact, there is more than one way to arrange the numbers in such a way in the circle. Determine all different arrangments. Note that we will consider two arrangments to be the same if one can be obtained from the other by a series of reflections and rotations

Step-by-step explanation:

Step 1: Re-state the problem in an easier way to set up a permutation problem

"The numbers 1,6,8,13,15,20 can be placed in the circle below, each exactly once, so that the sum of each pair of numbers adjacent in the circle is a multiple of seven"

The series above can be represented again, in form of:

7a + 1, 7b - 1, 7c + 1, 7d - 1, 7e + 1, 7f - 1

with a, b, c, d, e, f are non-negative integers.

=> 3 numbers are multiply of 7 plus 1

=> 3 numbers are multiple of 7 minus 1

Step 2: Perform the counting:

For the 1st number, there are 6 ways to select

To satisfy that each pair of numbers creates a multiple of 7, then:

For the 2nd number, there are 3 ways left to select

For the 3rd number, there are 2 ways left to select

For the 4rd number, there are 2 ways left to select

For the 5th number, there are 1 way left to select

For the 6th number, there are 1 way left to select

=> In total, the number of possible ways to select:

N = 6 x 3 x 2 x 2 x 1 x 1 = 72

However, these numbers are located around a circle, each option is counted twice.

=> The final number of possible ways:

N = 72/2 = 36

Hope this helps!

:)

l.jnl.j/n.m,lm,

step-by-step explanation:

answer: 94

step-by-step explanation: 79+91+92+94+94 = 450 divided by 5 (since there are 5 number that are being added) = 90