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• Unlimited access from just £6.99 per month   # The object of this exercise is to face the arrowhead in the opposite direction with minimum moves.

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Introduction

The object of this exercise is to face the arrowhead in the opposite direction with minimum moves.

Diagram 1          By observation we see the arrowhead is in the shape of equilateral triangle. It’s counters are round and arranged in such a way that, by increasing the base layer by one counter, so that the new base layer contains one more (b+1) with respect to previous one. Each time the base is increased, the arrowhead maintains it’s equilateral shape.

In any triangle we can find the minimum moves needed to face the triangle in the opposite direction.

We will see a pattern generating when we move the counters so that the arrow head faces in the opposite direction. We will investigate the structure in which these patterns are generated.

Middle

4 12                   78                   26

}                    4 13                   91                   30

}                    5 14                  105                  35

}                    5

15                  120                  40 From the above table we notice that the difference between each move is constant for three moves, then increases by one after every third move.

So we need to do a further Investigation by comparing the counters inside the hexagon and counters the hexagon (the corners of the triangle).

### And counters outside hexagon

TABLE 2        ## Base,b  Total number          [Counters          [Counters  =  Moves of counters, nbinside                outside

hexagon]           hexagon] 4         10                                            7       3                   =  4   (one above correct move) 5         15  (divisible by 3)                10        5                   =  5 6         21  (divisible by 3)                14        7                   =  7 7         28                                         19        9                   =  10   (one above correct move) 8         36  (divisible by 3)                24        12                 =  12 9         45  (divisible by 3)                30        15                 =   15 10       55                                         37        18                 =   19   (one above correct move) 11       66  (divisible by 3)                44        22                 =   22 12       78  (divisible by 3)                52        26                 =   26

13       91                                         61        30                 =   31   (one above correct move) 14      105 (divisible by 3)                70       35                  =   35 15      120 (divisible by 3)                80       40                  =   40   My deduction from the above table, is that the number of counters which are a multiple of 3, give the correct minimum number of moves. Other number of counters which are not multiple of 3, give 1 above the correct minimum number of moves.

From this information, I am going to try derive a formula to give the minimum number of moves.

### Deriving an equation for minimum number of moves

Base,b=1,2,3,4,…..

Number of counters, nb= b+(b-1)+(b-2)+(b-3)+…….+1

Number of moves=m

If nb is divisible by 3,then               (If nb mod 3 = 0)

m = nb/3

else

m = (nb-1)/3

Example                                                                Example

b =8                                                                      b =13

nb =8+7+6+5+4=3+2+1 =36                                  nb =13+12+11+10+9+8+7+6+5+4+3+2+1 = 91

nb is divisible by 3, thereforen is not divisible by 3,therefore

m =nb/3m = (nb-1)/3

m = 36/3 =12                                                          m = (91-1)3 = 30

note:=  nb = 1+2+3+4+…+(b-1)+ b

Therefore, we use ‘Arithmetic Progression’. Where first term, a =1, common difference, d =1

nb = (b/2) *(2a + (b-1)d) = [b *(b+1)] / 2   Therefore,

Conclusion

The next base is b’=3c+1,we have to prove equation (2)

m’= (b’-1)(b+2)  is also true.

6

At this stage one more base is added to the top triangle from the case where all 3 corner triangle bases were equal, but at this stage the top corner triangle has increased by one base.

Therefore adding c to both sides

m+c=b(b+1) + c

6

substituting 3c into bsince b=3c

m+c=b(b+1) + c

6

m+c=3c(3c+1) +c = 3c(3c+1)+6c

6                      6

m+c=3c(3c+3)

6

where b’=3c+1

But we know b’=3c+1 , since 3c=(3c+1)-1 and 3c+3=(3c+1)+2

Since 3c=b’–1

b’    –1

For the equation m+c=(bT-1)(bT+2)= m’

2

we get the multiples of the numerator

So this proves that if equation (1) works for b=3c, then equation (2) works for the next base

b’=3c+1

Assuming b’=3c+1 and equation m’=[(b-1)(b+2)]/6 are true.

Need to prove for the next base, b=b+1 is true for m=[b(b+1)]/6

Again the moves, m is formed from m’ by adding c+1 counters to the top triangle.

m’+c+1=(b-1)(b+2) + c+1

6

replacing bwith 3c+1m’+c+1=(3c+2)(c+1) = m

2

Using assumption b=3c+1 and b=b’+1

We get

b= b(b+1)

6

We have proved that if equation(2) works for any b=3c+1, then equation(3) will work for the b’.

Prove that  b=3c+2

If we look at the pattern of the cycle,3c, 3c+1, 3c+2, whilst in our induction we add c+1 to our equations ,according to this principle we assume the next cyclic move would be 3c+3.

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