The Skeleton Tower Task

The Skeleton Tower (A rather Halloweeny name, don't you think?) is a tower built of cubes in a pyramid fashion, with 4 stair steps extending from the center of the tower. In a tower that is 6 cubes high, there are 66 cubes altogether.

A tower 3 cubes high contains 15 cubes.

A tower 4 cubes high contains 28 cubes.

A tower 5 cubes high contains 45 cubes. 

The following table helped me in determining the number of cubes for each level. The highlighted portion was completed using mathematical equations only, without actually counting the cubes in a pyramid. 

Height	Cubes in all	Looking for Patterns
10	190	9 x 4 = 36 +1=37             *10x 19 = 190 153+37=190
9	153	8 x 4= 32 +1 = 33           *9 x 17= 153 33+ 120= 153
8	120	7 x 4=28 +1=29              *8 x 15= 120 91+29= 120
7	91	6 x 4=24 +1=25               *7 x 13= 91 66+25= 91
6	66	66-6=60          66-45=21 (the height x 4 +1) 60/4=15
5	45	45-5=40          45-28=17 (the height x 4 +1) 40/4=10
4	28	28-4=24          28-15=13 (the height x 4 +1) 24/4=6
3	15	15-3=12         15-6= 9 (the height x 4 +1) 12/4=3
2	6	6-2=4 4/4=1

The question now is whether or not there can be an equation to find the number of cubes in the pyramid at N cubes high. 

I noticed the pattern with an asterisk next to it in the table above. Obviously, 6 x 11 is 66. Initially, I thought the multiplication was just counting up odd numbers. Then I realized that 6 x 2 = 12, 7 x 2= 14, 9 x 2= 18, etc. So I wondered if I took N, multiplied it by 2, subtracted 1, then multiplied it by itself....would I get a correct answer? 

That would be [(N x 2)-1] x N.

Let's see if it is true if used for N = 4: 

[(4 x 2) - 1] x 4= [(8)-1] x 4= 7 x 4 = 28 (It works here, yay!) 

Now N= 10

[(10 x 2)-1] x 10= [(20)-1] x 10 = 19 x 10= 190 (Yay again!)

Using the table, I was able to identify patterns that would help formulate an equation to solve the problem. Oddly enough, the pattern that helped the most was just a side note when I noticed that all of the numbers in the not highlighted section were multiples of the height in cubes. 

Chicken Nuggets Task...Otherwise known as 'Now I am Hungry'

McDonald's sells chicken nuggets in boxes of 6, 9, or 20. Obviously one could purchase exactly 15 nuggets by buying a box of 6 and a box of 9. Using only combinations of boxes of 6, 9, and/or 20 nuggets:

Could you purchase exactly 17 nuggets?
     No, because two boxes of 6 would result in 12. Two boxes of 9 would result in 18, which is too much. A box of 6 and a box of 9 is already noted at a total of 15 nuggets.

How would you purchase exactly 53 nuggets?
   

What is the largest number for which it is impossible to purchase exactly that number of nuggets? 

     This question is difficult to explain, as numbers are infinite. I think the most important point here is that the number obviously cannot be prime, as that would eliminate the divisibility of it by 6, 9 and/ or 20. I did notice that continuing from 53 all the way to 100 resulted in some form of divisibility by the three numbers combined. Once I realized this, I went backwards to see if maybe that largest number was below 53. 

52: 2 boxes of 20, 2 boxes of 6

51: 1 box of 9, 7 boxes of 6

50: 1 box of 20, 5 boxes of 6

49: 2 boxes of 20, 1 box of 9

48: 2 boxes of 9, 5 boxes of 6

47: 1 box of 20, 3 boxes of 9

46: 2 boxes of 20, 1 box of 6

45: 5 boxes of 9

44: 1 box of 20, 4 boxes of 6

43: Hm. I am stumped here!  Perhaps the answer to this question is 43?

Let's say you could only buy the nuggets in boxes of 7, 11, and 17. What is the largest number for which it is impossible to purchase exactly that number of nuggets? 

   To figure this out, I looked to see if there was a pattern for the above numbers. It is possible that I am not thinking about this 'mathematically' enough, however I can't seem to determine a formula or pattern to finding 43 as the number above. The best I could come up with is using a hundreds chart, I marked out all numbers divisible by 7 (in black). I kept going with the chart, marking out every 11th, 7th, or 17th number from one that was already eliminated. The highest number I have not marked out is 37. 

In closing, there must be a more mathematically sound method of solving this problem that I have not yet discovered. As it stands, I feel fairly confident that I figured out the solution....albeit in a rather elementary fashion.