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.
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