Consecutive hot dogs: Difference between revisions

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==Extra extra credit==
==Extra extra credit==


For which total numbers of hot dogs is it possible to determine the number of contestants participating in such an event? Assume there must be at least three contestants?
For which total numbers of hot dogs is it possible to determine the number of contestants participating in such an event? Assume there must be at least three contestants and that everyone ate more than 1 hot dog.


==Help==
==Help==
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{{Answer | There were 11 contestants}}
{{Answer | There were 11 contestants}}


{{Solution | Our task is equivalent to finding all the ways in which 253 can be expressed as the sum of consecutive positive integers. In fact, there are three solutions: 126 + 127, but this is not the answer because we know there are at least three contestants.  253 can be expressed as 1 + 2 +...+ 22, but that is not the answer, since Alvin ate at least 2 hot dogs.  The only other solution is 17 + 18 + ... + 29, and this give 11 contestants.
{{Solution | Our task is equivalent to finding all the ways in which 253 can be expressed as the sum of consecutive positive integers. In fact, there are three solutions: 126 + 127, but this is not the answer because we know there are at least three contestants.  253 can be expressed as 1 + 2 +...+ 22, but that is not the answer, since Alvin ate at least 2 hot dogs.  The only other solution is 17 + 18 + ... + 29, and this give 11 contestants.


In general, with <m>c</m> contestants will be a solution with <m>h</m> total hot dogs, if and only if <m>c|2h</m> and <m>\frac{c}{2} < \frac{h}{c}</m>. To see why this is, and as such to answer the extra credit parts, is left as an exercise for the reader.}}
In general, with <m>c</m> contestants there will be a solution with <m>h</m> total hot dogs if and only if <m>c\mid 2h</m> and <m>\frac{c}{2} < \frac{h}{c}</m>. To see why this is, and as such to answer the extra credit parts, is left as an exercise for the reader.}}


[[Category: Number theory]]
[[Category: Number theory]]
[[Category: Discrete math]]
[[Category: Discrete math]]


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Current revision as of 05:49, 25 July 2017

Here is a puzzle I made up after hearing a nice number theory result from a colleague.

Puzzle

At the warm-up event for Oscar’s All Star Hot Dog Eating Contest, Alvin ate hot dog after hot dog until he could eat no more. Bob then one-upped him by eating one more hot dog than Alvin ate. Not to be outdone, Carl ate one more than Bob. This continued with each contestant eating one more dog than the previous contestant, until in the end, a total of 253 hot dogs had been eaten. How many contestants participated in the event?

Extra credit

After hearing about the success of the event, Nathan claimed that he ran an almost identical event with a different group of contestants. Again, each contestant ate one more hot dog than the previous contestant, but upon completion, 256 hot dogs had been consumed. How do you know that Nathan is lying?

Extra extra credit

For which total numbers of hot dogs is it possible to determine the number of contestants participating in such an event? Assume there must be at least three contestants and that everyone ate more than 1 hot dog.

Help

Hint
Although we know neither number, what can you say about the number of contestants times the average number of hot dogs consumed per contestant? You can assume nobody ate a fraction of hot dogs, and that there were no fractional contestants
Answer
There were 11 contestants
Solution
Our task is equivalent to finding all the ways in which 253 can be expressed as the sum of consecutive positive integers. In fact, there are three solutions: 126 + 127, but this is not the answer because we know there are at least three contestants. 253 can be expressed as 1 + 2 +...+ 22, but that is not the answer, since Alvin ate at least 2 hot dogs. The only other solution is 17 + 18 + ... + 29, and this give 11 contestants. In general, with <m>c</m> contestants there will be a solution with <m>h</m> total hot dogs if and only if <m>c\mid 2h</m> and <m>\frac{c}{2} < \frac{h}{c}</m>. To see why this is, and as such to answer the extra credit parts, is left as an exercise for the reader.