Factors: A teacher writes a number on a whiteboard. He then produces a hat....
- Thread starter Bryson
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Can you find numbers with following factors:
1, 2 & 3...................................
1, 2, 3 & 4...................................
1, 2, 3, 4 & 5...................................
1, 2, 3, 4, 5 & 6...................................
How did you find those?
Ok
Thank you. The principal does help. however if I were to multiply the lowest of the 24 numbers (ie 1-22) I will be left with an enormous number. Moreover, the question states that it has to be the lowest possible number that is applied.
The question is:
A teacher writes a number on a whiteboard. He then produces a hat which contains a set of 24 tickets numbered 1 to 24. Each of the 24 students in his class receives one randomly selected ticked and is asked whether the number on the board is divisible by the number on their ticket. All but two of the students answer "yes". The two students who say "no" have consecutive numbers. Assuming that all the students have done their divisions correctly, and that the number on the whiteboard is the smallest possible, what are the numbers of the two students who said "no" and what is the number on the whiteboard?
Thank you. The principal does help. however if I were to multiply the lowest of the 24 numbers (ie 1-22) I will be left with an enormous number. Moreover, the question states that it has to be the lowest possible number that is applied.
The question is:
A teacher writes a number on a whiteboard. He then produces a hat which contains a set of 24 tickets numbered 1 to 24. Each of the 24 students in his class receives one randomly selected ticked and is asked whether the number on the board is divisible by the number on their ticket. All but two of the students answer "yes". The two students who say "no" have consecutive numbers. Assuming that all the students have done their divisions correctly, and that the number on the whiteboard is the smallest possible, what are the numbers of the two students who said "no" and what is the number on the whiteboard?
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Deleted member 4993
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You did not say that in your original post. That is a horse of a different color!
Dr.Peterson
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As has been pointed out, there is a good reason we ask you to state the entire problem; this is a perfect example of that. What you initially asked (if I'm correctly reading around the various edits) is an entirely different question.
We're looking for a number N such that all of 1, 2, ..., 24 are divisors of N except for a pair of consecutive numbers.
I would start by thinking about how divisibility by different numbers is related. For example, suppose that 12 is not a divisor of N. Then what other numbers can't be divisors of N? First, 24 can't be a divisor, because if it were, then 12 would be also (since 12 is a divisor of 24). That would force us to have two non-consecutive non-divisors. So 12 can't be one of the two. On the other hand, ignoring that, either 3 or 4 also would have to be a non-divisor. Can you see why? This gets more subtle, but will be important.
Steven G
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I have a couple of things to say. 1st, a huge number, as you call it, might be the smallest such number that satisfies this problem. 2ndly, if you multiply numbers 1-22, which equals 22!, this number will not work either. Since 22! has factors of 8 and 3 (do you see why?), then the student who has number 24 would be one of the students who does say that their number, 24, divides N.
Here are some hints.
\(\displaystyle f(n) = \dfrac{24!}{n(n + 1)} = \dfrac{24!}{n^2 + n} \text { with } n \in \mathbb Z \text { and } 0 < n < 24.\)
Do you see that function's relevance?
There are 23 possible values of the function, and the function declines as n increases. You are asked to find the smallest value of f(n) that has a certain attribute. So if there are several possible values of n that result in f(n) having that attribute, you want to pick the largest such value. Make sense?
Furthermore, to have only two integers from 1 through 24 be divisors of f(n), either n or n + 1 must be a prime. Can you explain why?
\(\displaystyle f(n) = \dfrac{24!}{n(n + 1)} = \dfrac{24!}{n^2 + n} \text { with } n \in \mathbb Z \text { and } 0 < n < 24.\)
Do you see that function's relevance?
There are 23 possible values of the function, and the function declines as n increases. You are asked to find the smallest value of f(n) that has a certain attribute. So if there are several possible values of n that result in f(n) having that attribute, you want to pick the largest such value. Make sense?
Furthermore, to have only two integers from 1 through 24 be divisors of f(n), either n or n + 1 must be a prime. Can you explain why?