Finding all possible values

How many values can 7a+13b7a+13b take, if a,bZ0 and 0a+b100a,b\in\mathbb{Z}_{\geq 0} \text{ and } 0 \leq a+b \leq 100?

Try to generalize. How many values can ma+nbm\cdot a+n\cdot b take, if a,bZ0, m,nZ and 0a+bka,b\in\mathbb{Z}_{\geq 0},\text{ }m,n\in\mathbb{Z} \text{ and } 0 \leq a+b \leq k, for some kNk\in\mathbb{N}?

Note by Tim Vermeulen
6 years, 3 months ago

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This is a very interesting problem, and it's not immediately clear how to proceed, or what the answer is.

I've recently created a problem which uses this as the main approach (and just so happened to chance upon your discussion).

Calvin Lin Staff - 6 years, 3 months ago

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Is it about counting the terms when a polynomial is raised to a certain power? Because that's where I got the idea from.

Tim Vermeulen - 6 years, 3 months ago

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That's one place where this can arise.

The original place that I was looking at, was in modifying the Chicken Mcnugget Theorem (see Strong Induction), by adding the restriction that we are only allowed to buy a certain number of boxes so as to not appear like a glutton.

These two are related through the theory of generating functions.

Calvin Lin Staff - 6 years, 3 months ago

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@Calvin Lin One BIMC problem this year...................

Zi Song Yeoh - 6 years, 3 months ago

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Great Problem! One approach is to note that the first 12 multiples of 7 (7,14,21,...) form a complete set of residues mod 13. Adding up to 100 multiples of 13 gives most numbers up to 1300 (everything except 1,2,3,...71). Subtracting these numbers we have made from 1007+10013=2000 gives everything apart from 1,2,3,...71,1999,1998,...1929. It's possible to show these numbers are impossible, and to generalise the argument. For a and b coprime a result of Sylvester gives (a-1)(b-1) impossible numbers.

David Vaccaro - 6 years, 3 months ago

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You might want to read the question again. Note that you can't get above 1300 as a+b100a+b \leq 100 .

Note: You need to type your equations in the latex syntax using the brackets \ ( \ ). Otherwise, * gets interpreted in Markdown, which would italize your equations instead. I believe that your line 1007+10013 should have been 1007+10013 100 * 7 + 100*13 instead.

Calvin Lin Staff - 6 years, 3 months ago

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Yep. Sorry I thought it was 0<= a,b <= 100 rather then 0<=a+b<=100

David Vaccaro - 6 years, 3 months ago

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I don't completely get what your saying. Where does 7171 come from?

By the way, check your formatting, some things went wrong.

Tim Vermeulen - 6 years, 3 months ago

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71 is the largest number which can not be made from a combination of 7 and 13.

I misread your question and thought condition was that a and b were both less than or equal to 100, rather than the sum. I found all the values between 0 and 100(7+13) which were not possible.

If the sum a+b is bounded (0<=a+b<=100) then the number of possible values is just 101+100+99+...89=1235 corresponding to the number of the 13s (the larger value) it is possible to put with 0,1,...12 lots of 7s. (You never need 13 or more lots of 7 because it would be better to convert them to 7 lots of 13.) All of these are different because the multiples of 7 will have different residues mod 13.

This should generalize for general k to give (k+1)+k+....(k+2-a) possible values

David Vaccaro - 6 years, 3 months ago

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@David Vaccaro That solutions seems correct, I solved it differently: for every 0b1000 \leq b \leq 100, there are 1313 possible values for aa, but I need to subtract 12+11++112+11+\dots + 1 because for b=100b=100 I counted 1313 values for aa while only a=0a=0 is valid due to the fact that a+b100a+b \leq 100. Similarly, for b=99b=99, 1313 values for aa were counted while only 2 are valid, etcetera. So, the answer is 13101(12+11++1)=123513 \cdot 101 - (12 + 11 + \dots + 1) = 1235.

Generalization: denote max(m,n)\max(m,n) by qq. The total values ma+nb m \cdot a + n \cdot b can take if a+bk a+b \leq k then is

q(k+1)((q1)+(q2)++1)=q(k+1)q(q1)2=q((k+1)q12)=q(q+2k+3)2. \begin{aligned} q \cdot (k+1) - \left( (q-1) + (q-2) + \dots + 1 \right) &= q \cdot (k+1) - \frac{q(q-1)}{2}\\ &= q \left( (k+1) - \frac{q-1}{2} \right)\\ &= \frac{q(-q+2k+3)}{2}. \end{aligned}

And indeed:

(k+1)+k++(k+2a)=(k+1)(k+2)2(k+2a)(k+1a)2=(k2+3k+2)(k2+3k+22qk3q+q2)2=q2+2qk+3q2=q(q+2k+3)2. \begin{aligned} (k+1) + k + \dots + (k+2-a) &= \frac{(k+1)(k+2)}{2} - \frac{(k+2-a)(k+1-a)}{2}\\ &= \frac{(k^2+3k+2) - (k^2+3k+2-2qk-3q+q^2)}{2}\\ &= \frac{-q^2+2qk+3q}{2}\\ &= \frac{q(-q+2k+3)}{2}. \end{aligned}

Tim Vermeulen - 6 years, 3 months ago

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@Tim Vermeulen That looks good- my a is your q so we agree.

Of course we need to also consider case when m and n are not coprime, where we just need to cancel a common factor.

Also if k is smaller than q we just have k(k+1)/2 possible values, since in my way of thinking of things we can have any of the k possible amounts of m without redundancy.

David Vaccaro - 6 years, 3 months ago

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@David Vaccaro Yes, I had thought about the fact that they need to be coprime, but I forgot to mention it. I had not thought about your last point, but that seems correct.

What I think is much harder, is what the answer would be with more than 2 terms: for instance, 7a+13b+29c7a + 13b + 29c with a+b+c100a+b+c \leq 100. I can't find any way to even begin solving this.

Tim Vermeulen - 6 years, 3 months ago

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