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A country has three denominations of coins, worth 7, 10, and 53 units of value. What is the maximum number of units of currency which one cannot have if they are only carrying these three kinds of coins?

I think the problem is asking for the number of nonnegative integers that cannot be represented as $7a+10b+53c$ for $a,b,c \ge 0$. However, I'm not sure how to continue from here.

MathMagician
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    See Frobenius coins. – lulu Jun 22 '22 at 16:57
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    Locally I recommend this. As it happens $(7-1)(10-1)-1=53$ units is the largest amount that cannot be made coins worth $7$ or $10$ alone, so you need to find the next largest "gap". I'm sure you can manage that with the argument in Robjohn's answer. – Jyrki Lahtonen Jun 22 '22 at 17:12
  • Maybe it's like this: some amounts cannot be realized. What is the largest of them ( what we know: since $\gcd(7, 10, 53) = 1$ all numbers largest that a certain limit can be realized. So the set of numbers that can't be realized is finite ( non-void...$1$ can't be realized), so it has a largest element. Find that... – orangeskid Jun 22 '22 at 19:02

1 Answers1

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The largest number that cannot be represented is $46$, because neither of $6, 16, 26, 36, 46$ (which are $46-k\cdot 10$) is divisible by $7$.

Conversely, the 7 consecutive numbers $47, ..., 47+6=53$ are all representable:

$\begin{align} \qquad47 &= 40 + 7 \\ 48 &= 20 + 7\cdot 4 \\ 49 &= 7\cdot 7 \\ 50 &= 50 \\ 51 &= 30 + 7\cdot 3 \\ 52 &= 10 + 7\cdot 6 \\ 53 &= 53 \end{align}$

thus, all following numbers are also representable by adding $7$s to an appropriate number from above, like: $54=7+46, 55=7+48,...$

As an aside:

  • The largest number that is uniquely represented is $99=5\cdot10 + 7\cdot7$.

  • The smallest number, that has more than one representation (up to order) is $60 = 6\cdot 10 = 53+7$.

emacs drives me nuts
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