If $a$ and $b$ must be nonnegative integers, what is the largest integer $n$ such that $13a + 18b = n$ has no solutions?
I have tried doing the postage stamp method, but this doesn't work with such large numbers. Any help?
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2$12*17-1.$ Look up the Frobenius problem and it says if gcd of coin values u,v is 1 it is $uv-u-v$ i.e. $(u-1)(v-1)-1.$ – coffeemath Oct 28 '15 at 00:13
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@coffeemath Ok thank you! One more question. If a and b are positive integers only, how would you solve it then? – rd360 Oct 28 '15 at 00:14
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If a,b must be positive only, make new variables by subtracting 1 from a,b. So in your example u=a-1 v=b-1 so 13a+18b=n iff 13u+18v=n+13+18. Maybe I should have subtracted 13+18 from n here... – coffeemath Oct 28 '15 at 00:17
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Does this answer your question? The Frobenius Coin Problem – Jam Jan 28 '20 at 21:17
2 Answers
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The general solution to the equation
$$13a+18b=n$$
where $a,b$ are integers is
$$a=7n+18t$$
$$b=-5n-13t$$
Because of $a\ge 0$,$b\ge 0$ we get $\frac{-5n}{13}\ge t\ge \frac{-7n}{18}$
The largest $n$ such that no $t$ satisfying this inequality is $n=203$ which I checked with PARI/GP :
? for(n=1,10000,if(floor(-5*n/13)<ceil(-7*n/18),print1(n," ")))
1 2 3 4 5 6 7 8 9 10 11 12 14 15 16 17 19 20 21 22 23 24 25 27 28 29 30 32 33 34
35 37 38 40 41 42 43 45 46 47 48 50 51 53 55 56 58 59 60 61 63 64 66 68 69 71 7
3 74 76 77 79 81 82 84 86 87 89 92 94 95 97 99 100 102 105 107 110 112 113 115 1
18 120 123 125 128 131 133 136 138 141 146 149 151 154 159 164 167 172 177 185 1
90 203
Peter
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