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Is there any general solution of this? using 2 integers, what is the minimum number formed after which we can make any number using those 2 integers?

so it says 3a + 5b = m

now we know 4, 7 do not occur, but how can we say we can make any integer m>11 !!

I know 11 is the minimum, but how did it come,

what if it wasnt 3's and 5's say it was 4's and 5's?

is there any general method for this? I cant find it, please help.

abstractnature
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  • Look up numerical semigroups. Mind you, this question is a duplicate. Meaning that it will be closed soon. Follow the latter link! In his answer Robjohn explains the general situation (two coprime integers). – Jyrki Lahtonen Feb 08 '14 at 06:18
  • There is a standard theorem that says if $m$ and $n$ are relatively prime, then any number $\ge (m-1)(n-1)$ can be expressed, and $(m-1)(n-1)-1$ cannot be. For $(3,5)$ or any particular pair, this is easy to show. The general case is not particularly hard, but takes a while to write out. – André Nicolas Feb 08 '14 at 06:22
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    To all: it should be clear that the question is interesting only with the requirement $a,b\ge0$ that OP forgot to state. I assume that is the intention as otherwise the statement about 4 and 7 missing doesn't make sense. – Jyrki Lahtonen Feb 08 '14 at 06:22
  • Its not a duplicate, as I wanted to ask the number from which onwards I can make all the integers, not the set of numbers formed as asked in the duplicate mentioned. If you still think its wrong to ask please remove it. I have already a bad experience on this site anyways from OPs, I answer and if by any chance i committed a careless mistake in my answer I get a negative rating, as if I am done something wrong or OP is too selfish to just know whats the correct answer. – abstractnature Jun 27 '17 at 08:54

2 Answers2

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If you can do $m=12 \to 26$, then you can certainly do them all, since you just repeat adding a multiple of $15$, the LCM of $3$ and $5$.

So

$$12 = 3+3+3+3 \\ 13 = 5+5+3 \\ 14 = 3+3+3+5 \\ 15 = 5+5+5 \\ 16 = 3+3+5+5 \\ 17 = 3+3+3+5 \\ 18 = 3+3+3+3+3+3 \\ 19 = 5+5+3+3+3 \\ 20 = 5+5+5+5 \\ 21 = 3+3+3+3+3+3+3 \\ 22 = 5+5+3+3+3+3 \\ 23 = 5+3+3+3+3+3+3 \\ 24 = 3+3+3+3+3+3+3+3 \\ 25 = 5+5+5+5+5 \\ 26 = 5+3+3+3+3+3+3+3.$$

Now just add three $5$'s or five $3$'s to taste for each group of $15$ natural numbers.

John
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If we are allowed take any integer value of $a,b$ we can from any integer $n=n(3\cdot2-5)$ or $n(5\cdot2-3\cdot3)$

If we enforce $a,b\ge0$

Any integer multiple of $3$ can be written as $3a$ where $a$ is an integer

For any integer $m=3n+1$ we can write $=3(n-3)+5\cdot2$ for $n\ge3\implies m\ge10$

For any integer $m=3n+2$ we can write $=3(n-1)+5$ for $n\ge1\implies m\ge5$

lab bhattacharjee
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