<s>[INST] How can we make any integer m>11 using 3's and 5's only?? 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.
 [/INST] 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.
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