In order that a positive integer X be a sum of 5 and 6 cent stamps, there must exist positive integers m and n such that 5m+ 6n= X. That's a "Diophantine equation". It is clear that 6- 5= 1 so 6X+ 5(-X)= X. That is, m= -X and n= X are solutions. Of course, -X is not a positive integer. But if we add 6A to m subtract 5A from n, for A any integer, 5(-X+ 6A)+ 6(X- 5A)= -5X+ 30A+ 6X- 30A = X for all A. We want or . But we must also have or . That is, , recalling that A must be an integer. In particular, that means, in particular, that if X/5 and X/6 differ by at least 1, there exist such an integer. So the first requirement we have is that from which, multiplying on both sides by 5(6)= 30, we can always form such a combination if . That does not mean that every number below thatcan'tbe formed by 5 and 6 cent stamps but gives us an upper bound so we can start just working down from that:

29= 4*6+ 5 so 29 cents is 4 six cent and 1 five cent stamp.

28= 6+ 22 and 22 is not a multiple of 5

28= 12+ 16 and 16 is not a multiple of 5

28= 18+ 10 so 28 cents is 3 six cent stamps and 2 five cent stamps.

27= 6+ 21 and 21 is not a multiple of 5

27= 12+ 15 so 27 cents is 2 six cent stamps and 3 five cent stamps.

26= 6+ 20 so 26 cents is 1 six cent stamp and 4 five cent stamps.

25= 5 five cent stamps.

24= 6 six cent stamps.

23= 6+ 17 and 17 is not a multiple of 5

23= 12+ 11 and 11 is not a multiple of 5

23= 18+ 5 so 23 is 3 six cent stamps and 1 five cent stamps.

22= 6+ 16 and 16 is not a multiple of 5

22= 12+ 10 so 22 is 2 six cent stamps and 2 five cent stamps

21= 6+ 15 so 21 is 1 six cent stamp and 3 five cent stamps

20 is 4 five cent stamps

19= 6+ 13 and 13 is not a multiple of 5

19= 12+ 17 and 17 is not a multiple of 5

19= 18+ 1 and 1 is not a multiple of 5

Whew! We have determined that every number larger than 19 can be found as a combination of 5 and 6.