Find the smallest number m such that postage of exactly n cents can be made using only 3-cent and 8-cent stamps for all n>=m. Prove your claim using simple and complete induction.
I am confused about this question.
If n is the amount of cents made from the two stamps, wouldn't the smallest m be 0 for all n?
What the question is asking is this:
Find the least natural number m such that for all natural numbers n >= m, there are natural numbers j and k such that n = 3j+8k. And prove that the m you found is indeed the the least natural number m such that for all natural numbers n >= m, there are natural numbers j and k such that n = 3j+8k.
suppose we could make 3 consecutive integers in such a fashion: m, m+1, and m+2.
adding a 3-cent stamp to each configuration would then give m+3, m+4, and m+5.
can you see that we can use induction to show we can make all integers ≥ m this way?
what are the smallest 3 consecutive integers we can make?
i think i know it
14 = 2(3)+1(8)
15 = 5(3)+0(8)
16 = 0(3)+2(8)
17 = 3(3)+1(8)
18 = 6(3)+0(8)
19 = 1(3)+2(8)
20 = 4(3)+1(8)
21 = 7(3)+0(8)
22 = 2(3)+2(8)
23 = 5(3)+1(8)
24 = 8(3)+0(8)
25 = 3(3)+2(8)
26 = 6(3)+1(8)
(i also see a pattern where the multiplier for 8 is 1,0,2,1,0,2... and for 3 its, +3,-5,+3, +3,-5,+3, +3,-5,+3, +3,-5,+3)
with a combination of 3 and 8 cent stamps to get one more, you can either remove a 8-cent stamp and put in 3 3-cent stamps or remove 5 3-cent stamps and put in 2 8-cent stamps. Therefore the smallest number is 14.
Is this the right way to get 14, or do I have to show some formal way, since here i am just assuming it?
so
if i set it up like
3j+8k = n
then
3j'+8k'=n+1
then two cases:
case 1:
if k>=1
remove a 8-cent stamp and put in 3 3-cent stamps
so
k'=k-1
j'=j+3
then k' and j' are in N
and
3j'+8k'=n+1
case 2:
if k=0
3j>=14
j>=14/3
j>=5
remove 5 3-cent stamps and put in 2 8-cent stamps
so
k'=k+2
j'=j-5
then k' and j' are in N
and
3j'+8k'=n+1
Is this is basic informal correct method?