Answer
a) 4,8, 11, 12, 15, 16, 19, 20, 22, 23, 24, 26, 27, 28, and all values
greater than or equal to 30
b) Solution is in (STEP BY STEP ) process
c) Solution is in (STEP BY STEP ) process
Work Step by Step
a)
-- 4,8, 11, 12, 15, 16, 19, 20, 22, 23, 24, 26, 27, 28, and all values
greater than or equal to 30
b)
--Let P(n) be the statement that
we can form n cents of postage using just 4-cent and 11-cent
stamps.
- We want to prove that P(n) is true for all n ≥ 30.
For the basis step, 30 = 11 + 11 + 4 + 4.
-Assume that
we can form k cents of postage (the inductive hypothesis);
we will show how to form k + 1 cents of postage.
- If the k cents included an 11-cent stamp, then replace it by three 4-
cent stamps.
Otherwise,
-k cents was formed from just 4-cent
stamps. Because k ≥ 30, there must be at least eight 4-cent
stamps involved.
-Replace eight 4-cent stamps by three 11-cent
stamps, and we have formed k + 1 cents in postage.
c)
P(n) is the same as in part (b). To prove that P(n) is true for all
n ≥ 30, we check for the basis step that
30 = 11+11+4+4,
31 = 11+4+4+4+4+4,
32 = 4+4+4+4+4+4+4+4,
33 = 11+11+11.
For the inductive step, assume the inductive
hypothesis, that P(j) is true for all j with 30 ≤ j ≤ k,
where k is an arbitrary integer greater than or equal to 33.
We want to show that P(k + 1) is true. Because k − 3 ≥ 30,
we know that P(k − 3) is true, that is, that we can form k − 3
cents of postage.
-Put one more 4-cent stamp on the envelope,
and we have formed k +1 cents of postage. In this proof, our
inductive hypothesis was that P(j) was true for all values of
j between 30 and k inclusive, rather than just that P(30) was
true.
