Question

Compute 90, 91, 92, 93, 94, and 95. 90 = 1 91 = 9 92 = 81 93 = 729 94 = 6561 95 = 59049 Make a conjecture about the units digit of 9n, where n is a positive integer. If n is even, the units digit of 9n is 1 . If n is odd, the units digit of 9n is 9 . Use strong mathematical induction to prove your conjecture. Submit your proof as a free response. (Submit a file with a maximum size of 1 MB.)

Answer 1

Answer:

Step-by-step explanation:

First we compute $9^n$, for $n = 0,1,2,3,4,5$

$9^0=1,9^1=9,9^2=81,9^3=729,9^4=6561,9^5=59049$

We have the unit digit of $9^n$, is either 1 or 9

Therefore, the conjecture can be state as follows

Conjecture; the unit of digit of $9^n$, is 1, when n is even and 9 when n is odd

Let the property p(n) be the formula;

"The unit digit of $9^n$, is 1 when n is even and 9 when n is odd"↔ P(n)

To show p(n) is true , we will use strong mathematical inductive

Show tgat p(0) and p(1) are true

We have  to show the unit digit of $9^0$ is 1 and $9^1$ is 9

Since any integer with zero power is one and hence

$9^0=1$ and $9^1 = 9$

Therefore, p(0) and p(1) are true

Show that for al integer $k\geq 1$, if  p(i) is true for all integer i from o through k, then p(k+1) is also true

let k be any integer with $k \geq 1$ and suppose that for all integer i with $0 \leq i \leq \leq k$

the unit digit of $9^i$ is 1 when is even and 9 when n is odd

we must show that $9^{k +1}$ equals 1 when n is even and 9 when n is odd

Case I; (k +1 is odd)

in this case k is even and so, by inductive hypothesis, the unit digit of $9^k$ is 1 and hence there is some non negative integer q such that

$9^k = 10q + 1$

now, $9^{k+1}=9^k(9)$

= (10q + 1)9

= 90q + 9

= 10.9q + 9

Note that for any non negative integer q , 9q is also an integer and this implies the unit digit of $9^{k+1}$ is 9

Case II

(k +1 is even)

in this case k is odd and so, by inductive hypothesis, the unit digit of $9^k$ is 9 and hence there is some non negative integer q such that

$9^k=10q+9$

Now,

$9^{k+1}=9^k.9\\=(10q+9)9\\=90q+81\\=10(9q+8)+1$

Note that any non negative integer q, 9q + 8 is also an integer and this implies the unit of $9^{k+1}$ is 1

Since we have proved both the basis and the inductive step of the strong mathematical induction, we conclude that the given statement is true