these questions are for discrete math pls do both 3 and 4 and questions we have

Question

these questions are for discrete math
pls do both 3 and 4 and questions we have to use the prove from question 3
i request to do both questions with discription

3. Prove that e k e N+.Ym E N. k = 2m + 1 11 1 (10" + 1). 10] 4. Giveu a pusiive integur T, we cau write th deciml represenlation o as 10 where d' is the ill digit and is an integer from 0 to 9 inclusive, dk0 (i.e., the left-most digit cannot be 0) or 0, 0 aud: Deline the alterruating su For example Prove that if nis divisible hy 1 then n is divisible by 11. Hint: Use what you proved in Question 3. 8180928-8 219 081

Explanation / Answer

(Note: Please check the WHOLE solution.)

3. k = 2m + 1

This means k is odd.

10k + 1 = (11 - 1)k + 1

= 11k - 11k-1 + 11k-2 - .....+111 - 1k + 1

(Note, since k is odd, the sign of 1k is negative)

= 11k - 11k-1 + 11k-2 - .....+111

Since all terms contain 11, 11 | 10k + 1.

4. We have n = d0 * 100 + d1 * 101 + d2 * 102 + d3 * 103 + ...... dk-1 * 10k-1 + dk * 10k

There are two cases: k is odd and k is even.

If k is odd, there are k+1 terms and we can pair them as below:

=> n = (d0 * 100 - d1 * 101 + 11 d1 * 101)  +  (d2 * 102 - d3 * 103 + 11 d3 * 103 + ...... (dk-1 * 10k-1 - dk * 10k + 11 dk * 10k)

=> n = (d0 * 100 - d1 * 101) + (d2 * 102 - d3 * 103) +...... (dk-1 * 10k-1 - dk * 10k) - (11 d1 * 101 + 11 d3 * 103 + ....11 dk * 10k)

=> n = sn - 11(d1 * 101 + d3 * 103 + ....dk * 10k)

If sn is divisible by 11, let sn = 11r where r is an integer.

=> n = 11r - 11(d1 * 101 + d3 * 103 + ....dk * 10k)

=> n = 11 (r - (d1 * 101 + d3 * 103 + ....dk * 10k))

Therefore n is divisible by 11.

If k is even, there are k+1 terms and we add a "k+2th term" to make sure pairing is possible

=> n = n = d0 * 100 + d1 * 101 + d2 * 102 + d3 * 103 + ...... dk-1 * 10k-1 + dk * 10k + dk+1 10k+1 - dk+1 10k+1 where dk+1 is some integer.

=> n = (d0 * 100 - d1 * 101 + 11 d1 * 101)  +  (d2 * 102 - d3 * 103 + 11 d3 * 103 + ...... (dk* 10k - dk+1 * 10k+1 + 11 dk+1 * 10k+1) - 10 dk+1 * 10k+1

=> n = (d0 * 100 - d1 * 101) + (d2 * 102 - d3 * 103) +...... (dk-1 * 10k-1 - dk * 10k) - (11 d1 * 101 + 11 d3 * 103 + ....11 dk * 10k) - 10 dk+1 * 10k+1

=> n = sn - 11(d1 * 101 + d3 * 103 + ....dk * 10k) - 10 dk+1 * 10k+1

Using the result of 3, we know that 11 | 10k+1 => 11 | 10 dk+1 * 10k+1 and let 10 dk+1 * 10k+1 = 11t.

If sn is divisible by 11, let sn = 11r where r is an integer.

=> n = 11r - 11(d1 * 101 + d3 * 103 + ....dk * 10k) - 11t

=> n = 11 (r - (d1 * 101 + d3 * 103 + ....dk * 10k) - t)

Therefore n is divisible by 11.

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