This is for discrete math PLEASE do question 3 and 4, because to do question 4 u

Question

This is for discrete math PLEASE do question 3 and 4, because to do question 4 u need to use part of question 3.

I request to do Question 3 and question 4

3. Prove that vk e N+, Ym e N, k = 2m + 1 11 l (10k + 1). 4. Given a positive integer n, we can write the decimal representation of n as where di is the ith digit and is an integer from 0 to 9 inclusive, dk0 (ie, the left-most digit cannot be 0) for n > 0, k >0 and: n=dk . 10k + dk-1 . 10k-i+4k-2-10k-2 + +41 . 101 + do . 100 Define the alternating sum For example, 180928 8-2+9-0+8-1 Prove that if sn is divisible by 11 then n is divisible by 11. Hint: Use what you proved in Question 3.

Explanation / Answer

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 * dk-1 + dk * dk

=> 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.

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