Maths Question, HELP!?

Question 1. Show that 43^n + 83 x 92^(3n - 1) is divisible for all positive intergers n.





Question 2. A florist has to make a floral arrangement. She has 6 banksias, 5 wattles and 4 waratahs. All the flowers of each kind are different. In how many ways can the florist make a bunch of 10 flowers if she has to use at least 3 of each kind?

Maths Question, HELP!?
1) 43^n + 83 x 92^(3n - 1)



***You can solve this with recurrence:

1- Prove this is true for n=1

2- Prove that if it is true for n, it is also true for n+1

Then it is true for any strictly positive integer



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For n =1

P(1) = 43^1 + 83*92^(3-1)= 702,555

702,555 = 7*100,365

So it is true for n=1



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if it is true for n, then P(n) = 43^n+83*92^(3n-1) = 7q

where q is an integer.



Let us know verify it is true for n+1:



P(n+1) = 43^(n+1) + 83*92^(3n+3-1)

P(n+1)= 43^(n+1) + 83*92^(3n-1)*92^2



we know that: 83*92^(3n-1) = 7p - 43^n

Let us replace that in the previous formula:



P(n+1) = 43^(n+1) + (7p - 43^n)*92^2

P(n+1) = 7p*92^2 + 43^(n+1) - 43^n*92^2

P(n+1) = 7p*92^2 +43^n(43-92^2)

P(n+1) = 7p*92^2 - 43^n*8,421

8,421 = 7*1,203, so:

P(n+1) = 7(p*92^2 -43^n*1,203)

P(n+1) is divisible by 7.



so P(n) is divisible by any integer n%26gt;0





Question 2. A florist has to make a floral arrangement. She has 6 banksias, 5 wattles and 4 waratahs. All the flowers of each kind are different. In how many ways can the florist make a bunch of 10 flowers if she has to use at least 3 of each kind?



Pick a banksia = 6

Pick a banksia = 5

Pick a banksia = 4



Pick a wattle = 5

Pick a wattle = 4

Pick a wattle = 3



Pick a waratah = 4

Pick a waratah = 3

Pick a waratah = 2



Remaining flowers = 3 banksias, 2 wattles and 1 waratah = 6



Pick a banksia / wattle /waratah = 6



So, total ways



6 x 5 x 4 x 5 x 4 x 3 x 4 x 3 x 2 x 6 = 1036800
Reply:And provide CORRECT answer (1036800???) too! Report It
Reply:Q1. Divisible by what, I don't see? By 7 as Mika has supposed? Well, that can be done directly, which is much shorter than inductively:

43 = 42 + 1 = 7*6 + 1≡ 1 (mod 7);

83 = 84 - 1 = 7*12 - 1 ≡ -1 (mod 7);

92 = 91 +1 = 7*13 + 1 ≡ 1 (mod 7), so

43^n + 83 x 92^(3n - 1) ≡ 1^n - 1*1^(3n-1) ≡ 1 - 1 ≡ 0 (mod 7).



Q2. Sum 10 can be obtained by 3 ways:

10 = 3 + 3 + 4 = 3 + 4 + 3 = 4 + 3 + 3, then the answer is

(6C3)*(5C3)*(4C4) + (6C3)*(5C4)*(4C3) + (6C4)*(5C3)*(4C3) =

200 + 400 + 600 = 1200, much less (3!3!4! = 6*6*24 times) than Mika has found.

Here nCr is the number of combinations of "n" things, taken "r" at a time.