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





**************


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.

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