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Zielflug [23.3K]
3 years ago
6

Prove that given any 17 integers, there exist nine of them whose sum is divisible by 9.

Mathematics
2 answers:
vaieri [72.5K]3 years ago
7 0

This is a simple consequence of the Zero-sum problem: https://en.wikipedia.org/wiki/Zero-sum_problem

yaroslaw [1]3 years ago
7 0

Answer:

Step-by-step explanation:

let x,x+1,x+2,x+3,x+4,x+5,x+6,x+7,x+8

be 9 integers

x+x+1+x+2+x+3+x+4+x+5+x+6+x+7+x+8=9x+36=9(x+4)

which is divisible by 9

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V_F=\frac{343}{3}\pi+294\pi=\frac{343}{3}\pi+\frac{294\cdot3}{3}\pi=\frac{343}{3}\pi+\frac{882}{3}\pi=\boxed{\frac{1225}{3}\pi\ (ft^3)}\\\\\approx\frac{1225}{3}\cdot3.14}=\frac{3846.5}{3}\approx\boxed{1,282.17\ (ft^3)}\leftarrow answer
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Neporo4naja [7]
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Answer:

<h3>2min/customer</h3>

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Given

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Rate = 16min/8customers

Rate = 2min/customer

Hence the rate at which the customers entering the store in minutes per customer is 2min/customer.

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Hope this helps!
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