# Cans

How many cans must be put in the bottom row if we want 182 cans arrange in 13 rows above so that each subsequent row has always been one tin less? How many cans will be in the top row?

Result

a =  20
b =  8

#### Solution:

$s = 182 \ \\ n = 13 \ \\ d = -1 \ \\ \ \\ s = n(a+b)/2 \ \\ s = n(a+a+(n-1)d)/2 \ \\ \ \\ 182 = 13(2a-n+1)/2 \ \\ 28 = 2a-n+1 \ \\ 28 = 2a-12 \ \\ 40 = 2a \ \\ a = 20$
$b=20-1 \cdot \ (13-1)=8$

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