The total value of the final 84 coins on the table is $10.71
The given parameters;
The sum of multiples of 2 between 2 and 84 inclusive:
[tex]= \frac{84 - 2}{2} + 1 = 42[/tex]
The maximum number of Dimes in the first replacement = 42
The sum of multiples of 3 between 3 and 84 inclusive:
[tex]= \frac{84 - 3}{3} + 1 = 28[/tex]
The maximum number of Nickle in the second replacement = 28
The sum of multiples of 4 between 4 and 84 inclusive:
[tex]= \frac{84 - 4}{4} + 1 = 21[/tex]
The maximum number of Penny in the third replacement = 21
The sum of multiples of will 2 reduce as we introduce multiples of 3 and multiples of 4 because of overlap in multiples of 6 for 2 & 3, and multiples 4 for 2 $ 4.
We are going to remove this overlap in the sum of multiples of 2.
The sum of multiples of 6 between 6 and 84 inclusive:
[tex]= \frac{84 - 6}{6} + 1= 14[/tex]
The sum of multiples of 4 between 4 and 84 inclusive = 21
The new sum of multiples of 2 only = 42 - (14 + 21) = 7
The total number of dime = 7
The sum of multiples of 3 will reduce after we introduce multiples of 4 because of overlap in multiples of 12;
The sum of multiples of 12 between 12 and 84 inclusive:
[tex]= \frac{84-12}{12} + 1= 7[/tex]
The new sum of multiples of 3 only = 28 - (7) = 21
The total number of Nickle = 21
There is no replacement for multiples of 4 because it is the last.
The total number of Penny = sum of multiples of 4 = 21
The total value of the final coins on the table is calculated as;
the sum of quarters not replaced = 84 - (7 + 21 + 21) = 35 quarters
total value of the coins = 35 quarters + 7 Dimes + 21 Nickels + 21 Penny
total value of the coins = 35($0.25) + 7($0.1) + 21($0.05) + 21($0.01)
total value of the coins = $10.71
Thus, the total value of the final 84 coins on the table is $10.71
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