Problem solving question need help!

[faa]

Hey everyone! I am struggling with this question, can someone help explain how they worked it out if you get it?

thank you!!

"One thousand and one pennies are arranged in a row on a table. Every second coin
is replaced with a nickel. Then every third coin is replaced with a dime. Finally,
every fourth coin is replaced with a quarter. What is the total value of coins left on
the table?"

 

[soroban]

Hello, faa!

One thousand and one pennies are arranged in a row on a table.
Every second coin is replaced with a nickel.
Then every third coin is replaced with a dime.
Finally, every fourth coin is replaced with a quarter.
What is the total value of coins left on the table?

Apply the procedure to the first 12 coins.

We have: .$\displaystyle P\;N\;D\;Q\;\;P\;D\;P\;Q\;\;D\;N\;P\;Q$
. . And this pattern repeats throughout the row of coins.

Among 12 consecutive coins, there are: .$\displaystyle \begin{Bmatrix}4\text{ P's} \\ 2\text{ N's} \\ 3\text{ D's} \\ 3\text{ Q's} \end{Bmatrix}$
$\displaystyle \dfrac{1001}{12} \:=\:83,\,\text{rem.}5$


\(\displaystyle \begin{array}{cccccc}
\text{83 sets of 4 P's:} & 83\cdot4 &=& 332\text{ P's} &=& \;332\rlap{/}c \\
\text{83 sets of 2 N's:} & 83\cdot2 &=& 166\text{ N's} &=& \;830\rlap{/}c \\
\text{83 sets of 3 D's:} & 83\cdot3 &=& 249\text{ D's} &=& 2490\rlap{/}c \\
\text{83 sets of 3 Q's:} & 83\cdot3 &=& 249\text{ Q's} &=& 6225\rlap{/}c \\ \hline
&&&\text{Total:} && 9877\rlap{/}c \end{array}\)

The final 5 coins are: .$\displaystyle P\;N\;D\;Q\;P \;=\;42\rlap{/}c$


Total value: .$\displaystyle 9877\rlap{/}c + 42\rlap{/}c \;=\;9919\rlap{/}c \;=\;\$99.19$

 

[faa]

Hi Soroban!

Thank you so much!! Great help, I see how it works out now, thanks for laying it out clearly!