Is it possible to have 60 coins, made up of pennies, dimes and quarters that add up to $5.00? Justify your answer.
anyone who can fully explain this one please.
I was unable to prove this.
Let $\displaystyle n_1,\ n_{10},\ n_{25} \in \mathbb{N}$ be the number of pennies, dimes and quarters in a solution if it exists.
Then we have:
$\displaystyle n_1+10n_{10}+25n_{25}=500$
and
$\displaystyle n_1+n_{10}+n_{25}=60.$
Assume you know $\displaystyle n_1$ and solve for $\displaystyle n_{25}$.
Is this solution consistent with $\displaystyle n_{25}$ being in $\displaystyle \mathbb{N}$?
RonL
They're two separate equations - the 500 isn't "changing" into 60. The first equation deals with the value of the coins. $5.00 is the total value, so that's been changed to 500 cents. n_1 is the number of pennies, so it simply has a value of n_1 cents. n_10 is the number of dimes, so it has a value of 10 * n_10, and likewise for n_25 and quarters.
The second equation deals with the total number of coins, which you know to be 60, and you know it is the sum of the number of pennies, dimes, and quarters, hence n_1 + n_10 + n_25 = 60.
Expand $\displaystyle \left( {\sum\limits_{k = 0}^{60} {x^k } } \right)\left( {\sum\limits_{k = 0}^{60} {x^{10k} } } \right)\left( {\sum\limits_{k = 0}^{60} {x^{25k} } } \right)$ and look at the coefficient of $\displaystyle x^{500}$ which is 127. That is the number of ways to make up $5 from sixty coins all of which are pennies, dimes, or quarters.
Hello, rcmango!
Is it possible to have 60 coins, made up of pennies, dimes and quarters
that add up to $5.00? Justify your answer.
Let $\displaystyle P$ = number of pennies,
. . .$\displaystyle D$ = number of dimes,
. . .$\displaystyle Q$ = number of quarters.
where $\displaystyle P,D,Q$ are nonnegative integers.
There are 60 coins: .$\displaystyle P + D + Q \:=\:60\;\;{\color{blue}[1]}$
Their value is 500 cents: .$\displaystyle P + 10D + 25Q \:=\:500\;\;{\color{blue}[2]}$
Subtract [1] from [2]: .$\displaystyle 9D + 24Q \:=\:440$
So we have: .$\displaystyle 3(3D + 8Q) \;=\;440$
The left side is a multiple of 3 . . . The right side is not.
. . There are no solutions in integers.
Therefore, the task is impossible.