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number theory

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George is about to get a certain amount of change less than one dollar from the cash register. If he gets the most quarters possible and the rest in pennies, he would need to receive 3 pennies to meet the amount. If he gets the most dimes possible and the rest in pennies, he would need to receive 8 pennies to meet the amount. What is the sum, in cents, of the possible amounts of change that he is trying to get?

Dec 11, 2018

#1
+1

78 cents

He can receive  3 quarters and 3 pennies

Or

He can receive 7 dimes and 8 pennies   Dec 11, 2018
#2
+3

Could also be 28 cents

1 quarter   3 pennies

2 dimes 8 pennies

Dec 11, 2018
#3
+10

George is about to get a certain amount of change less than one dollar from the cash register.

If he gets the most quarters possible and the rest in pennies, he would need to receive 3 pennies to meet the amount.

If he gets the most dimes possible and the rest in pennies, he would need to receive 8 pennies to meet the amount.

What is the sum, in cents, of the possible amounts of change that he is trying to get?

$$\text{Let n = quarters}\\ \text{Let m = dimes}$$

Formula:

$$\begin{array}{|rclrcl|} \hline \mathbf{25n+3} &=& \mathbf{10m+8} \\ 10m &=& 25n-5 \\\\ m &=& \dfrac{25n-5}{10} \\\\ &=& \dfrac{20n+5n-5}{10} \\\\ &=& \dfrac{20n}{10}+\dfrac{5n-5}{10} ~|~ \text{cut off whole parts} \\\\ m &=& 2n+ \underbrace{\dfrac{5n-5}{10}}_{=a} \qquad m = 2n+a \qquad (1) \\\\ & a &= \dfrac{5n-5}{10} \\\\ &10a &= 5n-5 \\\\ &5n &= 10a+5 \quad | \quad :5 \\\\ &\mathbf{n} & \mathbf{=} \mathbf{2a+1} \qquad (2) \\\\ m &=& 2n +a \\ &=& 2(2a+1) +a \\\\ \mathbf{m} & \mathbf{=}& \mathbf{5a+2} \qquad (3) \\ \hline \end{array}$$

conclusion:

$$\begin{array}{|rcll|} \hline \mathbf{n} & \mathbf{=}& \mathbf{2a+1} \qquad a \in \mathbb{Z} \\ \mathbf{m} & \mathbf{=}& \mathbf{5a+2} \qquad a \in \mathbb{Z} \\ \hline \end{array}$$

solution:

$$\begin{array}{|c|r|r|r|} \hline a & \text{quarters} & \text{dimes} & \text{pennies} \\ \hline 0 & 1 & 2& 28 \\ 1 & 3 & 7& 78 \\ 2 & 5 & 12 & 128 > 1~ ! \\ \hline \end{array}$$ Dec 11, 2018