AMC8 2023

AMC8 2023 · Q14

AMC8 2023 · Q14. It mainly tests Money / coins, Diophantine equations (integer solutions).

Nicolas is planning to send a package to his friend Anton, who is a stamp collector. To pay for the postage, Nicolas would like to cover the package with a large number of stamps. Suppose he has a collection of $5$-cent, $10$-cent, and $25$-cent stamps, with exactly $20$ of each type. What is the greatest number of stamps Nicolas can use to make exactly $\$7.10$ in postage? (Note: The amount $\$7.10$ corresponds to $7$ dollars and $10$ cents. One dollar is worth $100$ cents.)

Nicolas计划寄包裹给他的朋友Anton，后者是集邮爱好者。为了支付邮费，Nicolas想用大量邮票覆盖包裹。假设他有5美分、10美分和25美分邮票各正好20张。Nicolas能用多少张邮票来正好凑成$\$7.10$的邮费？（注：$\$7.10$对应7美元10美分。1美元=100美分。）

(A) \ 45 \ 45

(B) \ 46 \ 46

(D) \ 54 \ 54

(E) \ 55 \ 55

Answer

Correct choice: (E)

正确答案：（E）

Solution

Let's use the most stamps to make $7.10.$ We have $20$ of each stamp, $5$-cent (nickels), $10$-cent (dimes), and $25$-cent (quarters). If we want the highest number of stamps, we must have the highest number of the smaller value stamps (like the coins above). We can use $20$ nickels and $20$ dimes to bring our total cost to $7.10 - 3.00 = 4.10$. However, when we try to use quarters, the $25$ cents don’t fit evenly, so we have to give back $15$ cents to make the quarter amount $4.25$. The most efficient way to do this is to give back a $10$-cent (dime) stamp and a $5$-cent (nickel) stamp to have $38$ stamps used so far. Now, we just use $\frac{425}{25} = 17$ quarters to get a grand total of $38 + 17 = \boxed{\textbf{(E)}\ 55}$.

我们用最多邮票凑成$7.10$。有20张5美分（镍币）、20张10美分（角币）和25美分（25分币）。要用最多邮票，必须用最多小面值邮票。用20张5美分和20张10美分，总价值$20\times0.05 + 20\times0.10 = 1 + 2 = 3$美元，还需$4.10$。但用25美分邮票时，25美分不整除，所以调整为用17张25美分凑$4.25$，多出15美分，用1张10美分和1张5美分退回。这样用了20+19+17=56张？等，解法中说38+17=55。实际：20镍 + 19角 = 1+1.9=2.9，还需4.2，用17季度=4.25，总7.15，退1镍1角15美分，正好7.10，总20-1 +19-1 +17=19+18+17=54？解法说55，可能是20镍20角=3，还需4.1，调整为用17季度4.25，退15即1角1镍，用19镍19角17季度=19*5+19*10+17*25=95+190+425=710美分，是55张。\boxed{\textbf{(E)}\ 55}

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