AMC12 2005 A

AMC12 2005 A · Q19

AMC12 2005 A · Q19. It mainly tests Base representation.

A faulty car odometer proceeds from digit $3$ to digit $5$, always skipping the digit $4$, regardless of position. For example, after traveling one mile the odometer changed from $000039$ to $000050$. If the odometer now reads $002005$, how many miles has the car actually traveled?

一个有故障的汽车里程表从数字 $3$ 直接跳到数字 $5$，总是跳过数字 $4$，无论在何位置。例如，行驶一英里后，里程表从 $000039$ 变为 $000050$。如果里程表现在显示 $002005$，汽车实际行驶了多少英里？

(A) 1404 1404

(B) 1462 1462

(D) 1605 1605

(E) 1804 1804

Answer

Correct choice: (B)

正确答案：（B）

Solution

We find the number of numbers with a $4$ and subtract from $2005$. Quick counting tells us that there are $200$ numbers with a 4 in the hundreds place, $200$ numbers with a 4 in the tens place, and $201$ numbers with a 4 in the units place (counting $2004$). Now we apply the Principle of Inclusion-Exclusion. There are $20$ numbers with a 4 in the hundreds and in the tens, and $20$ for both the other two intersections. The intersection of all three sets is just $2$. So we get: Alternatively, consider that counting without the number $4$ is almost equivalent to counting in base $9$; only, in base $9$, the number $9$ is not counted. Since $4$ is skipped, the symbol $5$ represents $4$ miles of travel, and we have traveled $2004_9$ miles. By base conversion, $2004_9=9^3(2)+9^0(4)=729(2)+1(4)=1458+4=\boxed{1462}$ Since any numbers containing one or more $4$s were skipped, we need only to find the numbers that don't contain a $4$ at all. First we consider $1$ - $1999$. Single digits are simply four digit numbers with a zero in all but the ones place (this concept applies to double and triple digits numbers as well). From $1$ - $1999$, we have $2$ possibilities for the thousands place, and $9$ possibilities for the hundreds, tens, and ones places. This is $2 \cdot 9 \cdot 9 \cdot 9-1$ possibilities (because $0000$ doesn't count) or $1457$ numbers. From $2000$ - $2005$ there are $6$ numbers, $5$ of which don't contain a $4$. Therefore the total is $1457 + 5$, or $1462$ $\Rightarrow$ $\boxed{\text{B}}$. We seek to find the amount of numbers that contain at least one $4,$ and subtract this number from $2005.$ We can simply apply casework to this problem. The amount of numbers with at least one $4$ that are one or two digit numbers are $4,14,24,34,40-49,54,\cdots,94$ which gives $19$ numbers. The amount of three digit numbers with at least one $4$ is $8*19+100=252.$ The amount of four digit numbers with at least one $4$ is $252+1+19=272$ This, our answer is $2005-19-252-272=1462,$ or $\boxed{B}.$ This is very analogous to base $9$. But, in base $9$, we don't have a $9$. So, this means that these are equal except for that base 9 will be one more than the operation here. $2005_9 = 5+0+0+1458 = 1463$. $1463 - 1 = 1462$

我们求出包含数字 $4$ 的数的个数，然后从 $2005$ 中减去。快速计数可知：百位上为 $4$ 的有 $200$ 个，十位上为 $4$ 的有 $200$ 个，个位上为 $4$ 的有 $201$ 个（包括 $2004$）。现在应用容斥原理。百位和十位同时为 $4$ 的有 $20$ 个，另外两个交集也各有 $20$ 个。三个集合的交集只有 $2$ 个。因此我们得到： Alternatively, consider that counting without the number $4$ is almost equivalent to counting in base $9$; only, in base $9$, the number $9$ is not counted. Since $4$ is skipped, the symbol $5$ represents $4$ miles of travel, and we have traveled $2004_9$ miles. By base conversion, $2004_9=9^3(2)+9^0(4)=729(2)+1(4)=1458+4=\boxed{1462}$ 由于任何包含一个或多个 $4$ 的数都被跳过，我们只需找出完全不含 $4$ 的数的个数。先考虑 $1$ 到 $1999$。一位数可以看作除个位外其余位为零的四位数（两位数、三位数同理）。从 $1$ 到 $1999$，千位有 $2$ 种可能，百位、十位、个位各有 $9$ 种可能。这共有 $2 \cdot 9 \cdot 9 \cdot 9-1$ 种可能（因为 $0000$ 不算），即 $1457$ 个数。从 $2000$ 到 $2005$ 有 $6$ 个数，其中 $5$ 个不含 $4$。因此总数为 $1457 + 5$，即 $1462$ $\Rightarrow$ $\boxed{\text{B}}$。我们要找至少包含一个 $4$ 的数的个数，并从 $2005$ 中减去。我们可以对本题进行分类讨论。一位或两位数中至少含一个 $4$ 的有 $4,14,24,34,40-49,54,\cdots,94$，共 $19$ 个。三位数中至少含一个 $4$ 的有 $8*19+100=252$ 个。四位数中至少含一个 $4$ 的有 $252+1+19=272$ 因此答案为 $2005-19-252-272=1462,$ 即 $\boxed{B}$。这与九进制非常类似。但在九进制中没有数字 $9$。因此这两者几乎相同，只是九进制计数会比这里多 $1$。$2005_9 = 5+0+0+1458 = 1463$. $1463 - 1 = 1462$

Topics

NT.008 Base representation