2005 AMC 12A Problems/Problem 19
Problem
A faulty car odometer proceeds from digit 3 to digit 5, always skipping the digit 4, regardless of position. If the odometer now reads 002005, how many miles has the car actually traveled?
Solutions
Solution 1
We find the number of numbers with a and subtract from
. Quick counting tells us that there are
numbers with a 4 in the hundreds place,
numbers with a 4 in the tens place, and
numbers with a 4 in the units place (counting
). Now we apply the Principle of Inclusion-Exclusion. There are
numbers with a 4 in the hundreds and in the tens, and
for both the other two intersections. The intersection of all three sets is just
. So we get:

Solution 2
Alternatively, consider that counting without the number is almost equivalent to counting in base
; only, in base
, the number
is not counted. Since
is skipped, the symbol
represents
miles of travel, and we have traveled
miles. By basic conversion,
Solution 3
Since any numbers containing one or more s were skipped, we need only to find the numbers that don't contain a
at all. First we consider
-
. 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
-
, we have
possibilities for the thousands place, and
possibilities for the hundreds, tens, and ones places. This is
possibilities (because
doesn't count) or
numbers. From
-
there are
numbers,
of which don't contain a
. Therefore the total is
, or
.
See also
2005 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.