2003 AMC 10A Problems/Problem 25
Problem
Let be a
-digit number, and let
and
be the quotient and the remainder, respectively, when
is divided by
. For how many values of
is
divisible by
?
Solution
Solution 1
When a -digit number is divided by
, the first
digits become the quotient,
, and the last
digits become the remainder,
.
Therefore, can be any integer from
to
inclusive, and
can be any integer from
to
inclusive.
For each of the possible values of
, there are at least
possible values of
such that
.
Since there is "extra" possible value of
that is congruent to
, each of the
values of
that are congruent to
have
more possible value of
such that
.
Therefore, the number of possible values of such that
is
.
Solution 2
Let equal
, where
through
are digits. Therefore,
We now take :
The divisor trick for 11 is as follows:
"Let be an
digit integer. If
is divisible by
, then
is also divisible by 11."
Therefore, the five digit number is divisible by 11. The 5-digit multiples of 11 range from
to
. There are
divisors of 11 between those inclusive.
See Also
2003 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Final Question | |
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 10 Problems and Solutions |