2003 AMC 10A Problems/Problem 25
Contents
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
?
Solutions
Simple Solution
implies that
and therefore
, so
. Then,
can range from
to
for a total of
numbers.
<3
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
. Another way to think about it is the number of possible values of q when r, the remainder, is
. In this case, q itself has to be a multiple of
.
. Then we'll need to subtract
from
since we only want multiples of
greater than
Therefore, the number of possible values of such that
is
.
~ Minor Edit by PlainOldNumberTheory
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
."
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.
Note
The part labeled "divisor trick" actually follows from the same observation we made in the previous step: , therefore
and
for all
.
For a
digit number
we get
, as claimed.
Also note that in the "divisor trick" we actually want to assign the signs backwards - if we make sure that the last sign is a , the result will have the same remainder modulo
as the original number.
Solution 3
Since is a quotient and
is a remainder when
is divided by
. So we have
. Since we are counting choices where
is divisible by
, we have
for some
. This means that
is the sum of two multiples of
and would thus itself be a multiple of
. Then we can count all the five digit multiples of
as in Solution 2. (This solution is essentally the same as Solution 2, but it does not necessarily involve mods and so could potentially be faster.)
Solution 4
Defining and
in terms of floor functions and
would yield:
and
. Since
,
. Simplifying gets us
(
is always true, since floor function always yields an integer, and 99 is divisble by 11 w/o any remainder). After we come to this conclusion, it becomes easy to solve the rest of the problem (
).
~hw21
Video Solution 1
https://youtu.be/OpGHj-B0_hg?t=672
~IceMatrix
Video Solution 2
~savannahsolver
See Also
2003 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last 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 |
2003 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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.