2003 AMC 10A Problems/Problem 25
Contents
[hide]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 want to assign the signs backward - 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 essentially 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 divisible 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 |
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