Difference between revisions of "2023 AMC 10A Problems/Problem 12"
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*The number <math>N</math> is divisible by <math>7</math>. | *The number <math>N</math> is divisible by <math>7</math>. | ||
− | *The number formed by reversing the digits of <math>N</math> is | + | *The number formed by reversing the digits of <math>N</math> is divisible by <math>5</math>. |
<math>\textbf{(A) } 13 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 15 \qquad \textbf{(D) } 16 \qquad \textbf{(E) } 17</math> | <math>\textbf{(A) } 13 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 15 \qquad \textbf{(D) } 16 \qquad \textbf{(E) } 17</math> |
Revision as of 00:50, 10 November 2023
Contents
[hide]Problem
How many three-digit positive integers satisfy the following properties?
- The number
is divisible by
.
- The number formed by reversing the digits of
is divisible by
.
Solution 1
Multiples of always end in
or
and since it is a three-digit number (otherwise it would be a two-digit number), it cannot start with 0. All possibilities have to be in the range from
to
inclusive.
.
.
~walmartbrian ~Shontai ~andliu766 ~andyluo
Solution 2 (solution 1 but more thorough)
Let We know that
is divisible by
, so
is either
or
. However, since
is the first digit of the three-digit number
, it can not be
, so therefore,
. Thus,
There are no further restrictions on digits
and
aside from
being divisible by
.
The smallest possible is
. The next smallest
is
, then
, and so on, all the way up to
. Thus, our set of possible
is
. Dividing by
for each of the terms will not affect the cardinality of this set, so we do so and get
. We subtract
from each of the terms, again leaving the cardinality unchanged. We end up with
, which has a cardinality of
. Therefore, our answer is
~ Technodoggo
Solution 3 (modular arithmetic)
We first proceed as in the above solution, up to .
We then use modular arithmetic:
We know that . We then look at each possible value of
:
If , then
must be
.
If , then
must be
or
.
If , then
must be
.
If , then
must be
or
.
If , then
must be
.
If , then
must be
.
If , then
must be
or
.
If , then
must be
.
If , then
must be
or
.
If , then
must be
.
Each of these cases are unique, so there are a total of
~ Technodoggo
See Also
2023 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.