2023 AMC 10A Problems/Problem 12
Contents
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 will always end in
or
, and since the number has to be a three-digit number (otherwise it would be a two-digit number), it cannot start with 0, narrowing our choices to 3-digit numbers starting with
. 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
Solution 4
Initially I thought of finding that there are 142 such numbers divisible by 7 since 1000 divided by 7 gives 142 with a remainder. But it's not relevant!
The key point is that when reversed the number must start with a 0 or a 5 based on the second restriction. But numbers can't start with a 0.
So the problem is simply counting the number of multiples of 7 in the 500s.
7 x 70 = 490 so the first multiple is 7 x 72
7 x 80 = 560 so the first multiple more than 599 is 7 x 86 (since 7 x 6 = 42 and 560 + 42 is in the 600s)
Now we just have to count 7x72, 7x73, 7x74, ..., 7x85
We have a set that numbers 85-71 =
~Dilip
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 |
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