Difference between revisions of "2023 AMC 10A Problems/Problem 12"
m (→Solution 1) |
m (→Solution 4) |
||
(21 intermediate revisions by 14 users not shown) | |||
Line 10: | Line 10: | ||
==Solution 1== | ==Solution 1== | ||
− | Multiples of <math>5</math> will always end in <math>0</math> or <math>5</math>, and since the | + | Multiples of <math>5</math> will always end in <math>0</math> or <math>5</math>, and since the numbers have to be a three-digit numbers (otherwise it would be a two-digit number), it cannot start with 0, narrowing our choices to 3-digit numbers starting with <math>5</math>. Since the numbers must be divisible by 7, all possibilities have to be in the range from <math>7 \cdot 72</math> to <math>7 \cdot 85</math> inclusive. |
<math>85 - 72 + 1 = 14</math>. <math>\boxed{\textbf{(B) } 14}</math>. | <math>85 - 72 + 1 = 14</math>. <math>\boxed{\textbf{(B) } 14}</math>. | ||
− | ~walmartbrian ~Shontai ~andliu766 ~andyluo | + | (Add 1 to include 72) |
+ | |||
+ | ~walmartbrian ~Shontai ~andliu766 ~andyluo ~ESAOPS | ||
==Solution 2 (solution 1 but more thorough)== | ==Solution 2 (solution 1 but more thorough)== | ||
Line 64: | Line 66: | ||
==Solution 4== | ==Solution 4== | ||
− | + | The key point is that when reversed, the number must start with a <math>0</math> or a <math>5</math> based on the second restriction. But numbers can't start with a <math>0</math>. | |
+ | |||
+ | So the problem is simply counting the number of multiples of <math>7</math> in the <math>500</math>s. | ||
+ | |||
+ | <math>7 \times 72 = 504</math>, so the first multiple is <math>7 \times 72</math>. | ||
+ | |||
+ | <math>7 \times 85 = 595</math>, so the last multiple is <math>7 \times 85</math>. | ||
+ | |||
+ | Now, we just have to count <math>7\times 72, 7\times 73, 7\times 74,\cdots, 7\times 85</math>. | ||
+ | |||
+ | We have a set that numbers <math>85-71=\boxed{\textbf{(B) 14}}</math> | ||
+ | |||
+ | ~Dilip ~boppitybop ~ESAOPS (LaTeX) | ||
+ | |||
+ | |||
+ | ==Video Solution ⚡️ 2 min solution ⚡️== | ||
+ | https://youtu.be/YdaQIdxyBSg | ||
+ | |||
+ | <i> ~Education, the Study of Everything </i> | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/UYHCNlRDZBo | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
− | + | ==Video Solution== | |
+ | https://www.youtube.com/watch?v=4uKo5NR2o9Y | ||
− | + | -paixiao | |
− | + | ==Video Solution == | |
+ | https://youtu.be/N2lyYRMuZuk?si=6B-mTB070UP2yuDF&t=435 | ||
− | + | ~Math-X | |
− | + | ==Video Solution == | |
+ | https://www.youtube.com/watch?v=Mg6JUanYNJY | ||
− | + | ==Note== | |
+ | According to the official answer key, choice (B) is correct. However, some have argued that it is ambiguous whether the number <math>560</math> should be included in the count, since its reversal, <math>065</math>, has a leading zero. It is assumed that <math>065</math> denotes the two-digit number <math>65</math>, which is divisible by <math>5</math>, but MAA should have clarified what happens when a number with trailing zeros is reversed. | ||
− | ~ | + | ~A_MatheMagician ~ESAOPS ~sdpandit |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2023|ab=A|num-b=11|num-a=13}} | {{AMC10 box|year=2023|ab=A|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 13:46, 14 August 2024
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 numbers have to be a three-digit numbers (otherwise it would be a two-digit number), it cannot start with 0, narrowing our choices to 3-digit numbers starting with
. Since the numbers must be divisible by 7, all possibilities have to be in the range from
to
inclusive.
.
.
(Add 1 to include 72)
~walmartbrian ~Shontai ~andliu766 ~andyluo ~ESAOPS
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
The key point is that when reversed, the number must start with a or a
based on the second restriction. But numbers can't start with a
.
So the problem is simply counting the number of multiples of in the
s.
, so the first multiple is
.
, so the last multiple is
.
Now, we just have to count .
We have a set that numbers
~Dilip ~boppitybop ~ESAOPS (LaTeX)
Video Solution ⚡️ 2 min solution ⚡️
~Education, the Study of Everything
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
https://www.youtube.com/watch?v=4uKo5NR2o9Y
-paixiao
Video Solution
https://youtu.be/N2lyYRMuZuk?si=6B-mTB070UP2yuDF&t=435
~Math-X
Video Solution
https://www.youtube.com/watch?v=Mg6JUanYNJY
Note
According to the official answer key, choice (B) is correct. However, some have argued that it is ambiguous whether the number should be included in the count, since its reversal,
, has a leading zero. It is assumed that
denotes the two-digit number
, which is divisible by
, but MAA should have clarified what happens when a number with trailing zeros is reversed.
~A_MatheMagician ~ESAOPS ~sdpandit
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.