Difference between revisions of "2017 AMC 10A Problems/Problem 25"
m (→Solution 5 (Fastest)) |
Batmanstark (talk | contribs) |
||
Line 89: | Line 89: | ||
- Minor edits for clarity by dstanz5 | - Minor edits for clarity by dstanz5 | ||
+ | == Solution 5 (Bashing!) | ||
+ | |||
+ | Basically, we see that we just have to list out <math>81</math> numbers that satisfy the condition of divisible by <math>11</math> so we can basically list out the possible numbers and group them out by the hundred's digit. While, listing out, we notice that some of the numbers can be removed from the count because some numbers would have the same permutations, when we notice this we can remove these. | ||
+ | |||
+ | Using the remaining numbers, we just have to count the number of ways to arrange each of the possible cases. | ||
+ | |||
+ | These cases would be: | ||
+ | |||
+ | Case <math>1</math>: all distinct digits without <math>0</math>'s | ||
+ | |||
+ | There are <math>28</math> from the remaining list that satisfy this. As each has <math>6</math> permutations we get <math>28 \times 6 = 168</math> ways. | ||
+ | |||
+ | Case <math>2</math>: All distinct digits, but with <math>0</math>: | ||
+ | |||
+ | There are <math>4</math> numbers from our remaining list, and as the digits are distinct, there are <math>6</math> ways - <math>2</math> ways = <math>4</math> ways because we've to remove the permutations of when the first digit is <math>0</math> which is <math>2</math> ways. This gives <math>4\times4=16</math> ways. | ||
+ | |||
+ | Case <math>3</math>: <math>1</math> common digit, <math>2</math> different digits, no <math>0</math>: | ||
+ | |||
+ | There are <math>8</math> numbers from our remaining list, and as there are 2 common digits, we get the number of ways to arrange as <math>\dfrac{3!}{2!}=3</math> ways. This gives <math>3\times8=24</math> ways. | ||
+ | |||
+ | Case <math>4</math>: <math>1</math> common digit, <math>2</math> different digits, the other digit is <math>0</math>: | ||
+ | |||
+ | There are <math>9</math> numbers from our remaining list that work and there are <math>3-1=2</math> ways, to not include <math>0</math> as first digit. This gives <math>2\times9=18</math> ways. | ||
+ | |||
+ | Adding the cases up gives us <math>168+16+24+18=\boxed{B. 226}</math> | ||
+ | |||
+ | |||
+ | From our list, we see that there | ||
==Video Solution== | ==Video Solution== | ||
Two different variations on solving it. | Two different variations on solving it. |
Revision as of 00:42, 7 November 2022
Contents
Problem
How many integers between and
, inclusive, have the property that some permutation of its digits is a multiple of
between
and
For example, both
and
have this property.
Solution 1
There are 81 multiples of 11 between and
inclusive. Some have digits repeated twice, making 3 permutations.
Others that have no repeated digits have 6 permutations, but switching the hundreds and units digits also yield a multiple of 11. Switching shows we have overcounted by a factor of 2, so assign permutations to each multiple.
There are now 81*3 = 243 permutations, but we have overcounted*. Some multiples of 11 have as a digit. Since
cannot be the digit of the hundreds place, we must subtract a permutation for each.
There are 110, 220, 330 ... 990, yielding 9 extra permutations
Also, there are 209, 308, 407...902, yielding 8 more permutations.
Now, just subtract these 17 from the total (243) to get 226.
- If short on time, observe that 226 is the only answer choice less than 243, and therefore is the only feasible answer.
Solution 2
We note that we only have to consider multiples of and see how many valid permutations each has. We can do casework on the number of repeating digits that the multiple of
has:
All three digits are the same.
By inspection, we find that there are no multiples of
here.
Two of the digits are the same, and the third is different.
There are
multiples of
without a zero that have this property:
,
,
,
,
,
,
,
.
Each contributes
valid permutations, so there are
permutations in this subcase.
There are
multiples of
with a zero that have this property:
,
,
,
,
,
,
,
,
.
Each one contributes
valid permutations (the first digit can't be zero), so there are
permutations in this subcase.
All the digits are different.
Since there are
multiples of
between
and
, there are
multiples of
remaining in this case. However,
of them contain a zero, namely
,
,
,
,
,
,
, and
. Each of those multiples of
contributes
valid permutations, but we overcounted by a factor of
; every permutation of
, for example, is also a permutation of
. Therefore, there are
. Therefore, there are
remaining multiples of
without a
in this case. Each one contributes
valid permutations, but once again, we overcounted by a factor of
(note that if a number ABC is a multiple of
, then so is CBA). Therefore, there are
valid permutations in this subcase.
Adding up all the permutations from all the cases, we have .
Solution 3
We can first overcount and then subtract.
We know that there are multiples of
.
We can then multiply by for each permutation of these multiples. (Yet some multiples do not have six distinct permutations.)
Now divide by , because if a number
with digits
,
, and
is a multiple of
, then
is also a multiple of
so we have counted the same permutations twice.
Basically, each multiple of has its own
permutations (say
has
and
whereas
has
and
). We know that each multiple of
has at least
permutations because it cannot have
repeating digits.
Hence we have permutations without subtracting for overcounting.
Now note that we overcounted cases in which we have
's at the start of each number. So, in theory, we could just answer
and then move on.
If we want to solve it, then we continue.
We overcounted cases where the middle digit of the number is and the last digit is
.
Note that we assigned each multiple of three permutations.
The last digit is gives
possibilities where we overcounted by
permutation for each of
.
The middle digit is gives
possibilities where we overcount by
.
and
Subtracting gives
.
Now, we may ask if there is further overlap (i.e if two of and
and
were multiples of
). Thankfully, using divisibility rules, this can never happen, as taking the divisibility rule mod
and adding, we get that
,
, or
is congruent to
. Since
are digits, this can never happen as none of them can equal
and they can't equal
as they are the leading digit of a three-digit number in each of the cases.
Solution 4 (1 but quicker)
The smallest multiple of above
is
, while the largest multiple of
less than
is
. This means there are
multiples of
between
and
.
As there are permutations for each multiple, we have
. However, we have overcounted, as numbers like
shouldn't be counted. Looking at the answer choices, we notice there is only one that is less than
, and so we have our answer as
.
-¢
Solution 5 (Fastest)
The lowest multiple of 11 between 100 and 999 is 110, and the largest is 990. We have
multiples of 11, so we know there are at most
answers.
However, there must be overcount in these 243, so obviously, the answer is
, since it is the only answer less than
.
- Little - Minor edits for clarity by dstanz5
== Solution 5 (Bashing!)
Basically, we see that we just have to list out numbers that satisfy the condition of divisible by
so we can basically list out the possible numbers and group them out by the hundred's digit. While, listing out, we notice that some of the numbers can be removed from the count because some numbers would have the same permutations, when we notice this we can remove these.
Using the remaining numbers, we just have to count the number of ways to arrange each of the possible cases.
These cases would be:
Case : all distinct digits without
's
There are from the remaining list that satisfy this. As each has
permutations we get
ways.
Case : All distinct digits, but with
:
There are numbers from our remaining list, and as the digits are distinct, there are
ways -
ways =
ways because we've to remove the permutations of when the first digit is
which is
ways. This gives
ways.
Case :
common digit,
different digits, no
:
There are numbers from our remaining list, and as there are 2 common digits, we get the number of ways to arrange as
ways. This gives
ways.
Case :
common digit,
different digits, the other digit is
:
There are numbers from our remaining list that work and there are
ways, to not include
as first digit. This gives
ways.
Adding the cases up gives us
From our list, we see that there
Video Solution
Two different variations on solving it. https://youtu.be/z5KNZEwmrWM
https://youtu.be/MBcHwu30MX4 -Video Solution by Richard Rusczyk
~savannahsolver
See Also
2017 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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.