2017 AMC 10A Problems/Problem 25
Contents
Problem
How many integers between 100 and 999, inclusive, have the property that some permutation of its digits is a multiple of 11 between 100 and 999 ? For example, both 121 and 211 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 (process of elimination on multiple choices)
Taken from solution three, we notice that there are a total of multiples of between and , and each of them have at most permutation (and thus is a permutations of numbers), giving us a maximum of 
valid numbers.
However, if can be divided by , so can , which is distinct if 
. And if 
then and have the same permutations. Either way, we have doubled counted.
This reduces the number of permutations to 
.
Furthermore, if 
or 
(which turn out to be equivalent conditions! for example ), not all (inverse) permutations are distinct (
) or valid (
). (There are 9 of these.)
Similarly, for 
, not all (inverse) permutations are valid. (There are 8 of these.)
As long as you notice at least one example of one of these 3 cases, you may infer that the answer must be smaller than . This leaves us with only one possible answer: .
Video Solution
https://youtu.be/leCUVcplmZ0?si=kGq5U9TSKUP1Y30I
~ Pi Academy
Video solution 2
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. 
