2017 AMC 10A Problems/Problem 25

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.

$\mathrm{\textbf{(A)} \ }226\qquad \mathrm{\textbf{(B)} \ } 243 \qquad \mathrm{\textbf{(C)} \ } 270 \qquad \mathrm{\textbf{(D)} \ }469\qquad \mathrm{\textbf{(E)} \ } 486$

Solution 1

There are 81 multiples of 11. 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. Therefore, assign 3 permutations to each multiple.

There are now 81*3 = 243 permutations, but we have overcounted*. Some multiples of 11 have a zero, and 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), getting 226. $\boxed{\textbf{(A) } 226}$

Note: If short on time, note that 226 is the only answer choice less than 243, and therefore is the only feasible answer.

Solution 3

We note that we only have to consider multiples of $11$ and see how many valid permutations each has. We can do casework on the number of repeating digits that the multiple of $11$ has:

$\textbf{Case 1:}$ All three digits are the same. By inspection, we find that there are no multiples of $11$ here.

$\textbf{Case 2:}$ Two of the digits are the same, and the third is different.

$\textbf{Case 2a:}$ There are $8$ multiples of $11$ without a zero that have this property: $121$ , $242$ , $363$ , $484$ , $616$ , $737$ , $858$ , $979$ . Each contributes $3$ valid permutations, so there are $8 \cdot 3 = 24$ permutations in this subcase.

$\textbf{Case 2b:}$ There are $9$ multiples of $11$ with a zero that have this property: $110$ , $220$ , $330$ , $440$ , $550$ , $660$ , $770$ , $880$ , $990$ . Each one contributes $2$ valid permutations (the first digit can't be zero), so there are $9 \cdot 2 = 18$ permutations in this subcase.

$\textbf{Case 3:}$ All the digits are different. Since there are $\frac{990-110}{11}+1 = 81$ multiples of $11$ between $100$ and $999$ , there are $81-8-9 = 64$ multiples of $11$ remaining in this case. However, $8$ of them contain a zero, namely $209$ , $308$ , $407$ , $506$ , $605$ , $704$ , $803$ , and $902$ . Each of those multiples of $11$ contributes $2 \cdot 2=4$ valid permutations, but we overcounted by a factor of $2$ ; every permutation of $209$ , for example, is also a permutation of $902$ . Therefore, there are $8 \cdot 4 / 2 = 16$ . Therefore, there are $64-8=56$ remaining multiples of $11$ without a $0$ in this case. Each one contributes $3! = 6$ valid permutations, but once again, we overcounted by a factor of $2$ (note that if a number ABC is a multiple of $11$ , then so is CBA). Therefore, there are $56 \cdot 6 / 2 = 168$ valid permutations in this subcase.

Adding up all the permutations from all the cases, we have $24+18+16+168 = \boxed{\textbf{(A) } 226}$ .

Solution 4

We can first overcount and then subtract. We know that there are $81$ multiples of $11$ .

We can then multiply by $6$ for each permutation of these multiples. (Yet some multiples do not have six distinct permutations.)

Now divide by $2$ , because if a number $abc$ with digits $a$ , $b$ , and $c$ is a multiple of $11$ , then $cba$ is also a multiple of $11$ so we have counted the same permutations twice.

Basically, each multiple of $11$ has its own $3$ permutations (say $abc$ has $abc$ $acb$ and $bac$ whereas $cba$ has $cba$ $cab$ and $bca$ ). We know that each multiple of $11$ has at least $3$ permutations because it cannot have $3$ repeating digits.

Hence we have $243$ permutations without subtracting for overcounting. Now note that we overcounted cases in which we have $0$ 's at the start of each number. So, in theory, we could just answer $A$ and then move on.

If we want to solve it, then we continue.

We overcounted cases where the middle digit of the number is $0$ and the last digit is $0$ .

Note that we assigned each multiple of $11$ three permutations.

The last digit is $0$ gives $9$ possibilities where we overcounted by $1$ permutation for each of $110, 220, ... , 990$ .

The middle digit is $0$ gives $8$ possibilities where we overcount by $1$ . $605, 704, 803, 902$ and $506, 407, 308, 209$

Subtracting $17$ gives $\boxed{\textbf{(A) } 226}$ .

Now, we may ask if there is further overlap (i.e if two of $abc$ and $bac$ and $acb$ were multiples of $11$ ). Thankfully, using divisibility rules, this can never happen, as taking the divisibility rule mod $11$ and adding, we get that $2a$ , $2b$ , or $2c$ is congruent to $0\ (mod\ 11)$ . Since $a, b, c$ are digits, this can never happen as none of them can equal $11$ and they can't equal $0$ as they are the leading digit of a three-digit number in each of the cases.

Solution 5: A Slightly Adjusted Version of Solution 2

$\textbf{WARNING:}$ If you do not feel comfortable looking at a massive amount of casework, please skip the solution.

Recalling the divisibility rule for $11$ , if we have a number $ABC$ where $A$ , $B$ , $C$ are digits, then $11\mid -A+B-C$ .

Notice that for any three-digit positive integer $ABC$ , $-A+B-C<11$ , thus we have 2 possibilities: $-A+B-C=0$ and $-A+B-C=-11$ .

$\textbf{Case 1:}$ $-A+B-C=0\Longrightarrow A+C=B$

Subcase $a$ : $A\neq B\neq C\neq0$

We have these values for $A+C=B$ $1+2=3,1+3=4,1+4=5,...,1+8=9$ $2+3=5,2+4=6,...,2+7=9$ $3+4=7,3+5=8,3+6=9$ $4+5=9$ From which we get $7+5+3+1=16$ triples $(A,B,C)$ . Counting every permutation, we have $16\cdot3!=96$ possibilities.

Subcase $b$ : $A=C$ , $A,B,C\neq0$

We have $1+1=2,2+2=4,3+3=6,4+4=8$ From which we get $4\cdot3=12$ possibilities.

Subcase $c$ : $A=B,C=0$

We have $1+0=1,2+0=2,...,9+0=9$ Since $0$ can't be the hundreds digit, from here we get $9\cdot2=18$ possibilities. Summing up case $1$ , we have $96+12+18=126$ possibilities.

$\textbf{Case 2:}$ $-A+B-C=-11\Longrightarrow A+C-B=11$

Subcase $a$ : $A\neq B\neq C\neq0$

We have these values for $A+C-B=11$ $9+8-6=11,9+7-5=11,...9+3-1=11$ $8+7-4=11,8+6-3=11,8+5-2=11,8+4-1$ $...$ From which we get $(6+4+2)\cdot3!=72$ possibilities.

Subcase $b$ : $A=C$ , $A,B,C\neq0$

We have $9+9=7,8+8-5,7+7-3=11,6+6-1=11$ From which we get $4\cdot3=12$ possibilities.

Subcase $c$ : $B=0\Longrightarrow A+C+11$

We have $9+2=8+3=7+4=6+5=11$ From which we get $2\cdot2\cdot4=16$ possibilities. Summing up case $2$ , we have $72+12+16=100$ possibilities.

Adding the $2$ cases, we get a total of $126+100=226$ possibilities. $\boxed{\mathrm{(A)}}$

~ Nafer

Solution 6 (1 but quicker)

The smallest multiple of $11$ above $100$ is $110 = 11 \cdot 10$ , while the largest multiple of $11$ less than $999$ is $990 = 11 \cdot 90$ . This means there are $90 - 10 + 1 = 81$ multiples of $11$ between $100$ and $999$ .

As there are $3$ permutations for each multiple, we have $81 * 3 = 243$ . However, we have overcounted, as numbers like $099$ shouldn't be counted. Looking at the answer choices, we notice there is only one that is less than $243$ , and so we have our answer as $\boxed{\textbf{(A) } 226}$ .

-¢

Video Solution

Two different variations on solving it. https://youtu.be/z5KNZEwmrWM

https://youtu.be/MBcHwu30MX4 -Video Solution by Richard Rusczyk

https://youtu.be/Ly69GHOq9Yw

~savannahsolver

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