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

Let the three-digit number be $ACB$ :

If a number is divisible by $11$ , then the difference between the sums of alternating digits is a multiple of $11$ .

There are two cases: $A+B=C$ and $A+B=C+11$

We now proceed to break down the cases. Note: let $A \geq C$ so that we avoid counting the same permutations and having to subtract them later.

$\textbf{Case$ (Error compiling LaTeX. Unknown error_msg)1$}$ (Error compiling LaTeX. Unknown error_msg): $A+B=C$ .

$\textbf{Part 1}$ : $B=0, A=C$ , this case results in $110, 220, 330...990$ . There are two ways to arrange the digits in each of those numbers. $2 \cdot$ 9 = 18$$ (Error compiling LaTeX. Unknown error_msg)\textbf{Part $2$ } $:$ B=1, A+ $1$ =C $, this case results in$ 121, 231,... 891 $. There are$ 6 $ways to arrange the digits in all of those number except the first, and 3 ways for the first. This leads to$ 45 $cases.$ \textbf{Part $3$ } $:$ B=2, A+2=C $, this case results in$ 242, 352,... 792 $. There are$ 6 $ways to arrange the digits in all of those number except the first, and 3 ways for the first. This leads to$ 33 $cases.$ \textbf{Part $4$ } $:$ B=3, A+3=C $, this case results in$ 363, 473,...693 $. There are$ 6 $ways to arrange the digits in all of those number except the first, and 3 ways for the first. This leads to$ 21 $cases.$ \textbf{Part $5$ } $:$ B=4, A+4=C $, this case results in$ 484 $and$ 594 $. There are$ 6 $ways to arrange the digits in 594 and 3 ways for 484. This leads to$ 9$cases.

This case has$ (Error compiling LaTeX. Unknown error_msg)18+45+33+21+9=126 $subcases.$ \textbf{Case $2$ } $:$ A+B=C+11 $.$ \textbf{Part $1$ } $:$ C=0, A+B=11 $, this cases results in$ 209, 308, 407, 506 $. There are$ 4 $ways to arrange each of those cases. This leads to$ 16 $cases.$ \textbf{Part $2$ } $:$ C=1, A+B=12 $, this cases results in$ 319, 418,517,616 $. There are$ 6 $ways to arrange each of those cases, except the last. This leads to$ 21 $cases.$ \textbf{Part 3} $:$ C=2, A+B=13 $, this cases results in 429, 528, 627. There are$ 6 $ways to arrange each of those cases. This leads to$ 18$cases.

... If we continue this counting, we receive$ (Error compiling LaTeX. Unknown error_msg)16+21+18+15+12+9+6+3=100 $subcases.$ 100+126=\boxed{\textbf{(A) } $226$ }$

Solution 2

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 3

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 4: 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

2017 AMC 10A (Problems • Answer Key • Resources)
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