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.

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

Solution

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: $A+B=C$ $A+B=C+11$

We now proceed to break down the cases.

$\textbf{Case 1}$ : $A+B=C$ . This has $18+45+33+21+9=126$ cases.

$\textbf{Part 1}$ : $B=0$ $A=C$ , this case results in 110, 220, 330...990. There are two ways to arrange the digits in each of those numbers. $2 \cdot 9 = 18$

$\textbf{Part 2}$ : $B>0$ $B=1, A=C+1$ , 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 ways for the first. This leads to $45$ cases.

$\textbf{Part 3}$ : $B=2, A=C+2$ , 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 ways for the first. This leads to $33$ cases.

$\textbf{Part 4}$ : $B=3, A=C+3$ , 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 ways for the first. This leads to $21$ cases.

$\textbf{Part 5}$ : $B=4, A=C+4$ , this case results in 484 and 594. There are $6$ ways to arrange the digits in all of those number except the first, and 3 ways ways for the first. This leads to $9$ cases.

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

$\textbf{Part 1}$ : $C=0, A+B=11$ , this cases results in 209, 308, ...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, ...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, ...617. There are $6$ ways to arrange each of those cases. This leads to $18$ cases.

... If you continue this counting, you receive $16+21+18+15+12+9+6+3=100$ cases.

$100+126=\boxed{(A) 226}$

2017 AMC 10A (Problems • Answer Key • Resources)
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2017 AMC 10A Problems/Problem 25

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