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Difference between revisions of "2017 AMC 10A Problems/Problem 25"

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<math>\textbf{Case 3:}</math> All the digits are different.
 
<math>\textbf{Case 3:}</math> All the digits are different.
−
Since there are <math>\frac{990-110}{11}+1 = 81</math> multiples of 11 between 100 and 999, there are <math>81-8-9 = 64</math> multiples of 11 remaining in this case. However, 8 of them contain a zero; they are 209, 308, 407, 506, 605, 704, 803, and 902. Each of those multiples of 11 contributes <math>2 \cdot 2=4</math> 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 <math>8 \cdot 4 / 2 = 16</math>. Therefore, there are <math>64-8=56</math> remaining multiples of 11 without a 0 in this case. Each one contributes <math>3! = 6</math> 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 <math>56 \cdot 6 / 2 = 168</math> valid permutations in this subcase.
+
Since there are <math>\frac{990-110}{11}+1 = 81</math> multiples of 11 between 100 and 999, there are <math>81-8-9 = 64</math> 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 <math>2 \cdot 2=4</math> 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 <math>8 \cdot 4 / 2 = 16</math>. Therefore, there are <math>64-8=56</math> remaining multiples of 11 without a 0 in this case. Each one contributes <math>3! = 6</math> 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 <math>56 \cdot 6 / 2 = 168</math> valid permutations in this subcase.
  
 
Adding up all the permutations from all the cases, we have <math>24+18+16+168 = \boxed{\textbf{(A) } 226}</math>.
 
Adding up all the permutations from all the cases, we have <math>24+18+16+168 = \boxed{\textbf{(A) } 226}</math>.

Revision as of 19:31, 8 February 2017

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 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.


$\textbf{Case 1}$: $A+B=C$. 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.

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

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 $24+18+16+168 = \boxed{\textbf{(A) } 226}$.

See Also

2017 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

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