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

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==Problem==
 
==Problem==
−
How many integers between <math>100</math> and <math>999</math>, inclusive, have the property that some permutation of its digits is a multiple of <math>11</math> between <math>100</math> and <math>999?</math> For example, both <math>121</math> and <math>211</math> have this property.
+
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.
  
−
<math> \mathrm{\textbf{(A)} \ }226\qquad \mathrm{\textbf{(B)} \ } 243 \qquad \mathrm{\textbf{(C)} \ } 270 \qquad \mathrm{\textbf{(D)} \ }469\qquad \mathrm{\textbf{(E)} \ } 486</math>
+
<math>\textbf{(A) } 226 \qquad \textbf{(B) } 243 \qquad \textbf{(C) } 270 \qquad \textbf{(D) } 469 \qquad \textbf{(E) } 486</math>
  
 
==Solution 1==
 
==Solution 1==
 +
There are 81 multiples of 11 between <math>100</math> and <math>999</math> inclusive. Some have digits repeated twice, making 3 permutations.
  
−
Let the three-digit number be <math>ACB</math>:
+
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  <math>6 \div 2 = 3</math> permutations to each multiple.
  
−
If a number is divisible by <math>11</math>, then the difference between the sums of alternating digits is a multiple of <math>11</math>.
+
There are now 81*3 = 243 permutations, but we have overcounted*. Some multiples of 11 have <math>0</math> as a digit. Since <math>0</math> cannot be the digit of the hundreds place, we must subtract a permutation for each.  
  
−
There are two cases:
+
There are 110, 220, 330 ... 990, yielding 9 extra permutations
−
<math>A+B=C</math> and <math>A+B=C+11</math>
 
  
−
We now proceed to break down the cases. Note: let <math>A \geq C</math> so that we avoid counting the same permutations and having to subtract them later.
+
Also, there are 209, 308, 407...902, yielding 8 more permutations.
  
 +
Now, just subtract these 17 from the total (243) to get 226. <math>\boxed{\textbf{(A) } 226}</math>
  
−
<math>\textbf{Case 1}</math>: <math>A+B=C</math>.
+
*If short on time, observe that 226 is the only answer choice less than 243, and therefore is the only feasible answer.
−
 
 
−
 
 
−
 
 
−
<math>\textbf{Part 1}</math>: <math>B=0</math>
 
−
<math>A=C</math>, this case results in 110, 220, 330...990. There are two ways to arrange the digits in each of those numbers.
 
−
<math>2 \cdot 9 = 18</math>
 
−
 
 
−
<math>\textbf{Part 2}</math>: <math>B>0</math>
 
−
<math>B=1, A+1=C</math>, this case results in 121, 231,... 891. There are <math>6</math> ways to arrange the digits in all of those number except the first, and 3 ways for the first. This leads to <math>45</math> cases.
 
−
 
 
−
<math>\textbf{Part 3}</math>: <math>B=2, A+2=C</math>, this case results in 242, 352,... 792. There are <math>6</math> ways to arrange the digits in all of those number except the first, and 3 ways for the first. This leads to <math>33</math> cases.
 
−
 
 
−
<math>\textbf{Part 4}</math>: <math>B=3, A+3=C</math>, this case results in 363, 473,...693. There are <math>6</math> ways to arrange the digits in all of those number except the first, and 3 ways for the first. This leads to <math>21</math> cases.
 
−
 
 
−
<math>\textbf{Part 5}</math>: <math>B=4, A+4=C</math>, this case results in 484 and 594. There are <math>6</math> ways to arrange the digits in 594 and 3 ways for 484. This leads to <math>9</math> cases.
 
−
 
 
−
This case has <math>18+45+33+21+9=126</math> subcases.
 
−
 
 
−
 
 
−
 
 
−
<math>\textbf{Case 2}</math>: <math>A+B=C+11</math>.
 
−
 
 
−
 
 
−
<math>\textbf{Part 1}</math>: <math>C=0, A+B=11</math>, this cases results in 209, 308, ...506. There are <math>4</math> ways to arrange each of those cases. This leads to <math>16</math> cases.
 
−
 
 
−
<math>\textbf{Part 2}</math>: <math>C=1, A+B=12</math>, this cases results in 319, 418, ...616. There are <math>6</math> ways to arrange each of those cases, except the last. This leads to <math>21</math> cases.
 
−
 
 
−
<math>\textbf{Part 3}</math>: <math>C=2, A+B=13</math>, this cases results in 429, 528, ...617. There are <math>6</math> ways to arrange each of those cases. This leads to <math>18</math> cases.
 
−
 
 
−
...
 
−
If we continue this counting, we receive <math>16+21+18+15+12+9+6+3=100</math> subcases.
 
−
 
 
−
<math>100+126=\boxed{\textbf{(A) } 226}</math>
 
−
 
 
−
~Mathguy1492
 
  
 
==Solution 2==
 
==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:
+
We note that we only have to consider multiples of <math>11</math> and see how many valid permutations each has. We can do casework on the number of repeating digits that the multiple of <math>11</math> has:
  
 
<math>\textbf{Case 1:}</math> All three digits are the same.  
 
<math>\textbf{Case 1:}</math> All three digits are the same.  
−
By inspection, we find that there are no multiples of 11 here.
+
By inspection, we find that there are no multiples of <math>11</math> here.
  
 
<math>\textbf{Case 2:}</math> Two of the digits are the same, and the third is different.
 
<math>\textbf{Case 2:}</math> Two of the digits are the same, and the third is different.
  
 
<math>\textbf{Case 2a:}</math>
 
<math>\textbf{Case 2a:}</math>
−
There are 8 multiples of 11 without a zero that have this property:
+
There are <math>8</math> multiples of <math>11</math> without a zero that have this property:
−
121, 242, 363, 484, 616, 737, 858, 979.
+
<math>121</math>, <math>242</math>, <math>363</math>, <math>484</math>, <math>616</math>, <math>737</math>, <math>858</math>, <math>979</math>.
−
Each contributes 3 valid permutations, so there are <math>8 \cdot 3 = 24</math> permutations in this subcase.
+
Each contributes <math>3</math> valid permutations, so there are <math>8 \cdot 3 = 24</math> permutations in this subcase.
  
 
<math>\textbf{Case 2b:}</math>
 
<math>\textbf{Case 2b:}</math>
−
There are 9 multiples of 11 with a zero that have this property:
+
There are <math>9</math> multiples of <math>11</math> with a zero that have this property:
−
110, 220, 330, 440, 550, 660, 770, 880, 990.
+
<math>110</math>, <math>220</math>, <math>330</math>, <math>440</math>, <math>550</math>, <math>660</math>, <math>770</math>, <math>880</math>, <math>990</math>.
−
Each one contributes 2 valid permutations (the first digit can't be zero), so there are <math>9 \cdot 2 = 18</math> permutations in this subcase.
+
Each one contributes <math>2</math> valid permutations (the first digit can't be zero), so there are <math>9 \cdot 2 = 18</math> permutations in this subcase.
  
 
<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, 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.
+
Since there are <math>\frac{990-110}{11}+1 = 81</math> multiples of <math>11</math> between <math>100</math> and <math>999</math>, there are <math>81-8-9 = 64</math> multiples of <math>11</math> remaining in this case. However, <math>8</math> of them contain a zero, namely <math>209</math>, <math>308</math>, <math>407</math>, <math>506</math>, <math>605</math>, <math>704</math>, <math>803</math>, and <math>902</math>. Each of those multiples of <math>11</math> contributes <math>2 \cdot 2=4</math> valid permutations, but we overcounted by a factor of <math>2</math>; every permutation of <math>209</math>, for example, is also a permutation of <math>902</math>. Therefore, there are <math>8 \cdot 4 / 2 = 16</math>. Therefore, there are <math>64-8=56</math> remaining multiples of <math>11</math> without a <math>0</math> in this case. Each one contributes <math>3! = 6</math> valid permutations, but once again, we overcounted by a factor of <math>2</math> (note that if a number ABC is a multiple of <math>11</math>, 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>.
  
−
==Solution 3 (Shorter and Not Casework)==
+
==Solution 3 ==
  
−
We can overcount and then subtract.
+
We can first overcount and then subtract.
−
We know there are <math>81</math> multiples of <math>11</math>.
+
We know that there are <math>81</math> multiples of <math>11</math>.
  
−
We can multiply by <math>6</math> for each permutation of these multiples. (Yet some multiples don't have 6)
+
We can then multiply by <math>6</math> for each permutation of these multiples. (Yet some multiples do not have six distinct permutations.)
  
−
Now divide by <math>2</math>, because if a number <math>abc</math> with digits <math>a</math>, <math>b</math>, and <math>c</math> is a multiple of 11, then <math>cba</math> is also a multiple of 11 so we have counted the same permutations twice.  
+
Now divide by <math>2</math>, because if a number <math>abc</math> with digits <math>a</math>, <math>b</math>, and <math>c</math> is a multiple of <math>11</math>, then <math>cba</math> is also a multiple of <math>11</math> so we have counted the same permutations twice.  
  
−
Basically, each multiple of 11 has its own 3 permutations (say <math>abc</math> has <math>abc</math> <math>acb</math> and <math>bac</math> whereas <math>cba</math> has <math>cba</math> <math>cab</math> and <math>bca</math>). We know that each multiple of 11 has at least 3 permutations because it cannot have 3 repeating digits.
+
Basically, each multiple of <math>11</math> has its own <math>3</math> permutations (say <math>abc</math> has <math>abc</math> <math>acb</math> and <math>bac</math> whereas <math>cba</math> has <math>cba</math> <math>cab</math> and <math>bca</math>). We know that each multiple of <math>11</math> has at least <math>3</math> permutations because it cannot have <math>3</math> repeating digits.
  
−
Hence we have <math>243</math> permutations without subtracting for over counting.
+
Hence we have <math>243</math> permutations without subtracting for overcounting.
−
Now note that we overcounted cases is which we have 0's at the start of each number. So in theory we could just answer <math>A</math> and move on.
+
Now note that we overcounted cases in which we have <math>0</math>'s at the start of each number. So, in theory, we could just answer <math>A</math> and then move on.
  
 
If we want to solve it, then we continue.
 
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.
+
We overcounted cases where the middle digit of the number is <math>0</math> and the last digit is <math>0</math>.
  
−
Note that we assigned each multiple of 11 3 permutations.
+
Note that we assigned each multiple of <math>11</math> three permutations.
  
 
The last digit is <math>0</math> gives <math>9</math> possibilities where we overcounted by <math>1</math> permutation for each of <math>110, 220, ... , 990</math>.
 
The last digit is <math>0</math> gives <math>9</math> possibilities where we overcounted by <math>1</math> permutation for each of <math>110, 220, ... , 990</math>.
  
−
The middle digit is 0 gives 8 possibilities where we overcount by 1.
+
The middle digit is <math>0</math> gives <math>8</math> possibilities where we overcount by <math>1</math>.
 
<math>605, 704, 803, 902</math> and <math>506, 407, 308, 209</math>
 
<math>605, 704, 803, 902</math> and <math>506, 407, 308, 209</math>
  
 
Subtracting <math>17</math> gives <math>\boxed{\textbf{(A) } 226}</math>.
 
Subtracting <math>17</math> gives <math>\boxed{\textbf{(A) } 226}</math>.
  
−
Now, we may ask if there is further overlap (I.e if two of <math>abc</math> and <math>bac</math> and <math>acb</math> were multiples of <math>11</math>) Thankfully, using divisibility rules, this can never happen as taking the divisibility rule mod 11 and adding we get that <math>2a</math>, <math>2b</math>, or <math>2c</math>  is congruent to <math>0\ (mod\ 11)</math>. Since <math>a, b, c</math> 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 3 digit number in each of the cases
+
Now, we may ask if there is further overlap (i.e if two of <math>abc</math> and <math>bac</math> and <math>acb</math> were multiples of <math>11</math>). Thankfully, using divisibility rules, this can never happen, as taking the divisibility rule mod <math>11</math> and adding, we get that <math>2a</math>, <math>2b</math>, or <math>2c</math>  is congruent to <math>0\ (mod\ 11)</math>. Since <math>a, b, c</math> are digits, this can never happen as none of them can equal <math>11</math> and they can't equal <math>0</math> 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 <math>81</math> multiples of <math>11</math> between <math>100</math> and <math>999</math>, and each of them have at most <math>6</math> permutation (and thus is a permutations of <math>6</math> numbers), giving us a maximum of <math>486</math> valid numbers.
 +
 
 +
However, if <math>abc</math> can be divided by <math>11</math>, so can <math>cba</math>, which is distinct if <math> c \neq a</math>. And if <math>c = a</math> then <math>abc</math> and <math>cba</math> have the same permutations. Either way, we have doubled counted.
 +
This reduces the number of permutations to <math>486/2 =  243</math>.
 +
 
 +
Furthermore, if <math>a=b</math> or <math>c=0</math> (which turn out to be equivalent conditions! for example <math>220</math>), not all (inverse) permutations are distinct (<math>\mathbf{2}20 = 2\mathbf{2}0</math>) or valid (<math>022</math>). (There are 9 of these.)
 +
 
 +
Similarly, for <math>a0b</math>, 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 <math>243</math>. This leaves us with only one possible answer: <math>\boxed{\textbf{(A) } 226}</math>.
 +
 
 +
==Video Solution==
 +
https://youtu.be/leCUVcplmZ0?si=kGq5U9TSKUP1Y30I
 +
 
 +
~ Pi Academy
  
−
~fuzz1(slightly tweaked by squareman)
+
==Video solution 2==
 +
Two different variations on solving it.
 +
https://youtu.be/z5KNZEwmrWM
 +
 
 +
https://youtu.be/MBcHwu30MX4
 +
-Video Solution by Richard Rusczyk
 +
 
 +
https://youtu.be/Ly69GHOq9Yw
 +
 
 +
~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=A|num-b=24|after=Last Problem}}
 
{{AMC10 box|year=2017|ab=A|num-b=24|after=Last Problem}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
[[Category:Intermediate Combinatorics Problems]]

Latest revision as of 19:39, 17 October 2024

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.

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

Solution 1

There are 81 multiples of 11 between $100$ and $999$ 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 $6 \div 2 = 3$ permutations to each multiple.

There are now 81*3 = 243 permutations, but we have overcounted*. Some multiples of 11 have $0$ as a digit. Since $0$ 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. $\boxed{\textbf{(A) } 226}$

  • 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 $11$ 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 (process of elimination on multiple choices)

Taken from solution three, we notice that there are a total of $81$ multiples of $11$ between $100$ and $999$, and each of them have at most $6$ permutation (and thus is a permutations of $6$ numbers), giving us a maximum of $486$ valid numbers.

However, if $abc$ can be divided by $11$, so can $cba$, which is distinct if $c \neq a$. And if $c = a$ then $abc$ and $cba$ have the same permutations. Either way, we have doubled counted. This reduces the number of permutations to $486/2 = 243$.

Furthermore, if $a=b$ or $c=0$ (which turn out to be equivalent conditions! for example $220$), not all (inverse) permutations are distinct ($\mathbf{2}20 = 2\mathbf{2}0$) or valid ($022$). (There are 9 of these.)

Similarly, for $a0b$, 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 $243$. This leaves us with only one possible answer: $\boxed{\textbf{(A) } 226}$.

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

https://youtu.be/Ly69GHOq9Yw

~savannahsolver

See Also

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

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