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

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

@@ Line 1: / Line 1: @@
 ==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==
-There are 81 multiples of 11. Some have digits repeated twice, making 3 permutations.
+There are 81 multiples of 11 between <math>100</math> and <math>999</math> 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. Therefore, assign 3 permutations to each multiple.
+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.
-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 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 110, 220, 330 ... 990, yielding 9 extra permutations
@@ Line 15: / Line 17: @@
 Also, there are 209, 308, 407...902, yielding 8 more permutations.
-Now, just subtract these 17 from the total (243), getting 226. <math>\boxed{\textbf{(A) } 226}</math>
+Now, just subtract these 17 from the total (243) to get 226. <math>\boxed{\textbf{(A) } 226}</math>
-*Note: If short on time, note that 226 is the only answer choice less than 243, and therefore is the only feasible answer.
+*If short on time, observe that 226 is the only answer choice less than 243, and therefore is the only feasible answer.
-===Solution 2===
+==Solution 2==
-Let the three-digit number be <math>ACB</math>:
-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 two cases:
-<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.
-<math>\textbf{Case 1}</math>: <math>A+B=C</math>.
-<math>\textbf{Part 1}</math>: <math>B=0, A=C</math>, this case results in <math>110, 220, 330...990</math>. 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=1, A+1=C</math>, this case results in <math>121, 231,... 891</math>. There are <math>6</math> ways to arrange the digits in all of those number except the first, and <math>3</math> 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 <math>242, 352,... 792</math>. 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 <math>363, 473,...693</math>. 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 <math>484</math> and <math>594</math>. 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 <math>209, 308, 407, 506</math>. 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 <math>319, 418,517,616</math>. 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 <math>429, 528, 627</math>. 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>
-==Solution 3==
 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:
@@ Line 89: / Line 45: @@
 Adding up all the permutations from all the cases, we have <math>24+18+16+168 = \boxed{\textbf{(A) } 226}</math>.
-==Solution 4 ==
+==Solution 3 ==
 We can first overcount and then subtract.
@@ Line 118: / Line 74: @@
 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 5: A Slightly Adjusted Version of Solution 2 ==
+==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>.
-<math>\textbf{WARNING:}</math> If you do not feel comfortable looking at a massive amount of casework, please skip the solution.
+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.)
-Recalling the divisibility rule for <math>11</math>, if we have a number <math>ABC</math> where <math>A</math>, <math>B</math>, <math>C</math> are digits, then <math>11\mid -A+B-C</math>.
+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>.
-Notice that for any three-digit positive integer <math>ABC</math>, <math>-A+B-C<11</math>, thus we have 2 possibilities: <math>-A+B-C=0</math> and <math>-A+B-C=-11</math>.
+==Video Solution==
+https://youtu.be/leCUVcplmZ0?si=kGq5U9TSKUP1Y30I
-<math>\textbf{Case 1:}</math> <math>-A+B-C=0\Longrightarrow A+C=B</math>
-Subcase <math>a</math>: <math>A\neq B\neq C\neq0</math>
+~ Pi Academy
-We have these values for <math>A+C=B</math>
+==Video solution 2==
-<cmath>1+2=3,1+3=4,1+4=5,...,1+8=9</cmath>
-<cmath>2+3=5,2+4=6,...,2+7=9</cmath>
-<cmath>3+4=7,3+5=8,3+6=9</cmath>
-<cmath>4+5=9</cmath>
-From which we get <math>7+5+3+1=16</math> triples <math>(A,B,C)</math>. Counting every permutation, we have <math>16\cdot3!=96</math> possibilities.
-Subcase <math>b</math>: <math>A=C</math>, <math>A,B,C\neq0</math>
-We have
-<cmath>1+1=2,2+2=4,3+3=6,4+4=8</cmath>
-From which we get <math>4\cdot3=12</math> possibilities.
-Subcase <math>c</math>: <math>A=B,C=0</math>
-We have
-<cmath>1+0=1,2+0=2,...,9+0=9</cmath>
-Since <math>0</math> can't be the hundreds digit, from here we get <math>9\cdot2=18</math> possibilities. Summing up case <math>1</math>, we have <math>96+12+18=126</math> possibilities.
-<math>\textbf{Case 2:}</math> <math>-A+B-C=-11\Longrightarrow A+C-B=11</math>
-Subcase <math>a</math>: <math>A\neq B\neq C\neq0</math>
-We have these values for <math>A+C-B=11</math>
-<cmath>9+8-6=11,9+7-5=11,...9+3-1=11</cmath>
-<cmath>8+7-4=11,8+6-3=11,8+5-2=11,8+4-1</cmath>
-<cmath>...</cmath>
-From which we get <math>(6+4+2)\cdot3!=72</math> possibilities.
-Subcase <math>b</math>: <math>A=C</math>, <math>A,B,C\neq0</math>
-We have
-<cmath>9+9=7,8+8-5,7+7-3=11,6+6-1=11</cmath>
-From which we get <math>4\cdot3=12</math> possibilities.
-Subcase <math>c</math>: <math>B=0\Longrightarrow A+C+11</math>
-We have
-<cmath>9+2=8+3=7+4=6+5=11</cmath>
-From which we get <math>2\cdot2\cdot4=16</math> possibilities. Summing up case <math>2</math>, we have <math>72+12+16=100</math> possibilities.
-Adding the <math>2</math> cases, we get a total of <math>126+100=226</math> possibilities. <math>\boxed{\mathrm{(A)}}</math>
-~ Nafer
-==Video Solution==
 Two different variations on solving it.
 https://youtu.be/z5KNZEwmrWM
@@ Line 183: / Line 97: @@
 https://youtu.be/MBcHwu30MX4
 -Video Solution by Richard Rusczyk
+https://youtu.be/Ly69GHOq9Yw
+~savannahsolver
 ==See Also==

2017 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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