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

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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 <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.
  
<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>
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<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. Some have digits repeated twice, making 3 permutations.
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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
 
There are 110, 220, 330 ... 990, yielding 9 extra permutations
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Also, there are 209, 308, 407...902, yielding 8 more permutations.
 
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>
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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 3==
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==Solution 2==
  
 
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:
 
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:
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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 4 ==
+
==Solution 3 ==
  
 
We can first overcount and then subtract.
 
We can first overcount and then subtract.
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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.
 
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 ==
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==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.
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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>.
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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>.
 
 
 
<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>
 
 
 
We have these values for <math>A+C=B</math>
 
<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
 
== Solution 6 (1 but quicker) ==
 
 
 
The smallest multiple of <math>11</math> above <math>100</math> is <math>110 = 11 \cdot 10</math>, while the largest multiple of <math>11</math> less than <math>999</math> is <math>990 = 11 \cdot 90</math>. This means there are <math>90 - 10 + 1 = 81</math> multiples of <math>11</math> between <math>100</math> and <math>999</math>.
 
 
 
As there are <math>3</math> permutations for each multiple, we have <math>81 * 3 = 243</math>. However, we have overcounted, as numbers like <math>099</math> shouldn't be counted. Looking at the answer choices, we notice there is only one that is less than <math>243</math>, and so we have our answer as <math>\boxed{\textbf{(A) } 226}</math>.
 
 
 
 
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 16:02, 4 July 2023

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

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 (ProblemsAnswer KeyResources)
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