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

Revision as of 17:14, 4 April 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.

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

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 (when you are running out of time)

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

@@ Line 78: / Line 78: @@
 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.) Hence, the answer must be smaller than <math>243</math>. This leaves us with only one possible answer: <math>\boxed{\textbf{(A) } 226}</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==

2017 AMC 10A (Problems • Answer Key • Resources)
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