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

Revision as of 13:46, 24 July 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.

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

Solution 1

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. Note: let $A \geq C$ so that we avoid counting the same permutations and having to subtract them later.

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

$\textbf{Part 1}$ : $B=0$ $A=C$ , this case results in 110, 220, 330...990. There are two ways to arrange the digits in each of those numbers. $2 \cdot 9 = 18$

$\textbf{Part 2}$ : $B>0$ $B=1, A+1=C$ , 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 for the first. This leads to $45$ cases.

$\textbf{Part 3}$ : $B=2, A+2=C$ , 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 for the first. This leads to $33$ cases.

$\textbf{Part 4}$ : $B=3, A+3=C$ , 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 for the first. This leads to $21$ cases.

$\textbf{Part 5}$ : $B=4, A+4=C$ , this case results in 484 and 594. There are $6$ ways to arrange the digits in 594 and 3 ways for 484. This leads to $9$ cases.

This case has $18+45+33+21+9=126$ subcases.

$\textbf{Case 2}$ : $A+B=C+11$ .

$\textbf{Part 1}$ : $C=0, A+B=11$ , 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 we continue this counting, we receive $16+21+18+15+12+9+6+3=100$ subcases.

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

~Mathguy1492

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 overcount and then subtract. We know there are $81$ multiples of $11$ .

We can multiply by $6$ for each permutation of these multiples. (Yet some multiples don't have 6)

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 over counting. 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 $A$ and 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 3 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 3 digit number in each of the cases.

Solution 4: The Quickest and Most Absurd

Notice that (D) and (E) are the odd ones out, as both are in the 400s range, and the majority of the answers (3/5), are in the 200s range, so we can omit them. We now look at (A), (B), and (C). We quickly omit (C), as it is divisible by 10, and therefore the odd one out. We now have narrowed down to (A) and (B). At this point, you might flip a coin or just guess (A), but we can use another strategic elimination (as follows). According to a list of the probabilities of each answer choice, on average, for any AMC question, (A) is generally more likely than (B), so it pays to guess $\text{(A)}$ at this point.

@@ Line 105: / Line 105: @@
 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.
+==Solution 4: The Quickest and Most Absurd==
+Notice that (D) and (E) are the odd ones out, as both are in the 400s range, and the majority of the answers (3/5), are in the 200s range, so we can omit them. We now look at (A), (B), and (C). We quickly omit (C), as it is divisible by 10, and therefore the odd one out. We now have narrowed down to (A) and (B). At this point, you might flip a coin or just guess (A), but we can use another strategic elimination (as follows). According to a list of the probabilities of each answer choice, on average, for any AMC question, (A) is generally more likely than (B), so it pays to guess <math>\text{(A)}</math> at this point.
 ==See Also==
 {{AMC10 box|year=2017|ab=A|num-b=24|after=Last Problem}}
 {{MAA Notice}}

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

Page

Toolbox

Search