Difference between revisions of "2017 AMC 10A Problems/Problem 25"
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==Solution 1== | ==Solution 1== | ||
+ | There are 81 multiples of 11. 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. | |
− | + | 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 | + | 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. | |
+ | Now, just subtract these 17 from the total (243), getting 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. | |
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==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>. | ||
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==Solution 3 == | ==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 | + | 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 overcounting. | Hence we have <math>243</math> 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 <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 | + | 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 ( | + | 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 (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>. | ||
+ | |||
+ | -¢ | ||
− | ==Solution | + | ==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= | + | ==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]] | [[Category:Intermediate Combinatorics Problems]] |
Latest revision as of 22:11, 8 June 2021
Contents
Problem
How many integers between and , inclusive, have the property that some permutation of its digits is a multiple of between and For example, both and have this property.
Solution 1
There are 81 multiples of 11. 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.
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 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), getting 226.
- Note: If short on time, note 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 and see how many valid permutations each has. We can do casework on the number of repeating digits that the multiple of has:
All three digits are the same. By inspection, we find that there are no multiples of here.
Two of the digits are the same, and the third is different.
There are multiples of without a zero that have this property: , , , , , , , . Each contributes valid permutations, so there are permutations in this subcase.
There are multiples of with a zero that have this property: , , , , , , , , . Each one contributes valid permutations (the first digit can't be zero), so there are permutations in this subcase.
All the digits are different. Since there are multiples of between and , there are multiples of remaining in this case. However, of them contain a zero, namely , , , , , , , and . Each of those multiples of contributes valid permutations, but we overcounted by a factor of ; every permutation of , for example, is also a permutation of . Therefore, there are . Therefore, there are remaining multiples of without a in this case. Each one contributes valid permutations, but once again, we overcounted by a factor of (note that if a number ABC is a multiple of , then so is CBA). Therefore, there are valid permutations in this subcase.
Adding up all the permutations from all the cases, we have .
Solution 3
We can first overcount and then subtract. We know that there are multiples of .
We can then multiply by for each permutation of these multiples. (Yet some multiples do not have six distinct permutations.)
Now divide by , because if a number with digits , , and is a multiple of , then is also a multiple of so we have counted the same permutations twice.
Basically, each multiple of has its own permutations (say has and whereas has and ). We know that each multiple of has at least permutations because it cannot have repeating digits.
Hence we have permutations without subtracting for overcounting. Now note that we overcounted cases in which we have 's at the start of each number. So, in theory, we could just answer and then move on.
If we want to solve it, then we continue.
We overcounted cases where the middle digit of the number is and the last digit is .
Note that we assigned each multiple of three permutations.
The last digit is gives possibilities where we overcounted by permutation for each of .
The middle digit is gives possibilities where we overcount by . and
Subtracting gives .
Now, we may ask if there is further overlap (i.e if two of and and were multiples of ). Thankfully, using divisibility rules, this can never happen, as taking the divisibility rule mod and adding, we get that , , or is congruent to . Since are digits, this can never happen as none of them can equal and they can't equal as they are the leading digit of a three-digit number in each of the cases.
Solution 4 (1 but quicker)
The smallest multiple of above is , while the largest multiple of less than is . This means there are multiples of between and .
As there are permutations for each multiple, we have . However, we have overcounted, as numbers like shouldn't be counted. Looking at the answer choices, we notice there is only one that is less than , and so we have our answer as .
-¢
Video Solution
Two different variations on solving it. https://youtu.be/z5KNZEwmrWM
https://youtu.be/MBcHwu30MX4 -Video Solution by Richard Rusczyk
~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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