Difference between revisions of "2016 AMC 10B Problems/Problem 24"
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<math>\textbf{(A)}\ 9\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 20</math> | <math>\textbf{(A)}\ 9\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 20</math> | ||
− | ==Solution== | + | ==Solution 1== |
− | The numbers are <math>10a+b, 10b+c,</math> and <math>10c+d</math>. Note that only <math>d</math> can be zero | + | The numbers are <math>10a+b, 10b+c,</math> and <math>10c+d</math>. Note that only <math>d</math> can be zero, the numbers <math>ab</math>, <math>bc</math>, and <math>cd</math> cannot start with a zero, and <math>a\le b\le c</math>. |
To form the sequence, we need <math>(10c+d)-(10b+c)=(10b+c)-(10a+b)</math>. This can be rearranged as <math>10(c-2b+a)=2c-b-d</math>. Notice that since the left-hand side is a multiple of <math>10</math>, the right-hand side can only be <math>0</math> or <math>10</math>. (A value of <math>-10</math> would contradict <math>a\le b\le c</math>.) Therefore we have two cases: <math>a+c-2b=1</math> and <math>a+c-2b=0</math>. | To form the sequence, we need <math>(10c+d)-(10b+c)=(10b+c)-(10a+b)</math>. This can be rearranged as <math>10(c-2b+a)=2c-b-d</math>. Notice that since the left-hand side is a multiple of <math>10</math>, the right-hand side can only be <math>0</math> or <math>10</math>. (A value of <math>-10</math> would contradict <math>a\le b\le c</math>.) Therefore we have two cases: <math>a+c-2b=1</math> and <math>a+c-2b=0</math>. | ||
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There is no solution for <math>c=6</math>. | There is no solution for <math>c=6</math>. | ||
Added together, this gives us <math>8</math> answers for Case 1. | Added together, this gives us <math>8</math> answers for Case 1. | ||
+ | |||
+ | |||
===Case 2=== | ===Case 2=== | ||
This means that the digits themselves are in an arithmetic sequence. This gives us <math>9</math> answers, | This means that the digits themselves are in an arithmetic sequence. This gives us <math>9</math> answers, | ||
<cmath>1234, 1357, 2345, 2468, 3456, 3579, 4567, 5678, 6789.</cmath> | <cmath>1234, 1357, 2345, 2468, 3456, 3579, 4567, 5678, 6789.</cmath> | ||
− | |||
Adding the two cases together, we find the answer to be <math>8+9=</math> <math>\boxed{\textbf{(D) }17}</math>. | Adding the two cases together, we find the answer to be <math>8+9=</math> <math>\boxed{\textbf{(D) }17}</math>. | ||
+ | |||
+ | ==Solution 2 (Brute Force, when you have lots of time)== | ||
+ | Looking at the answer options, all the numbers are pretty small so it is easy to make a list. | ||
+ | |||
+ | <math>12|23|34 \rightarrow 1234</math> | ||
+ | |||
+ | <math>13|35|57 \rightarrow 1357</math> | ||
+ | |||
+ | <math>14|48|82 \rightarrow 1482</math> | ||
+ | |||
+ | |||
+ | <math>23|34|45 \rightarrow 2345</math> | ||
+ | |||
+ | <math>24|46|68 \rightarrow 2468</math> | ||
+ | |||
+ | <math>24|47|70 \rightarrow 2470</math> | ||
+ | |||
+ | <math>25|59|93 \rightarrow 2593</math> | ||
+ | |||
+ | |||
+ | <math>34|45|56 \rightarrow 3456</math> | ||
+ | |||
+ | <math>35|57|79 \rightarrow 3579</math> | ||
+ | |||
+ | <math>35|58|81 \rightarrow 3581</math> | ||
+ | |||
+ | |||
+ | <math>45|56|67 \rightarrow 4567</math> | ||
+ | |||
+ | <math>46|69|92 \rightarrow 4692</math> | ||
+ | |||
+ | |||
+ | <math>56|67|78 \rightarrow 5678</math> | ||
+ | |||
+ | <math>56|68|80 \rightarrow 5680</math> | ||
+ | |||
+ | |||
+ | <math>67|78|89 \rightarrow 6789</math> | ||
+ | |||
+ | <math>67|79|91 \rightarrow 6791</math> | ||
+ | |||
+ | |||
+ | <math>88|89|90 \rightarrow 8890</math> | ||
+ | |||
+ | Counting all the cases we get our answer of <math>17</math> which is <math>\boxed{D}</math> | ||
+ | -srisainandan6 | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=B|num-b=23|num-a=25}} | {{AMC10 box|year=2016|ab=B|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 13:03, 27 January 2021
Contents
Problem
How many four-digit integers , with , have the property that the three two-digit integers form an increasing arithmetic sequence? One such number is , where , , , and .
Solution 1
The numbers are and . Note that only can be zero, the numbers , , and cannot start with a zero, and .
To form the sequence, we need . This can be rearranged as . Notice that since the left-hand side is a multiple of , the right-hand side can only be or . (A value of would contradict .) Therefore we have two cases: and .
Case 1
If , then , so . This gives . If , then , so . This gives . If , then , so , giving . There is no solution for . Added together, this gives us answers for Case 1.
Case 2
This means that the digits themselves are in an arithmetic sequence. This gives us answers, Adding the two cases together, we find the answer to be .
Solution 2 (Brute Force, when you have lots of time)
Looking at the answer options, all the numbers are pretty small so it is easy to make a list.
Counting all the cases we get our answer of which is -srisainandan6
See Also
2016 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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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