Difference between revisions of "2016 AMC 10B Problems/Problem 24"
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==Solution== | ==Solution== | ||
− | 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>. | ||
− | + | '''Case 1''' | |
+ | |||
If <math>c=9</math>, then <math>b+d=8,\ 2b-a=8</math>, so <math>5\le b\le 8</math>. This gives <math>2593, 4692, 6791, 8890</math>. | If <math>c=9</math>, then <math>b+d=8,\ 2b-a=8</math>, so <math>5\le b\le 8</math>. This gives <math>2593, 4692, 6791, 8890</math>. | ||
If <math>c=8</math>, then <math>b+d=6,\ 2b-a=7</math>, so <math>4\le b\le 6</math>. This gives <math>1482, 3581, 5680</math>. | If <math>c=8</math>, then <math>b+d=6,\ 2b-a=7</math>, so <math>4\le b\le 6</math>. This gives <math>1482, 3581, 5680</math>. | ||
Line 17: | Line 18: | ||
Added together, this gives us <math>8</math> answers for Case 1. | Added together, this gives us <math>8</math> answers for Case 1. | ||
− | + | ||
− | This means that the digits themselves are in arithmetic sequence. This gives us <math>9</math> answers, < | + | |
− | + | '''Case 2''' | |
+ | |||
+ | 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> | ||
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 | ||
+ | |||
+ | ==Solution 3== | ||
+ | Let <math>x</math> be the difference between the numbers <math>ab</math>, <math>bc</math>, and <math>cd</math>. We then have | ||
+ | <cmath>10a + b = 10b + c - x</cmath> and | ||
+ | <cmath>10b + c = 10c + d - x.</cmath> | ||
+ | |||
+ | Subtracting the second equation from the first and then simplifying, we are left with: | ||
+ | <cmath>10a + 8c + d = 19b.</cmath> | ||
+ | |||
+ | Notice that <math>b > c</math>. Because the values of <math>a</math> and <math>d</math> are irrelevant compared to the other numbers, we can just find pairs of <math>(b, c)</math> such that <math>a > 0</math>. Trying out each value of <math>b</math> from <math>2</math> to <math>9</math> and summing the number of pairs yields <math>1 + 2 + 4 + 4 + 3 + 2 + 1 + 0 = \boxed{\textbf{(D) }17}</math> | ||
+ | |||
+ | - cappucher | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=UhPxvZ6V4Zs | ||
==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 20:48, 1 August 2024
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
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
Solution 3
Let be the difference between the numbers
,
, and
. We then have
and
Subtracting the second equation from the first and then simplifying, we are left with:
Notice that . Because the values of
and
are irrelevant compared to the other numbers, we can just find pairs of
such that
. Trying out each value of
from
to
and summing the number of pairs yields
- cappucher
Video Solution
https://www.youtube.com/watch?v=UhPxvZ6V4Zs
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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