Difference between revisions of "AoPS Wiki talk:Problem of the Day/June 15, 2011"
(→Solution: Formatting Solution 1 and added a Solution 2) |
|||
Line 2: | Line 2: | ||
{{:AoPSWiki:Problem of the Day/June 15, 2011}} | {{:AoPSWiki:Problem of the Day/June 15, 2011}} | ||
==Solution== | ==Solution== | ||
− | We can solve this problem by a bit of trial and error. | + | ===Solution 1=== |
+ | We can solve this problem by a bit of trial and error. | ||
+ | |||
+ | We can guess she rode <math>5</math> days and we get <math>7+10+13+16+19=(13)(5)=65</math> since the mean is clearly <math>13</math> and there are <math>5</math> terms. | ||
+ | |||
+ | That's a bit too small. | ||
+ | |||
+ | We can add <math>22</math> to <math>65</math> and get <math>87</math>. That's still too small. | ||
+ | |||
+ | Now, we add <math>25</math> to get <math>112</math>, the answer we want. | ||
+ | |||
+ | We now count how many numbers are in the following list: <math>7, 10, 13, 16, 19, 22, 25</math>. | ||
+ | |||
+ | Adding <math>2</math> to the list gives us <math>9, 12, 15, 18, 21, 24, 27</math>. | ||
+ | |||
+ | Dividing by <math>3</math> gives us <math>3, 4, 5, 6, 7, 8, 9</math>. Subtracting <math>2</math> gives us <math>1, 2, 3, 4, 5, 6, 7</math>. | ||
+ | |||
+ | Our list has <math>7</math> numbers. Since she started on a Monday, we must add <math>6</math> days. Our answer is <math>\boxed{Sunday}</math> | ||
+ | ===Solution 2=== | ||
+ | On the first day, Jenny rode <math>7</math> miles. On the second day, she rode <math>7+3=10</math> miles. On the third day, she rode <math>10+3=13</math> miles. | ||
+ | |||
+ | This is the sequence <math>7,10,13,...</math> which is an arithmetic sequence: first term <math>7</math>, common difference <math>3</math>. | ||
+ | |||
+ | We are trying to find the number of terms <math>n</math> such that the <math>n\text{th}</math> partial sum of the sequence is <math>112</math>. | ||
+ | |||
+ | The formula for the sum of a partial sequence is <math>\frac{n}{2}[2a+(n-1)d]</math>, where <math>a</math> is the first term, <math>n</math> is the number of terms, and <math>d</math> is the common difference. (Try to derive it!) | ||
+ | |||
+ | Let <math>a=7</math> and <math>d=3.</math> Then we have: | ||
+ | |||
+ | <math>\frac{n}{2}[14+3(n-1)]=112</math> | ||
+ | |||
+ | <math>n[14+3(n-1)]=224</math> | ||
+ | |||
+ | <math>14n+3n(n-1)=224</math> | ||
+ | |||
+ | <math>14n+3n^2-3n=224</math> | ||
+ | |||
+ | <math>3n^2+11n-224=0</math> | ||
+ | |||
+ | <math>(n-7)(3n+32)=0</math> | ||
+ | |||
+ | The second root is not an integer, so the workout lasted for <math>n=7</math> days. The <math>7\text{th}</math> day after Monday is <math>\boxed{\text{Sunday}}</math>. |
Revision as of 22:14, 14 June 2011
Contents
Problem
AoPSWiki:Problem of the Day/June 15, 2011
Solution
Solution 1
We can solve this problem by a bit of trial and error.
We can guess she rode days and we get
since the mean is clearly
and there are
terms.
That's a bit too small.
We can add to
and get
. That's still too small.
Now, we add to get
, the answer we want.
We now count how many numbers are in the following list: .
Adding to the list gives us
.
Dividing by gives us
. Subtracting
gives us
.
Our list has numbers. Since she started on a Monday, we must add
days. Our answer is
Solution 2
On the first day, Jenny rode miles. On the second day, she rode
miles. On the third day, she rode
miles.
This is the sequence which is an arithmetic sequence: first term
, common difference
.
We are trying to find the number of terms such that the
partial sum of the sequence is
.
The formula for the sum of a partial sequence is , where
is the first term,
is the number of terms, and
is the common difference. (Try to derive it!)
Let and
Then we have:
The second root is not an integer, so the workout lasted for days. The
day after Monday is
.