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Difference between revisions of "AoPS Wiki talk:Problem of the Day/June 15, 2011"
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==Solution==
 
==Solution==
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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 to 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 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>.
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 +
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>
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<math>n[14+3(n-1)]=224</math>
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<math>14n+3n(n-1)=224</math>
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 +
<math>14n+3n^2-3n=224</math>
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 +
<math>3n^2+11n-224=0</math>
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<math>(n-7)(3n+32)=0</math>
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 +
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

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 $5$ days and we get $7+10+13+16+19=(13)(5)=65$ since the mean is clearly $13$ and there are $5$ terms.

That's a bit too small.

We can add $22$ to $65$ and get $87$. That's still too small.

Now, we add $25$ to get $112$, the answer we want.

We now count how many numbers are in the following list: $7, 10, 13, 16, 19, 22, 25$.

Adding $2$ to the list gives us $9, 12, 15, 18, 21, 24, 27$.

Dividing by $3$ gives us $3, 4, 5, 6, 7, 8, 9$. Subtracting $2$ gives us $1, 2, 3, 4, 5, 6, 7$.

Our list has $7$ numbers. Since she started on a Monday, we must add $6$ days. Our answer is $\boxed{Sunday}$

Solution 2

On the first day, Jenny rode $7$ miles. On the second day, she rode $7+3=10$ miles. On the third day, she rode $10+3=13$ miles.

This is the sequence $7,10,13,...$ which is an arithmetic sequence: first term $7$, common difference $3$.

We are trying to find the number of terms $n$ such that the $n\text{th}$ partial sum of the sequence is $112$.

The formula for the sum of a partial sequence is $\frac{n}{2}[2a+(n-1)d]$, where $a$ is the first term, $n$ is the number of terms, and $d$ is the common difference. (Try to derive it!)

Let $a=7$ and $d=3.$ Then we have:

$\frac{n}{2}[14+3(n-1)]=112$

$n[14+3(n-1)]=224$

$14n+3n(n-1)=224$

$14n+3n^2-3n=224$

$3n^2+11n-224=0$

$(n-7)(3n+32)=0$

The second root is not an integer, so the workout lasted for $n=7$ days. The $7\text{th}$ day after Monday is $\boxed{\text{Sunday}}$.

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