View source for AoPS Wiki talk:Problem of the Day/June 15, 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 <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>, which is the total we want.

We started with <math>5</math> days and added <math>2</math> more, so there are <math>7</math> days in total.  We can add <math>7-1=6</math> days to her starting day, Monday, to find her final day.  Our answer is thus <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>.