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Difference between revisions of "AoPS Wiki talk:Problem of the Day/June 15, 2011"

m (→‎Solution 1: Simplified the list-counting method)
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Now, we add <math>25</math> to get <math>112</math>, which is the total we want.
 
Now, we add <math>25</math> to get <math>112</math>, which is the total we want.
  
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We started with <math>5</math> days and added <math>2</math> more, so there are <math>7</math> days in total.  Therefore, we must add <math>7-1=6</math> days to her starting day, Monday, to find the final day.  Our answer is thus <math>\boxed{Sunday}</math>
+
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===
 
===Solution 2===

Latest revision as of 11:55, 15 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 $112$, which is the total we want.

We started with $5$ days and added $2$ more, so there are $7$ days in total. We can add $7-1=6$ days to her starting day, Monday, to find her final day. Our answer is thus $\boxed{Sunday}$

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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