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 $5$ 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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AoPS Wiki talk:Problem of the Day/June 15, 2011

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Problem

Solution

Solution 1

Solution 2