Going Round in Circles

Lots of really great solutions to this problem were submitted. It is good to see a variety of methods used to solve the questions; have a look at the range below to see the different ways students have approached the task. Although there are several ways of working out the answers, some are more efficient than others, and some can be done mentally, whilst others require a calculator. Also, as several people noticed, you can apply the same method to the different problems. This is nice as it shows that you can solve many real-life problems with a few methods that you learn at school.

Some people used Anita's method; others used Luke's. Some students used a combination of the two, and others devised formulae as well. Some preferred to use remainders, and others worked with decimals or fractions. Have a look at the different ways, and think about the pros and cons of each of them.

The first question asked you to suggest other numbers that we could have chosen. John, from CGSB, gave some examples:

Monday: $35, 350, 3500, 35000, 21, 210, 2100, 21000$

Wednesday: $352, 3502, 37, 35002, 23, 212, 2102, 21002$

Sam, from Charters Sunningdale, explained that you could choose any number in the seven times table eg.$182, 14,406$ ... Lizzie and Becca pointed out that, for Wednesdays, you could choose any number that was $2$ more than any multiple of $7$.

Siddhartha, from the Garden International School, expressed this in an algebraic form:

Let $x$ equal the number of days since Monday.

For the day to be a Monday, the answer to $x\div7$ has to be a multiple of seven to be correct

i.e. the answer must be a whole number when divided by $7$.

Have a go at devising a formula for Wednesday.

Now, let us look at the next questions. James, from Wilson's School, explained how Anita could continue with her working. He also described how Luke's method helped him:

Going on from Anita's problem, she should continue adding on seventies until she's one $70$ away from going over $1000$, bringing her to the number $980$.

Then add on sevens for the days of the week which should make $994$.

Add on the extra six days to get to $1000$. Therefore in $1000$ days, it will be a Sunday (Monday + $6$ days is Sunday).

Luke's method helps him because he divides $1000$ by $7$ (because there are $7$ days in a week), which will give him the exact number of weeks in the $1000$ days. He got the figure $142.8571429$
He then multiplies $142$ (which is the number of whole weeks in $1000$ days) by $7$ to get $994$. This brings him to the nearest multiple of seven below $1000$.
Then do what Anita did by adding on the six days, and Luke will also know that Sunday is the $1000$th day.
However Luke could have rounded up the number $142.8571429$ to $143$, and multiplied that by $7$ to get $1001$. Then he could have just taken off the one day to get Sunday.

Several others got the correct answer, using Anita's method, Luke's method or a combination of the two: Patrick and William from St. Hughs, Christina from Tudhoe Grange, Nicole and Ann from York House School, and Natasha from Wilmslow High School.

Now let us look at the additional questions, starting with: "If it is November, what month will it be in $1000$ months?"

Shona from Tudhoe Grange applied Anita's method to this problem:

If it is November now, in $1000$ months' time it will be March.

I worked this out because in $12$ months' time it will be November.

It will also be November in $120$ months, $240$ months, $360$ months, $480$ months, $600$ months, $720$ months, $840$ months and $960$ months.

Then if you keep adding $12$ months at a time the highest number you can get without going over $1000$ months is $996$ months. So in $996$ months' time it will be November. Then if you add one month at a time then you will get to $1000$ months and it will therefore be March in $1000$ months' time.

Nicole and Ann, from York House School, used Luke's method:

To know what month it would be in $1000$ months if it is now November:

$1000\div12 = 83.333$.
That would be exactly the number of years in $1000$ months.
$83 \times 12$ would be how many months in $83$ whole years, which would be $996$.
There are $4$ more months until the $1000$th month. We know that the $996$th month is November, so $4$ more months would be March. (One month until January and then $3$ months into the next year).

William, from Toot Hill School, used Luke's method but with remainders:

$1000\div12$ months = $83$ remainder $4$

So, you add $4$ months on to November which leaves you with the answer March

Jenni began using Luke's method. She noticed:

$1000\div12= 83\frac{1}{3}$.

This means that there are $83$ whole more years to get back to November. Then another third of a year must be added on.

One third of a year is $4$ months ($12\times \frac{1}{3}$ or $\frac{12}{3}$).

Therefore, we add $4$ months to November, which gives March.

Other students who solved this problem correctly include: Thomas and Laura from Tudhoe Grange, Lizzie and Becca, and Elli from Aylesbury High.

Flora and Lydia, from Cottenham Village College, noticed that the same method can be applied to similar questions, but in a different context. They came up with a general set of formulae for "circular loop" problems:

We came up with a formula:
$x =$ the amount you want to add on. (in the months example, $1000$ months)
$y =$ the amount in the circular loop. (in the months example, $12$ months, which is the amount of months in a year)
$t =$ the starting point (in the example, November)
Integer part of $\frac{x}{y} = n$ (in the example, $83$ years)
$yn = z$ (in the example, $996$)
$x -z = p$ (in the example, $4$ months)
$t +p =$ the answer.

We will finish by briefly looking at the other questions; the methods shown above can be applied here too. Try the different methods for yourself.

"A railway line has $27$ stations on a circular loop. If I fall asleep and travel through $312$ stations, where will I end up in relation to where I started?"

Elli, from Aylesbury High, used Luke's method, but with fractions:

The first thing we do is determine how many loops we go through if we travel through $312$ stations.

$312\div27 = 11 \frac{5}{9}$

$\frac{5}{9}$ of $27$ stations is $15$

So we will be $15$ stations ahead, in relation to where we started.

Josh, from Wilmslow High School used the formulae described by Flora and Lydia above:

$X=312$
$Y=27$
$N=312\div27 = 11$
$Z=11\times27 = 297$
$P= 312 - 297 = 15$
Therefore the answer = $15$ Stations later

Natasha, also from Wilmslow High School got this right too, as did Lizzie and Becca, and Siddhartha, from the Garden International School.

Now for the next question: "If it is midday now, will it be light or dark in $539$ hours?"

Elli, from Aylesbury High, used Luke's method:

$539\div24 = 22.4583333$ This calculation gives us the exact number of days.
$0.458333$ of one day is $11$ hours.
Midday + $11$ hours = $11$pm.
Therefore it will be dark.

Others who were also right included: Laura from Tudhoe Grange, Nicole and Ann from York House School, Josh and Natasha from Wilmslow High School, and Siddhartha from the Garden International School.

The answer is indeed $11$pm. However, an interesting point is that there are some places in the world where it is still light at $11$pm at certain times of the year! So "light" could also be correct in certain special circumstances. Can you think of where these places may be?

Let us go running now ...

"If a running track is $400$ metres around, where will I be in relation to the start after running $6$ miles (approximately $9656$ metres)?"

Those who submitted the correct answer include: John from CGSB, Nicole and Ann from York House School, Luke from Charters, Elli from Aylesbury High, and Siddhartha from the Garden International School.

Elli showed us one way of working out the answer:

$9656$ metres $\div$ $400$ metres = $24.14$ circuits of the entire track.
$0.14 \times 400$ metres = $56$ metres.
So we'll be $56$ metres ahead

Can you use the other methods to work out the answer?

The final question was about a fairground wheel: "If I get on at the bottom of a fairground wheel and the wheel turns through 5000 degrees, whereabouts on the wheel will I be?"

Elli explained how she solved the problem, using Luke's method, but with fractions rather than decimals:

A fairground wheel is circular, and so going around once will take you through $360^\circ$.
$5000 \div 360 = 13 \frac{8}{9}$
So, if the wheel turns around $5000^\circ$, we go around in a circle $13$ and $\frac{8}{9}$ times.
$\frac{8}{9} \times 360 = 320^\circ$
So we will be $320^\circ$ around the wheel.

Laurence, Olivia, Abi, Jessie, Frank, Joe all attend Over Primary School. They used Luke's method to work out the correct answer:

$5000^\circ$ divided by $360^\circ$ (because there are $360^\circ$ in a full circle)
This makes $13.888$
Round down to $13$
Multiply $13$ by $360$ to get $4680^\circ$ which is the nearest multiple of $360$ below $5000$.
Find the difference between $4680$ and $5000$ which is $320^\circ$
So, after turning through $5000^\circ$ the wheel would end up $320^\circ$ from the bottom.

John, from CGSB, pointed out that the answer could also be $40^\circ$ from the bottom ($360-320=40$). Nicole and Ann, from York House School, noticed this too.

Amy from Charters used Anita's method:

A full turn is $360^\circ$, so after $3600^\circ$, you will also be at the bottom.

After $3960^\circ$ ($3600+360$), you will again be at the bottom.

You can keep on adding $360^\circ$, and still be at the bottom: $4320, 4680, 5040$ ...

So, you are $40^\circ$ from the bottom (or $320^\circ$, depending on the way you look at it).

Thank you to everyone who submitted solutions! If you want to have a go at a similar problem, take a look at Round and Round and Round which encourages similar ways of working with division and remainders on a calculator.

If you want to do an extension of this problem, try Days and Dates and Modular Arithmetic. Have fun!