### Consecutive Numbers

An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.

### Calendar Capers

Choose any three by three square of dates on a calendar page...

### Days and Dates

Investigate how you can work out what day of the week your birthday will be on next year, and the year after...

# How Many Miles to Go?

##### Stage: 3 Challenge Level:

We received very contrasting approaches to solving this problem. Rami from Pierre Laporte School approached this in stages:

We'll add 2 to that, to get "3s" (in the far right hand columns): 4633 & 175.3

We'll add 300 to that one, to get "4s" (in the far left hand columns): 4933 & 475.3

We add 20 to that one: 4953 & 495.3

Robert Stabler from Ardingly College Junior School used very similar reasoning:

In this problem I decided to alter the milometer until each single unit was the same as the trip meter's.

First I added 2 miles so they both ended in three.

I then added 20 so they both ended in 53.

Then I added 300 which made the milometer and trip meter end in 953.

They were now the same.

Andrei Lazanu from School No. 205 in Bucharest realised that the two numbers would contain the same digits when the number on the milometer was ten times larger than the number on the trip meter, and then went on to generalise. This is how he arrived at the solution:

When you go a distance x, both the milometer and the trip meter will go a distance x.

The trip meter will need to have a value that is ten times smaller than the milometer, so I have:

4631 + x = 10(173.3 + x)
4631 + x = 1733 + 10x
9x = 2898
x =322 (miles)

So, you have to travel another 322 miles for the two numbers to have the same digits in the same order.

In the second situation, I have:

4632 + x = 10(173.3 + x)
4632 + x = 1733 + 10x
9x = 2899
2899 isn't divisible by 9.

In general, if m is the value indicated by the milometer and t the value indicated by the trip meter and x the distance travelled after the measurement, the possibilities are:

m +x = t + x$\rightarrow$ m = t $\rightarrow$ the first trip of the car (very difficult and rare)
Any distance is possible.

m + x = 10(t + x) $\rightarrow$ m + x = 10t + 10x $\rightarrow$ 9x = m -10t
This is the most common possibility.
(m - 10t) must be divisible by 9.

Here are some possibilities that will lead to matching pairs:
m = 1653; t = 60.0 $\rightarrow$ x = 117 (miles)
m = 5330; t = 1.1 $\rightarrow$ x = 591 (miles)
m = 9442, t = 10.0 $\rightarrow$ x = 1038 (miles)

m + x = 100(t + x) $\rightarrow$ m + x = 100t + 100x $\rightarrow$ 99x = m -100t
(m - 100t) must be divisible by 99.

Here are two possibilities that will lead to matching pairs:
m = 596; t = 2.00 $\rightarrow$ x = 4 (miles)
m = 7758; t = 36.00 $\rightarrow$ x = 42 (miles)

m + x = 1000(t + x) $\rightarrow$ m + x = 1000t + 1000x = 999x = m - 1000t
This solution is very rare, as the result must be divisible by 999, t must be a small one-digit number.