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Find the largest integer which divides every member of the following sequence: 1^5-1, 2^5-2, 3^5-3, ... n^5-n.

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Stage: 4 Challenge Level: Challenge Level:2 Challenge Level:2

Well done to Oliver from Olchfa School, and others, for good problem-solving and reasoning.

$1905, 2587, 3951, 7020$ and $8725$ all have a remainder of $200$ when divided by $341$. Here's how we know that :

Whatever the largest divisor is going to be (call it $m$) all these numbers can be written in algebra as $am + r$ where $r$ is the remainder and '$a$' is the number of times $m$ goes into that number.

Now because all the numbers will have the same remainder, $r$, when divided by $m$, the difference between any two of those five numbers must be a multiple of $m$. In particular the difference from one number to the next one up must be a multiple of $m$.

So that means that I only need to find the largest number that divides exactly into $682, 1364, 3069$ and $1705$ (the differences).

The prime factors of each of those differences are :

$682$ is $2 . 11 . 31$

$1364$ is $2 . 2 . 11 . 31$

$3069$ is $3 . 3 . 11 . 31$

$1705$ is $5 . 11 . 31$

So the largest divisor of all four numbers will be $341$ ( $11 . 31$ ) and when this is tested on the original five numbers the remainder is $200$ each time.

That's it. But I was wondering if I needed all four differences, or could I have done less calculations, and the answer is that I did need all four differences, because otherwise if any of the original five were not used in the calculations that number could have been anything and plenty of numbers wouldn't work with the solution produced using only some of the original numbers. Hope that's clear!

I also wondered whether I'd done enough calculation - was four differences enough ?

When I have five original numbers there are $10$ differences possible : first with the second, third, fourth and fifth numbers (that's $4$), then second with the third, fourth and fifth (that's $3$ more), third with fourth and fifth, and fourth with fifth (that's $10$ in all)

I think that the answer is yes, four differences is always enough.

I imagined it like this :

Suppose there were just three original numbers, $a$, $b$ and $c$, in that order of size. I'll use the differences $a$ to $b$ and $b$ to $c$, but I wont need $a$ to $c$. That's because whatever the biggest divisor is for a to $b$, and also for $b$ to $c$, it will work for $a$ to $c$ as well.

Lots to think through here - nice solution.