Factorial one hundred (written 100!) has 24 noughts when written in full and that 1000! has 249 noughts? Convince yourself that the above is true. Perhaps your methodology will help you find the number of noughts in
10 000! and 100 000! or even 1 000 000!
Prove that if a^2+b^2 is a multiple of 3 then both a and b are multiples of 3.
Find the largest integer which divides every member of the
following sequence: 1^5-1, 2^5-2, 3^5-3, ... n^5-n.
Well done to Oliver from Olchfa School,
and others, for good problem-solving and reasoning.
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
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 $m$.
The prime factors of each of those differences are :
$1364$ 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
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
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