There's always One isn't there
Take any pair of numbers, say 9 and 14.
Take the larger number, 14, and count up by that amount :
Then divide each of the values by 9, your chosen smaller number, and look at the remainders.
Notice there's a one.
Now do the same again but using different numbers, say 7 and 12.
Counting in twelves and dividing each result by 7 :
Again somewhere in those remainders is a one.
Pick the pairs how you like, somewhere there'll always be a one - won't there?
What actually happens?
Why?
Try plenty of examples :
Maybe start with other numbers and take multiples but continue dividing by 9.
Next use other numbers again, taking multiples of each as before but now divide by 7.
Then again use different numbers, take multiples and divide by 8 - be sure to include 12 as on of the numbers you try multiples of.
What happens?
Why?
A number of people spotted that you cannot get a remainder of one if the two numbers you use have a common factor.
For example : 8 and 6 have 2 as a common factor.
Look at the multiples of 8 :
8, 16, 24, 32, 40, 48
And the remainders when we divide by 6 :
2, 4, 0, 2, 4, 0
Eight and six are both made from twos so taking away sixes is like taking away three twos from a collection of twos, and what's left over must be some twos! (or nothing at all)
OK, so if there's a common factor the remainder can never be one, but if the numbers don't have any common factor (coprime) does that mean there will always be a remainder of one somewhere ?
Dominic from St Paul's School says Yes :
If the numbers are a and b, and coprime, then ab is their lowest common multiple, the smallest multiple of a that is also a multiple of b.
The remainders, upon division by b, could be written $r_1, r_2, . . . . . , r_{b-1}$
Dominic is about to show that no two values from that set of remainders can be the same.
Just suppose two of the remainders, $r_k$ and $r_l$, were actually the same (with k< l, ie. k comes first).
Suppose that the remainder value when this happens is R then there is some integer value n
and another one, m, so that :
ka = nb + R and
la = mb + R
n is the number of times b goes into ka, and m is the number of times that b goes into la,
with R left as the remainder each time .
It follows, with that R the same in each expression, that la - ka = mb - nb
or (l - k) a = (m - n) b
and that would make (l - k)a divisible by b
However l - k is an integer value somewhere between 1 and b - 2 and we already know that ab is the smallest multiple of a that is divisible by b, so there is no l - k value to make (l - k)a divisible by b.
The conclusion is that no two remainders can be the same - they must all be different
And since there are b - 1 remainders, which have to take one of the values: 1,2,3...(b-1), and all be a different value, it follows that one of the remainders will be 1, which is what we wanted to prove.
Thank-you Dominic that's nice reasoning
Systematic working and recording of results help a lot here.
Conjectures are important, and should be encouraged, but along with a challenge to really explain why any claim might be true generally.
We are so familiar with numbers and what they do, or what we believe they do, that the challenge to account rigorously for the familiar can seem pedantic. Hopefully the problem expressed in this form will give students the pleasure of discerning real structure.