I was going through a backlog of emails in the train, when I came across this one from a friend (entry's title was the subject line)
1. Choose any number consist of 3 digits (e.g. 234)
2. Duplicate the number and write it all together (e.g. 234234)
3.Divide the last number by 7, then divide the result by 143.
4. you get the number that you choose in step 1.
Below my reply,
Let x1 be an integer between 100 and 999, written in base 10, with digits a, b and c, so that x2 = 100*a+10*b+c
Perform step 1 to get the number x2. Then x2 = 100000*a+10000*b+1000*c+100*a+10*b+c
Before I perform 3, i need the insurance that x2 is always a multiple of 7. For this I need to show that x2 = 0 [modulo 7]
The following reductions are done modulo 7 ( I do them to reduce the powers of ten )
10 = 3
100 = 9 = 2
1000 = 100*10 = 2*3 = 6
10000 = 1000*10 = 6*3 = 18 = 4
100000 = 10000*10 = 4*3 = 12 = 5
The following reduction are done module 7 ( I do them to reduce x2 )
= 5*a+4*b+6*c+2*a+3*b+c = 7*a+7*b+7*c = 0
So I have established that x2 is indeed a multiple of 7
We have x2 = 7*x3 where x3 is an integer. Now we want to show that x3 = 143*x1. It is enough to show that x2 = 7*143*x1 ( indeed if we know x2 = 7*x3 and x2 = 7*143*x1 , we have 7*x3=7*143*x1 and the result by dividing both sides by 7 ).
x2 = 7*143*x1 is equivalent to x2 = 1001*x1 . This last equality is obviously true due to
x2 = 100000*a+10000*b+1000*c+100*a+10*b+c = 1000 * ( 100*a+10*b+c ) + 100*a+10*b+c = 1001 * ( 100*a+10*b+c ) = 1001 * x1
My answer was obviously the kind of answer you would expect from a mathematician, but an even more interesting thing is noticing the pattern that made it work and then do with with 4 digit numbers: Take a number between 1000 and 9999, do the duplication thing (for instance 1234 gives 12341234) and impress yourself by dividing it by whatever prime factors 10001 happens to have. You can do it with n digits numbers if you divide the duplication by prime factors of 10^n+1.