# Arithmetical teaser

1. Jan 14, 2005

### Galileo

This one is actually quite simple once you see it.

Consider the following converation between man 1 and man 2:

Man 1: 'I have a puzzle for you. First, pick an integer from 100 to 999 in your head.'
Man 2: 'Uuhhm. Well.. okay. (thinks: I`ll take 501, 'cuz that me lucky number.')
Man 1: 'Okay, consider the number that you get by repeating the number in your head once. So if you had 123, you should have 123123.'
Man 2: 'Okay. (That'll be 501501)'
Man 1: 'Now divide that number by 13.'
Man 2: '..wait a sec.... okay. Fortunately it's an integer. (that's 38577)'
Man 1: 'Divide that number you got by 7'
Man 2: '..(that'll be 5511).. okay. Luckily another integer!'
Man 1: 'Now divide it by 11.'
Man 2: '(5511/11 is..eh 501). Hey, I get the number with which I started.'
Man 1: 'Really? I suppose you made a good choice at the start. My question is: How many integers from 100 to 999 have this property?'

So, how many integers are there from 100 to 999 which have this property?

2. Jan 14, 2005

### Gokul43201

Staff Emeritus
Surely I'm screwing up somewhere, but 1001 = 13*11*7, so they all do .

3. Jan 14, 2005

### NateTG

Hmmm,
$$\frac{1001}{7*11*13}=1$$
no idea at all. Guess, I'll just have to try them one by one.

4. Jan 14, 2005

### Fliption

The answer is all of them (900). This works because taking any number like 501 and making it 501501 is the same thing as this:

Assume the number you chose is "X". So X = 501

(X*1000)+X is the formula that gives you 501501.

1000X +X = 1001X

1001X/1001 = X

Works everytime regardless of X up to 999.

5. Jan 14, 2005

### Galileo

Mwa. I knew it was too easy. Oh well...

6. Jan 15, 2005

### vikasj007

well, yes it is quite easy, in fact i had read this in a book quite a few years back.

i think i should look for that book, maybe i can post some good questions from that book too.