Solve the Arithmetical Teaser: 100-999 Integer

[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?

 

[Gokul43201]

Surely I'm screwing up somewhere, but 1001 = 13*11*7, so they all do [/color].

 

[NateTG]

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

 

[Fliption]

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

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.

 

[Galileo]

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

 

[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.