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Number machine problem 1 (Posted on 2003-06-08) Difficulty: 1 of 5
I have a number machine. I say x gives y if when x goes in, y comes out. For any numbers x and y, by "xy" I mean x followed by y. Here are my two rules.

1x gives x.
For example, 13 gives 3.
If x gives y, then 2x gives yy.
For example, 213 gives 33 since 13 gives 3.

What is a number that gives itself?

See The Solution Submitted by Tim Axoy    
Rating: 3.7000 (10 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution re: Solution | Comment 2 of 12 |
(In reply to Solution by Jun)

I thought about that... but the number machine only seems to work on numbers beginning with 1_ or 21_

2x gives yy... so x = 2 and y = 2. But 2 doesn't seem to work.

Since 10^?+x will never equal x, we must be looking at the last sequence.

So... 2 __ = __ __

But putting the first sequence into the second one, you get

2(1 __)=(__)(__)

So 21 followed by a number equals the same number twice.

Since 121 gives 21, 2121 gives 2121.

This number must be 21, so 121 gives itself.



  Posted by Gamer on 2003-06-08 04:48:26

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