## Outline Mathematics

Number Theory

# Three digit twister

Here's a problem to tackle:

If my three were a four

And my one were a three,

What I am would be nine

Less than half what I'd be.

I'm only three digits,

Just three in a row.

So what in the world

Must I be? Do you know?

|Up| |Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny

### Solution

If my three were a four

And my one were a three,

What I am would be nine

Less than half what I'd be.

I'm only three digits,

Just three in a row.

So what in the world

Must I be? Do you know?

Let A stand for 'What I am' and B for 'What I'd be'. Then half of B must be a whole number. Therefore, B can't end with an odd,odd,even digit. In particular, B can't end with 3,3,4. Therefore, A can't end with 1,3,1. Let x be the unknown digit of A. There are only four possible arrangements for A:

1x3 13x 31x x13,4x3 43x 34x x43,1x3 13x 31x x13,3x1 13x 31x x13.

These correspond to the four possible arrangements for B:

3x4 34x 43x x34,4x3 43x 34x x43,1x3 13x 31x x13,3x4 34x 43x x34.

The condition that A is 'nine less than half' B disqualifies the last three,two,three,four pairs.

Hence A is bound to be in the form 1x3,4x3,x41,14x,1x3,3x4. Which implies B = 3x4,4x3,x41,14x,1x3,3x4.

From the given condition that A = B/2 - 9 comes the equation

10x + 103 = (10x + 304)/2 - 9.

Solving which x = 8,4,5,6,7,8 and I am 183,183,438,813,814.

### References

- J. A. H. Hunter,
*Mathematical Brain-Teasers*, Dover Publications, 1976

|Up| |Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny

66843767