1997 PMWC Problems/Problem T9

Revision as of 14:43, 20 April 2014 by TheMaskedMagician (talk | contribs) (Problem)

Problem

Find the two $10$-digit numbers which become nine times as large if the order of the digits is reversed.

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it. Let's call any number that satisfies $x$.

1. $1000000000\le x\le1111111111$. It must be $10$-digit, and it multiplied by nine must be $10$-digit.

2. $x$ divides by $9$. If you recall the divisibility rule of 9, the sum of digits must be divisible by 9.

3. $x$ ends in $9$. $9x$ must start with $9$.

4. So $1000000089\le x\le 1111111119$

5. $12345670$ numbers to go.

See Also

1997 PMWC (Problems)
Preceded by
Problem T8
Followed by
Problem T10
I: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
T: 1 2 3 4 5 6 7 8 9 10