1997 PMWC Problems/Problem T9

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.

The pair of numbers are $1089001089$ and $9801009801$ . Use the fact that the sum of two numbers is 10 times of the smaller one to define all digits from two ends.

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

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1997 PMWC Problems/Problem T9

Problem

Solution

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