Difference between revisions of "1997 PMWC Problems/Problem T9"
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5. <math>12345670</math> numbers to go. | 5. <math>12345670</math> numbers to go. | ||
| − | The | + | The pair of numbers are <math>1089001089</math> and <math>9801009801</math>. Use the fact that the sum of two numbers is 10 times of the smaller one to define all digits from two ends. |
==See Also== | ==See Also== | ||
Revision as of 23:53, 15 December 2015
Problem
Find the two 
-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 
.
1. 
. It must be -digit, and it multiplied by nine must be -digit.
2. divides by 
. If you recall the divisibility rule of 9, the sum of digits must be divisible by 9.
3. ends in . 
must start with .
4. So 
5. 
numbers to go.
The pair of numbers are 
and 
. Use the fact that the sum of two numbers is 10 times of the smaller one to define all digits from two ends.
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 | ||
