Difference between revisions of "1997 PMWC Problems/Problem T9"

Revision as of 05:24, 10 October 2013

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

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.

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 ==Solution==
 {{solution}}
+Let's call any number that satisfies <math>x</math>.
-. <math>1000000000\le x\le1111111111</math>
+. <math>1000000000\le x\le1111111111</math>. It must be <math>10</math>-digit, and it multiplied by nine must be <math>10</math>-digit.
+. <math>x</math> divides by <math>9</math>. If you recall the divisibility rule of 9, the sum of digits must be divisible by 9.
+. <math>x</math> ends in <math>9</math>. <math>9x</math> must start with <math>9</math>.
+. So <math>1000000089\le x\le 1111111119</math>
+. <math>12345670</math> numbers to go.
 ==See Also==