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

Revision as of 20:05, 16 February 2014

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

@@ Line 1: / Line 1: @@
 ==Problem==
 Find the two 10-digit numbers which become nine times as large if the order of the digits is reversed.
-==Solution==
-{{solution}}
-Let's call any number that satisfies <math>x</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==
-{{PMWC box|year=1997|num-b=T8|num-a=T10}}
-[[Category:Intermediate Number Theory Problems]]