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

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==Solution==
 
==Solution==
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{{solution}}
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The pair of numbers are <math>1089001089</math> and is <math>1098910989</math>.  
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Let's call any number that satisfies <math>x</math>.
 
  
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1. <math>1000000000\le x\le1111111111</math>. It must be <math>10</math>-digit, and it multiplied by nine must be <math>10</math>-digit.
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Notice that the sum of two numbers is 10 times of the smaller one. Let the smaller one be <math>a_9a_8a_7a_6a_5a_4a_3a_2a_1a_0</math>, the large one becomes <math>a_0a_1a_2a_3a_4a_5a_6a_7a_8a_9</math>. Then we have
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<math>a_9a_8a_7a_6a_5a_4a_3a_2a_1a_010</math> = <math>a_9a_8a_7a_6a_5a_4a_3a_2a_1a_0</math>+<math>a_0a_1a_2a_3a_4a_5a_6a_7a_8a_9</math>.
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It's obvious that <math>a_9=1</math> and <math>a_0=9</math>. Comparing the digits, we have <math>(a_8=0, a_1=8)</math>, <math>(a_7=8, a_2=0)</math>, <math>(a_6=9, a_3=1)</math>, and <math>(a_5=0, a_4=0)</math>.
  
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2. <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.
 
  
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3. <math>x</math> ends in <math>9</math>. <math>9x</math> must start with <math>9</math>.
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Side note:
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if its not clear how to obtain the digits
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4. So <math>1000000089\le x\le 1111111119</math>
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essentially we want to write an equation first and then compare the units,tens,hundreds digits etc.. on the left and right side. its easy to notice the digits must be one of {0,1,8,9} hence there aren't many cases to consider was well
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5. <math>12345670</math> numbers to go.
 
  
 
==See Also==
 
==See Also==

Latest revision as of 15:44, 6 August 2021

Problem

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

Solution

The pair of numbers are $1089001089$ and is $1098910989$.

Notice that the sum of two numbers is 10 times of the smaller one. Let the smaller one be $a_9a_8a_7a_6a_5a_4a_3a_2a_1a_0$, the large one becomes $a_0a_1a_2a_3a_4a_5a_6a_7a_8a_9$. Then we have $a_9a_8a_7a_6a_5a_4a_3a_2a_1a_010$ = $a_9a_8a_7a_6a_5a_4a_3a_2a_1a_0$+$a_0a_1a_2a_3a_4a_5a_6a_7a_8a_9$. It's obvious that $a_9=1$ and $a_0=9$. Comparing the digits, we have $(a_8=0, a_1=8)$, $(a_7=8, a_2=0)$, $(a_6=9, a_3=1)$, and $(a_5=0, a_4=0)$.


Side note: if its not clear how to obtain the digits essentially we want to write an equation first and then compare the units,tens,hundreds digits etc.. on the left and right side. its easy to notice the digits must be one of {0,1,8,9} hence there aren't many cases to consider was well

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
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