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

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==Solution==
 
==Solution==
The pair of numbers are <math>1089001089</math> and <math>9801009801</math>.  
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The pair of numbers are <math>1089001089</math> and <math>is </math>1098910989<math>.  
  
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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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
<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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</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>.
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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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)$.
  
 
==Mistake Above Fix==
 
==Mistake Above Fix==

Revision as of 20:58, 17 December 2020

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$ (Error compiling LaTeX. ! Missing $ inserted.)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)$.

Mistake Above Fix

The actual two numbers are $1089001089$, as mentioned above, but the second number is $1098910989$, not $9801009801$. Someone please fix.

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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