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

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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>.
 
<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>.
 
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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==Mistake Above Fix==
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The actual two numbers are <math>1089001089</math>, as mentioned above, but the second number is <math>1098910989</math>, not <math>9801009801</math>.
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Someone please fix.
  
 
==See Also==
 
==See Also==

Revision as of 22:33, 26 November 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 $9801009801$.

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 $(a_5=0, a_4=0)$.

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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