1997 PMWC Problems/Problem T9


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


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

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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