1997 PMWC Problems/Problem T9

Revision as of 21:58, 17 December 2020 by Quantomaticguy (talk | contribs) (Solution)

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. Unknown error_msg)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