1997 PMWC Problems/Problem T9

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

Mistake Above Fix

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

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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1997 PMWC Problems/Problem T9

Contents

Problem

Solution

Mistake Above Fix

See Also