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> | + | 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< | + | 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>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< | + | 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)$. |
==Mistake Above Fix== | ==Mistake Above Fix== |
Revision as of 20:58, 17 December 2020
Contents
[hide]Problem
Find the two -digit numbers which become nine times as large if the order of the digits is reversed.
Solution
The pair of numbers are and 1098910989$.
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_0a_0a_1a_2a_3a_4a_5a_6a_7a_8a_9a_9a_8a_7a_6a_5a_4a_3a_2a_1a_010a_9a_8a_7a_6a_5a_4a_3a_2a_1a_0a_0a_1a_2a_3a_4a_5a_6a_7a_8a_9a_9=1a_0=9(a_8=0, a_1=8)(a_7=8, a_2=0)(a_6=9, a_3=1)(a_5=0, a_4=0)$.
Mistake Above Fix
The actual two numbers are , as mentioned above, but the second number is , not . 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 |