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 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</math>, the large one becomes <math>a_0a_1a_2a_3a_4a_5a_6a_7a_8a_9</math>. Then we have | 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> = <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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+ | |||
+ | 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 <math>1089001089</math>, as mentioned above, but the second number is <math>1098910989</math>, not <math>9801009801</math>. | ||
+ | Someone please fix. | ||
==See Also== | ==See Also== |
Latest revision as of 22:01, 17 December 2020
Contents
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 is .
Notice that the sum of two numbers is 10 times of the smaller one. Let the smaller one be , the large one becomes . Then we have = +. It's obvious that and . Comparing the digits, we have , , , and .
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 , 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 |