Difference between revisions of "1997 PMWC Problems/Problem T9"
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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> | + | <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>. | ||
Revision as of 11:42, 31 August 2016
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 .
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 .
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 |