Difference between revisions of "1997 PMWC Problems/Problem T9"
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<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>. | ||
| + | |||
| + | ==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== | ||
Revision as of 21:33, 26 November 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 
.
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 
.
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 | ||
