Difference between revisions of "2010 UNCO Math Contest II Problems/Problem 6"
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== Solution == | == Solution == | ||
| + | What this problem is basically saying is that <math>10000a + 1000b + 100c + 100d + 3 = 50000 + 1000a+100b+10c+d</math>. This can be simplified to <math>9000a + 900b + 90c + 9d = 49997 \Rightarrow 9(1000a+100b+10c+d) = 49997 \Rightarrow 1000a+100b+10c+d = \frac{49997}{9}</math>. This has no integer solution. | ||
== See also == | == See also == | ||
Revision as of 21:21, 9 February 2015
Problem
is a 
-digit number 
. 
is a 
-digit number formed by augmenting with a 
on the right, i.e.
.
is another -digit number formed by placing a 
on the left , i.e. 
. If is
three times , what is the number ?
Solution
What this problem is basically saying is that 
. This can be simplified to 
. This has no integer solution.
See also
| 2010 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
| All UNCO Math Contest Problems and Solutions | ||
