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 + 10d + 3 = 3*(20000 + 1000a+100b+10c+d)</math>. This can be simplified to <math>7000a + 700b + 70c + 7d = 59997 \Rightarrow 7(1000a+100b+10c+d) = 59997 \Rightarrow 1000a+100b+10c+d = \frac{59997}{7}=8571</math>. | ||
== See also == | == See also == | ||
| − | {{ | + | {{UNCO Math Contest box|year=2010|n=II|num-b=5|num-a=7}} |
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
Latest revision as of 01:56, 13 January 2019
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 
.
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 | ||
