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.
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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 ==

Latest revision as of 02:56, 13 January 2019

Problem

$A$ is a $4$-digit number $abcd$. $B$ is a $5$-digit number formed by augmenting $A$ with a $3$ on the right, i.e. $B=abcd3$.

$C$ is another $5$-digit number formed by placing a $2$ on the left $A$, i.e. $C=2abcd$. If $B$ is three times $C$, what is the number $A$?


Solution

What this problem is basically saying is that $10000a + 1000b + 100c + 10d + 3 = 3*(20000 + 1000a+100b+10c+d)$. This can be simplified to $7000a + 700b + 70c + 7d = 59997 \Rightarrow 7(1000a+100b+10c+d) = 59997 \Rightarrow  1000a+100b+10c+d = \frac{59997}{7}=8571$.

See also

2010 UNCO Math Contest II (ProblemsAnswer KeyResources)
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
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