Question 784152
<font face="Times New Roman" size="+2">


CA is equal to 4 times AB, since 4 times anything must be even, CA must be even, and therefore A must be even.  A cannot be 3 or larger because the sum would then be a three digit number.  Hence, A is either 2 or 0.  Let A represent 2, then 4 times B is either 2 or 12.  If 4 times B is 2, then B would have to be 1/2 which is not a digit 0-9, so if A is 2, 4 times B must be 12, which means B is 3.  Then 4 times 2 is 8 plus the carry is 9 and C is 9.  And 4 times 23 is indeed 92.


What is not clear is whether the configuration AB implies that A cannot be zero.  I would assume not, however if it can be zero, then you have 4 times B ends in zero, so B must be 5 and then C is 2.  And 4 times 5 is 20.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>