The problem below (Moscow Puzzles #313) can be solved in more than 4 ways,
(each d1 by itself)  using different approaches.
Find the number t and the digit represented by k in:
[3*(230+t)]^2=492,k04
List your ways of solving it.
[3*(230+t)]^2=492,k04
My first idea:
[3*(230+t)]^2 = 492,004 + 100k
3*(230+t) = 701.429968 + (a little bit)
230+t = 233.8099893 + (a little bit)
t = 3.8099893 + (a little bit)
guess t=4
confirm
[3*(230+4)]^2 = 492,804
so k=8
Second way (pretty similar)
492,k04 is a perfect square
square root of 492004 = 701.4
square root of 493004 = 702.1
702^2 = 492804
so k=8 and t follows.

Posted by Jer
on 20171221 10:22:14 