perplexus.info :: Numbers : One problem

One problem - many ways to solve... (Posted on 2017-12-21)

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.

No Solution Yet Submitted by Ady TZIDON

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Simple ways with calculator

| Comment 1 of 6

[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 2017-12-21 10:22:14

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