I am stumped on my 13 year old borthers maths problem. :rolleyes: How sad is that?
x = a missing digit between 0 and 9
√(1xx2x) = xx3
I got this far:
:confused:Code:xx3
x xx3
-------
xx9
+ xxx0
xxx00
--------
11x2x9
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I am stumped on my 13 year old borthers maths problem. :rolleyes: How sad is that?
x = a missing digit between 0 and 9
√(1xx2x) = xx3
I got this far:
:confused:Code:xx3
x xx3
-------
xx9
+ xxx0
xxx00
--------
11x2x9
Whow about 123 *123= 15129
Is that a guess or did you use a mathematical formula to work it out?Quote:
Originally Posted by opus
Dim i As Long
For i = 103 To 993 Step 10
If CStr(i * i) Like "1##2#" Then
MsgBox "DONE: " & i
End If
Next i
You found the last digit (3).
Since the result is less than 20000, and Sqr(20000) is 141,....., the number has to 103,113,123 or 133. and for those I did try and error to find the one that looks like 1xx2x.
Not much math needed here.
You made me curious.
Instead of using xx3 * xx3 use a different Var for each digit, like:
ab3 * ab3
the result would be:
the 4th digit has to be a 2 , so b has to 2 (6*2=12) or 7 (6*7=42).Code:a b 3
* a b 3
________________________
(3a) (3b) 9 'each bracket makes one digit
(ba) (bb) (3b) 0
+ (aa) (ba) (3b) 0 0
_______________________
(aa) (2ba)(....)(6b) 9 ' (..) stands for (3a +bb+3b)
If b would be 7 , the 2 digit would be 2*7*a, which is more than 10, so the first digit would be aa +1 ore more, and that is not possible, since it has to be a 1.
so b is 2.
Since the first digit is 1, a*a must be 1, that makes a=1Code:a 2 3
* a 2 3
________________________
(3a) 6 9 'each bracket makes one digit
(2a) (4) 6 0
+(aa) (2a) (6) 0 0
_______________________
(aa) (4a) (....)(2) 9 ' (..) stands for (3a +bb+3b+1)
√(1xx2x) = xx3Quote:
Originally Posted by visualAd
√(10020 + 1101x) = 110x + 3
10020 + 1101x = (110x + 3)^2
10020 + 1101x = 12100x^2 + 330x + 9
12100x^2 - 7971x - 10011 = 0
x = (7971 +- √(7971^2 - 4*12100*-10011)) / (2*12100)
or something like that