Missing Numbers

[visualAd]

I am stumped on my 13 year old borthers maths problem. How sad is that?


x = a missing digit between 0 and 9

√(1xx2x) = xx3

I got this far:

     xx3
x    xx3
-------
     xx9
+   xxx0
   xxx00
--------
  11x2x9

[opus]

Whow about 123 *123= 15129

[visualAd]

Whow about 123 *123= 15129

Is that a guess or did you use a mathematical formula to work it out?

[wossname]

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

[opus]

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.

[opus]

You made me curious.

Instead of using xx3 * xx3 use a different Var for each digit, like:

ab3 * ab3
the result would be:

               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)

the 4th digit has to be a 2 , so b has to 2 (6*2=12) or 7 (6*7=42).
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.

                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)

Since the first digit is 1, a*a must be 1, that makes a=1

[alkatran]

I am stumped on my 13 year old borthers maths problem. How sad is that?


x = a missing digit between 0 and 9

√(1xx2x) = xx3

I got this far:

     xx3
x    xx3
-------
     xx9
+   xxx0
   xxx00
--------
  11x2x9

√(1xx2x) = xx3
√(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