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Non zero product (Posted on 2015-10-07) Difficulty: 3 of 5
Find the total number of 8-digit positive integers (each containing no zeros) such that the product of four leftmost digits is equal to 6 times the product of the four rightmost digits.

No Solution Yet Submitted by K Sengupta    
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Solution computer assisted calculation Comment 1 of 1
There are 225 possible products of four non-zero digits. They are shown here with the ways the digits can be arranged to produce the given product; for example, the 10 ways of producing 4 are the four positions the 4 can have among three 1's plus the 10 ways of permuting two 2's and two 1's.

   1   1
   2   4
   3   4
   4  10
   5   4
   6  16
   7   4
   8  20
   9  10
  10  12
  12  36
  14  12
  15  12
  16  31
  18  36
  20  24
  21  12
  24  64
  25   6
  27  16
  28  24
  30  36
  32  40
  35  12
  36  72
  40  40
  42  36
  45  24
  48  88
  49   6
  50  12
  54  52
  56  40
  60  60
  63  24
  64  44
  70  24
  72 112
  75  12
  80  48
  81  19
  84  60
  90  60
  96  96
  98  12
 100  18
 105  24
 108  88
 112  48
 120  84
 125   4
 126  60
 128  40
 135  28
 140  36
 144 132
 147  12
 150  24
 160  48
 162  52
 168  84
 175  12
 180  84
 189  28
 192  88
 196  18
 200  24
 210  48
 216 116
 224  48
 225  18
 240  84
 243  16
 245  12
 250   4
 252  84
 256  31
 270  60
 280  48
 288 120
 294  24
 300  24
 315  36
 320  40
 324  72
 336  84
 343   4
 350  12
 360  96
 375   4
 378  60
 384  64
 392  24
 400  18
 405  24
 420  48
 432 112
 441  18
 448  40
 450  24
 480  60
 486  36
 490  12
 500   4
 504  96
 512  20
 525  12
 540  60
 560  36
 567  24
 576  88
 588  24
 600  24
 625   1
 630  48
 640  24
 648  76
 672  60
 675  12
 686   4
 700  12
 720  72
 729  10
 735  12
 750   4
 756  60
 768  36
 784  18
 800  12
 810  36
 840  48
 864  76
 875   4
 882  24
 896  24
 900  18
 945  24
 960  36
 972  36
 980  12
1000   4
1008  72
1024  10
1029   4
1050  12
1080  52
1120  24
1125   4
1134  36
1152  48
1176  24
1200  12
1215  12
1225   6
1260  36
1280  12
1296  55
1323  12
1344  36
1350  12
1372   4
1400  12
1440  36
1458  16
1470  12
1512  52
1536  16
1568  12
1575  12
1600   6
1620  24
1680  24
1701  12
1715   4
1728  40
1764  18
1792  12
1800  12
1890  24
1920  12
1944  28
1960  12
2016  36
2025   6
2048   4
2058   4
2160  24
2187   4
2205  12
2240  12
2268  24
2304  18
2352  12
2401   1
2430  12
2520  24
2560   4
2592  24
2646  12
2688  12
2744   4
2835  12
2880  12
2916  10
3024  24
3072   4
3087   4
3136   6
3240  12
3402  12
3456  12
3528  12
3584   4
3645   4
3888  12
3969   6
4032  12
4096   1
4374   4
4536  12
4608   4
5103   4
5184   6
5832   4
6561   1

Then, the ways of each is multiplied by the ways of 6 times that value, if that exists:

1*16 + 4*36 +  4*36  + 10*64   + ...

The grand total is 225896.


DefDbl A-Z
Dim crlf$, nlzero(6561)


Private Sub Form_Load()
 Form1.Visible = True
 
 Text1.Text = ""
 crlf = Chr$(13) + Chr$(10)

 For a = 1 To 9
 For b = 1 To 9
 For c = 1 To 9
 For d = 1 To 9
   t = a * b * c * d
   nlzero(t) = nlzero(t) + 1
   If nlzero(t) = 1 Then dct = dct + 1
 Next
 Next
 Next
 Next
 
 
 
 Text1.Text = dct & crlf
 
 
 For i = 0 To 6561
  If nlzero(i) > 0 Then
   Text1.Text = Text1.Text & mform(i, "###0")
   Text1.Text = Text1.Text & mform(nlzero(i), "###0") & crlf
  End If
 Next
 Text1.Text = Text1.Text & crlf

 For low = 1 To 6561 / 6
   totnumber = totnumber + nlzero(low) * nlzero(6 * low)
 Next
 
 Text1.Text = Text1.Text & totnumber & crlf & " done"
  
End Sub

Function mform$(x, t$)
  a$ = Format$(x, t$)
  If Len(a$) < Len(t$) Then a$ = Space$(Len(t$) - Len(a$)) & a$
  mform$ = a$
End Function


  Posted by Charlie on 2015-10-07 20:02:02
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