Let an oddly even number be an even number that contains its greatest odd factor. For example, 1792=2^8*7 is an oddly even number because the greatest odd factor of 1792 is 7, which is in 1792. X and Y are both oddly even numbers. When you remove the first digit of X, you get Y. What are X and Y?
The first 63 oddly even numbers (all those below 2 million) are listed below together with their greatest odd factors:
16 1
128 1
384 3
512 1
1024 1
1536 3
1792 7
2176 17
2560 5
2912 91
3072 3
5120 5
7168 7
8192 1
9216 9
11264 11
13312 13
15360 15
15616 61
16384 1
17408 17
19456 19
21504 21
23552 23
25600 25
27648 27
28672 7
29696 29
31744 31
33792 33
35840 35
37376 73
37888 37
39936 39
41984 41
43392 339
57344 7
66560 65
90112 11
98304 3
131072 1
161792 79
229376 7
262144 1
270336 33
290912 9091
294912 9
393216 3
397312 97
443392 433
458752 7
491520 15
497152 971
589824 9
655360 5
786432 3
917504 7
1048576 1
1179648 9
1185792 579
1323008 323
1474560 45
1955840 955
The two that fit the puzzle's description of X and Y are:
443392 and 43392
from:
DefDbl A-Z
Dim crlf$
Dim fct(20, 1), oddlyEven$(1000)
Private Sub Form_Load()
Form1.Visible = True
Text1.Text = ""
crlf = Chr$(13) + Chr$(10)
For n = 2 To 2000000 Step 2
gof = n
While gof Mod 2 = 0
gof = gof / 2
Wend
g$ = LTrim(Str(gof))
ns$ = LTrim(Str(n))
If InStr(ns$, g) > 0 Then
Text1.Text = Text1.Text & Str(ns) & Str(g) & crlf
ct = ct + 1
oddlyEven(ct) = ns
End If
DoEvents
Next
Text1.Text = Text1.Text & crlf
For i = 1 To ct
For j = 1 To i - 1
If oddlyEven(j) = Mid(oddlyEven(i), 2) Then
Text1.Text = Text1.Text & oddlyEven(i) & " " & oddlyEven(j) & crlf
End If
Next
Next
Text1.Text = Text1.Text & crlf & ct & " done"
End Sub
Edited on October 1, 2017, 3:50 pm
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Posted by Charlie
on 2017-10-01 15:48:18 |
computer solution