Puzzle to Find a 10 Digit NumberDate: 10/10/2004 at 11:58:56 From: Ravi Subject: A Very Special Mystery Number I have to find a 10 digit number which uses each of the digits 0-9 and has ALL the following properties: 1) The first digit from the left(the billions digit) is divisible by 1. 2) The number formed by the first 2 digits from the left is divisible by 2. 3) The number formed by the first 3 digits from the left is divisible by 3. 4) The number formed by the first 4 digits from the left is divisible by 4. 5) The number formed by the first 5 digits from the left is divisible by 5. 6) The number formed by the first 6 digits from the left is divisible by 6. 7) The number formed by the first 7 digits from the left is divisible by 7. 8) The number formed by the first 8 didgits from the left is divisible by 8. 9) The number formed by the first 9 digits from the left is divisible by 9. 10) The number itself is divisible by 10. I figured out the answer, but it took me nearly two weeks. The first thing I found out was that the number ended in a 0 and that the fifth digit was a 5. I also knew that the second, fourth, and eighth digits were all even. From that I figured out that the first, third, and seventh digits would have to be odd. From there I just did guess and check. I was wondering if there is a better method than guess and check? Could you explain it to me step-by-step? Thanks a lot. Date: 10/16/2004 at 12:44:53 From: Doctor Ian Subject: Re: A Very Special Mystery Number Hi Ravi, >I have to find a 10 digit number which uses each of the digits 0-9 >and has ALL the folowing properties: > >1)The first digit from the left(the billions digit)is divisible by 1. This clue doesn't tell us anything at all, since _every_ digit will be divisible by 1. >2)The number formed by the first 2 digits from the left is divisible >by 2. > >3)The number formed by the first 3 digits fromt he left is divisible >by 3. > >4)The number formed by the first 4 digits from the left is divisible >by 4. > >5)The number formed by the first 5 digits from the left is divisible >by 5. This tells you that the 5th digit has to be a zero or a 5. So we have _ _ _ _ [05] _ _ _ _ _ That's not much, but it's a start. >6)The number formed by the first 6 digits from the left is divisible >by 6. > >7)The number formed by the first 7 digits from the left is divisible >by 7. > >8)The number formed by the first 8 didgits from the left is divisible >by 8. > >9)The number formed by the first 9 digits from the left is divisible >by 9. > >10)The number itself is divisible by 10. This tells you that the final digit has to be a 0, which means the 5th digit can't be zero. So we have _ _ _ _ 5 _ _ _ _ 0 Clue 2 tells us that the second digit is even, but we already know it can't be 0: _ [2468] _ _ 5 _ _ _ _ 0 Clue 4 tells us that the 4th digit is also even, and it can't be the same digit as digit 2. At this point, we have to start considering dependencies. That is, we make a tentative choice for the second digit, and it leads to tentative choices for the fourth digit: - 2 - 4 5 - - - - 0 - 2 - 6 5 - - - - 0 - 2 - 8 5 - - - - 0 - 4 - 2 5 - - - - 0 - 4 - 6 5 - - - - 0 - 4 - 8 5 - - - - 0 - 6 - 2 5 - - - - 0 - 6 - 4 5 - - - - 0 - 6 - 8 5 - - - - 0 - 8 - 2 5 - - - - 0 - 8 - 4 5 - - - - 0 - 8 - 6 5 - - - - 0 The 9th clue isn't much help, either: If we add up the digits other than zero, 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 the sum is divisible by 9. So _any_ arrangement of those digits will be divisible by 9. So let's recap: Clues Used: 2, 4, 5, 10 Useless: 1, 9 Unused: 3, 6, 7, 8 Let's look at clue 8. As before, the 8th digit has to be even, and it can't be any of the other digits. So we can expand our possibilities again: - 2 - 4 5 - - 6 - 0 - 2 - 4 5 - - 8 - 0 - 2 - 6 5 - - 4 - 0 - 2 - 6 5 - - 8 - 0 - 2 - 8 5 - - 4 - 0 - 2 - 8 5 - - 6 - 0 - 4 - 2 5 - - 6 - 0 - 4 - 2 5 - - 8 - 0 - 4 - 6 5 - - 2 - 0 - 4 - 6 5 - - 8 - 0 - 4 - 8 5 - - 2 - 0 - 4 - 8 5 - - 6 - 0 - 6 - 2 5 - - 4 - 0 - 6 - 2 5 - - 8 - 0 - 6 - 4 5 - - 2 - 0 - 6 - 4 5 - - 8 - 0 - 6 - 8 5 - - 2 - 0 - 6 - 8 5 - - 4 - 0 - 8 - 2 5 - - 4 - 0 - 8 - 2 5 - - 6 - 0 - 8 - 4 5 - - 2 - 0 - 8 - 4 5 - - 6 - 0 - 8 - 6 5 - - 2 - 0 - 8 - 6 5 - - 4 - 0 Note that in each case, there is just one even digit left, and that will have to go in the 6th place, since the first 6 digits are divisible by 6, which means they're divisible by 2, which means the 6th digit is even. So we can enumerate those possibilities: - 2 - 4 5 8 - 6 - 0 - 2 - 4 5 6 - 8 - 0 - 2 - 6 5 8 - 4 - 0 - 2 - 6 5 4 - 8 - 0 - 2 - 8 5 6 - 4 - 0 - 2 - 8 5 4 - 6 - 0 - 4 - 2 5 8 - 6 - 0 - 4 - 2 5 6 - 8 - 0 - 4 - 6 5 8 - 2 - 0 - 4 - 6 5 2 - 8 - 0 - 4 - 8 5 6 - 2 - 0 - 4 - 8 5 2 - 6 - 0 - 6 - 2 5 8 - 4 - 0 - 6 - 2 5 4 - 8 - 0 - 6 - 4 5 8 - 2 - 0 - 6 - 4 5 2 - 8 - 0 - 6 - 8 5 4 - 2 - 0 - 6 - 8 5 2 - 4 - 0 - 8 - 2 5 6 - 4 - 0 - 8 - 2 5 4 - 6 - 0 - 8 - 4 5 6 - 2 - 0 - 8 - 4 5 2 - 6 - 0 - 8 - 6 5 4 - 2 - 0 - 8 - 6 5 2 - 4 - 0 We can use clues 3 and 6 together at this point. The sum of the first 3 digits, and the sum of the first 6 digits, have to be divisible by 3. So that means the first 3 digits have to add up to something divisible by 3; and so do the first 6 digits. But this means that the _second_ 3 digits have to add up to something divisible by 3 as well! (Do you see why?) So we can rule out any possibilities where this isn't the case, e.g., - 2 - 4 5 8 - 6 - 0 4 + 5 + 8 = 17 No. So we can run through our table and get rid of the entries where the second 3 digits don't add up to a multiple of 3: - 2 - 4 5 8 - 6 - 0 4 + 5 + 8 = 17 No. - 2 - 4 5 6 - 8 - 0 4 + 5 + 6 = 15 Yes. - 2 - 6 5 8 - 4 - 0 6 + 5 + 8 = 19 No. - 2 - 6 5 4 - 8 - 0 6 + 5 + 4 = 15 Yes. - 2 - 8 5 6 - 4 - 0 8 + 5 + 6 = 19 No. - 2 - 8 5 4 - 6 - 0 8 + 5 + 4 = 17 No. - 4 - 2 5 8 - 6 - 0 2 + 5 + 8 = 15 Yes. - 4 - 2 5 6 - 8 - 0 2 + 5 + 6 = 13 No. - 4 - 6 5 8 - 2 - 0 - 4 - 6 5 2 - 8 - 0 and - 4 - 8 5 6 - 2 - 0 so - 4 - 8 5 2 - 6 - 0 on - 6 - 2 5 8 - 4 - 0 - 6 - 2 5 4 - 8 - 0 - 6 - 4 5 8 - 2 - 0 - 6 - 4 5 2 - 8 - 0 - 6 - 8 5 4 - 2 - 0 - 6 - 8 5 2 - 4 - 0 - 8 - 2 5 6 - 4 - 0 - 8 - 2 5 4 - 6 - 0 - 8 - 4 5 6 - 2 - 0 - 8 - 4 5 2 - 6 - 0 - 8 - 6 5 4 - 2 - 0 - 8 - 6 5 2 - 4 - 0 What next? Well, we can use clue 8 again. Why? Because 1000 is divisible by 8; so if a number is divisible by 8, the final three digits must be divisible by 8. For example, if we know that - 2 - 4 5 6 - 8 - 0 is divisible by 8, then 6 - 8 must be divisible by 8. In this case, 600 is divisible by 8, so - 8 has to be divisible by 8, and the '-' will have to be an odd digit. There are just a few cases, so we can check them: 18, 38, 58, 78, 98 n n n n n None of them are divisible by 8, so we can remove this entry from our table. What about something like this? - 4 - 2 5 8 - 6 - 0 Again, 800 is divisible by 8, so we need to check 16, 36, 56, 76, 96 y n y n y Three of these are divisible by 8, so this expands to - 4 - 2 5 8 1 6 - 0 - 4 - 2 5 8 5 6 - 0 - 4 - 2 5 8 9 6 - 0 If you do that for the remaining possibilities, you'll end up with only three digits left to be deduced. Actually, just two, since _any_ digit can go in the 9th place and satisfy clue 9. Now, that still leaves you with some guessing and checking to do; but not an overwhelming amount. Does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ Date: 12/24/2011 at 19:30:11 From: Dries Subject: Re: A Very Special Mystery Number I, too, had been thinking about this puzzle for quite some time. My brother decided to write a little program to find the solution. Since there might be other people looking for ways to solve this puzzle, it could be a good idea to provide the code to everybody out there. Credit goes to Brecht Vermeulen. Paste the following in visual studio.net and just let it run: Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click Dim getallen As ArrayList = New ArrayList For teller As Double = 1 To 9 getallen.Add(teller) Next For lengte As Integer = 1 To 9 Dim tempGetallen As ArrayList = New ArrayList(getallen) getallen = New ArrayList For aantal As Double = 0 To tempGetallen.Count - 1 Dim getal As Double = tempGetallen(aantal) For teller As Integer = 0 To 9 Dim tempString As String = System.Convert.ToString(getal) & teller If (System.Convert.ToDouble(tempString) Mod (lengte + 1) = 0) Then getallen.Add(System.Convert.ToDouble(tempString)) End If Next Next Next For teller As Integer = 1 To 9 Dim finalresult As ArrayList = New ArrayList(getallen) getallen = New ArrayList For i As Integer = 0 To finalresult.Count - 1 Dim getal As Double = finalresult(i) If System.Convert.ToString(getal).Contains(10 - teller) Then getallen.Add(getal) End If Next Next MsgBox("het getal is " & getallen(0)) End Sub |
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