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#1 2007-07-24 00:42:54

bobby_05
Member
Registered: 2007-07-24
Posts: 1

9 Flips. Interesting but tough.

9 Flips

abcdefghijkl
X               9
----------------
lkjihgfedcba

Suppose that N is a positive number written base 10, and that 9xN has the same digits as N but is in a reversed order. Then we shall say for short that N is a 9-Flip.

Find all 9-flips with 12 digits
Is it possible to say exactly how many 9-flips there are with precisely n digits?




I've managed to find a few by trial and error.
108910891089, 109999999989, 108900001089

But still cant reason out why.

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#2 2007-07-25 08:00:24

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

Re: 9 Flips. Interesting but tough.

Here's another.  109989109989 * 9 = 989901989901

...and another... 109890010989 * 9 = 989010098901

Last edited by John E. Franklin (2007-07-26 03:20:52)


igloo myrtilles fourmis

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#3 2007-07-25 08:07:52

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: 9 Flips. Interesting but tough.

Perhaps an easier solution will come when you write it as 9 additions rather than one multiplication.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#4 2007-07-25 09:50:53

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: 9 Flips. Interesting but tough.

I'd guess that patterns would be easier to spot if we used less digits.
I haven't got very far yet, but I've found that every 9-flip (regardless of its amount of digits) must be of the form 10???89.


Why did the vector cross the road?
It wanted to be normal.

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#5 2007-07-25 12:01:44

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: 9 Flips. Interesting but tough.

Let's make sure we're posting reasons as well as "theorems".  For sake of reading, put your conclusions in bold.

abcdefghijkl
               9
--------------
lkjihgfedcba

If a > 1, then the result would end up in a 13 digit number.  Thus, a = 0 or a = 1.  Obviously, if a = 0, then the first number is actually an 11 digit number.  Hence, a = 1.

Since a = 1, from the very right hand column it must be that l * 9 = 1 (mod 10).  The only solution is therefore l = 9.

If b > 1, then there will be a carry to the very left column.  This isn't possible since a = 1 and l = 9, so b <= 1.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#6 2007-07-25 20:41:23

mathsyperson
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Registered: 2005-06-22
Posts: 4,900

Re: 9 Flips. Interesting but tough.

Assume that b is 1. Then 9N >= 990000000000.
Therefore, k = l = 9.

However, 99x9 = 891, which means that the last two digits of 9N are 91. However, we've already stated that the first two digits of N are 11. Contradiction, ∴, b=0.

We now have the first two digits of N, which means we have the last two digits of 9N.
Trying all values of k, only one satisfies kl x 9 = ?01. ∴, k=8.


Why did the vector cross the road?
It wanted to be normal.

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#7 2007-07-25 22:33:41

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: 9 Flips. Interesting but tough.

Let’s continue. We now have N = 10cdefghij89

In order for the second digit of 9N to be 8, we need 9×c = 72 or 81. Thus c = 8 or 9. Thus j = 0 or 9.

N = 108defghi089 or 109defghi989.

Last edited by JaneFairfax (2007-07-26 00:16:09)

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#8 2007-07-25 23:36:30

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: 9 Flips. Interesting but tough.

Well …

If  c = 8, the possibilities are

1080efgh0089
1081efgh9089
1082efgh8089
…
1088efgh2089
1089efgh1089

Now we try

9 × 1080 = 9720
9 × 1081 = 9729
…
9 × 1088 = 9792
9 × 1098 = 9801

None of these will give a 9-flip, even allowing for carry-ons. For example, for the 1081 case, we need 9×that to start with 9809; there’s no way this can be done. As for the 1088 case, 9×that must start with 9802; the largest carry-on possible is 8 and 9792 + 8 = 9801 – not good enough. The only possibility is 9×1089 = 9801.

(All right, here’s a more detailed explanation. Suppose the first four digits of N are 1088. Then the last four digits of N must be 2089, so that the last four digits of 9N would be 8801. But then 9 × 1088 = 9792, so the first four digits of 9N can only be 9792, 9793, … 9801. Yet we need the first four digits to be 9802 – and it can’t be done. So N = 1088efgh2089 is not possible.)

∴ d = 9, wherefore i = 1: N = 1089efgh1089

Similary, if c = 9, repeating the above argument gives N = 1098efgh0989 or 1099efgh9989

So N has been narrowed down to
N = 1089efgh1089, 1098efgh0989 or 1099efgh9989.

Last edited by JaneFairfax (2007-07-26 00:56:54)

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#9 2007-07-26 00:20:49

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: 9 Flips. Interesting but tough.

I have extensively edited my two posts above because I keep finding flaws in my reasoning. Argh.gif

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#10 2007-07-26 00:48:02

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: 9 Flips. Interesting but tough.

I’ve found one that hasn’t yet been found in this thread: 109890010989.

9 × 109890010989 = 989010098901.

This is the only 9-flip beginning with 1098…

So the other 9-flips are of the form 1089efgh1089 or 1099efgh9989. I have to stop now and go and do my shopping. tongue

Last edited by JaneFairfax (2007-07-26 00:58:05)

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#11 2007-07-26 15:57:05

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

Re: 9 Flips. Interesting but tough.

These are the only five answers
that start with 1 and end in 9, as
computed with a C language computer program.
I jumped by tens, starting with 100000000009,
and ending with 111111111119 when my program
overflows the 12 digits into 13.


Beginning of Five Answers.

108900001089
980100009801

108910891089
980198019801

109890010989
989010098901

109989109989
989901989901

109999999989
989999999901

End of Five Answers.


igloo myrtilles fourmis

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#12 2007-07-27 00:06:49

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: 9 Flips. Interesting but tough.

Very interesting! From looking at that, it seems like the method for producing 9-flips goes something like this:

- Stop after any stage you like. Even here, you could stop and produce a valid 9-flip of 0.
- Put 10 at the start and 89 at the end.
- *Optionally place an amount of 9's after the 10 and the same amount before the 89. (If you want to finish the 9-flip, one in the middle is also allowed)
- Put 89 after the left 9's and 10 before the right 9's.
- Optionally place an amount of 0's after the 89 and the same amount before the 10. (If you want to finish, one is also allowed)
- Put 10 after the left 0's and 89 before the right 0's.
- Go back to the *stage and repeat until bored.

I have no proof that that works, but I'm pretty sure it does.


Why did the vector cross the road?
It wanted to be normal.

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#13 2007-07-27 04:30:11

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

Re: 9 Flips. Interesting but tough.

Mathsy, what do you mean "one is also allowed" ? 
"or if you want to finish, one is also allowed."
smile  Oh, I know, one for the amount of times you place a 9 or a 0 before and after.
One or More,  I think I read what you wrote now.


igloo myrtilles fourmis

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#14 2007-07-27 05:32:42

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: 9 Flips. Interesting but tough.

Yeah, I didn't explain that very well. Basically, 9-flips in construction are 2 separate numbers, and you just stick them together when you want to finish.

I'll try to make one to show you how it works.

First we need to start off with the 10 and 89. We have 10...89.
Now we put in some 9's, if we want. I'll put two on each end. 1099...9989.
Now we put a 89 and a 10 in there. 109989...109989.
Next come the 0's. I don't want to put any there though, so I'll skip this stage.
Now the process starts again, so we put another 10 and 89 in there. 10998910...89109989.
Next we put some more 9's. I want to finish here, so I have the option of using as many as I want to stick the two halves together. (Normally, because you put the same amount on each end, you would be restricted to an even amount overall) I'll use three.

So the final result is: 1099891099989109989.

Here are some more examples of 9-flips being constructed. Hopefully I've explained it well enough so you can follow.

10...89
109...989
10989...10989
109890010989
---

10...89
10...89
1089...1089
108900...001089
10890010...89001089
10890010989001089
---

10...89
1099989


Why did the vector cross the road?
It wanted to be normal.

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