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Consecutive

 Lemon Slice
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Consecutive
This puzzle is to find all occurrences of two consecutive numbers below 1000 which produce a palindromic product.
Cinelli
Cinelli

 The full Lemon
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Re: Consecutive
cinelli wrote:This puzzle is to find all occurrences of two consecutive numbers below 1000 which produce a palindromic product.
Cinelli
Rules clarification please.
Does 10x11 = 0110 qualify?
Or indeed the trivial 1x2 = 2?

 The full Lemon
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Re: Consecutive
In the absence of clarification, let's take it as anything goes. By brute force, having noted merely that the product (mod 10) of any two consecutive numbers ends in 2, 6 or 0 to narrow the problem space.
1 x 2 = 2
2 x 3 = 6
4 x 5 = 020
5 x 6 = 030
9 x 10 = 090
10 x 11 = 0110
16 x 17 = 272
74 x 75 = 05550
77 x 78 = 6006
110 x 111 = 012210
264 x 265 = 069960
504 x 505 = 0254520
538 x 539 = 289982
605 x 606 = 0366630
919 x 920 = 0845480
1 x 2 = 2
2 x 3 = 6
4 x 5 = 020
5 x 6 = 030
9 x 10 = 090
10 x 11 = 0110
16 x 17 = 272
74 x 75 = 05550
77 x 78 = 6006
110 x 111 = 012210
264 x 265 = 069960
504 x 505 = 0254520
538 x 539 = 289982
605 x 606 = 0366630
919 x 920 = 0845480

 Lemon Quarter
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Re: Consecutive
If leading zeros are allowed, you could also have:
24 x 25 = 00600
99 x 100 = 009900
100 x 101 = 0010100
999 x 1000 = 000999000
Since all negative numbers are below 1000, There are an infinite number of solutions. There will be at least two per power of 10, e.g.,
9999 x 10000 = 000099990000
10000 x 10001 = 0000100010000
Even if leading zeros are not allowed, I suspect that the number of solutions is infinite if you allow negative numbers.
Julian F. G. W.
24 x 25 = 00600
99 x 100 = 009900
100 x 101 = 0010100
999 x 1000 = 000999000
Since all negative numbers are below 1000, There are an infinite number of solutions. There will be at least two per power of 10, e.g.,
9999 x 10000 = 000099990000
10000 x 10001 = 0000100010000
Even if leading zeros are not allowed, I suspect that the number of solutions is infinite if you allow negative numbers.
Julian F. G. W.

 Lemon Quarter
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Re: Consecutive
One of my answers is not valid as one of the numbers is not below 1000:
999 x 1000 = 000999000.
However, I have thought of four more:
0 x 1 = 0
1 x 0 = 0
0 x 1i = 0
1i x 0 = 0
I would not think that a negative product would be palindromic due to the minus sign so there would be no other solution involving imaginary numbers.
I trust that base 10 should be assumed in the absence of controrary information.
Infinity is a concept, not a number, although the product would be symmetrical as well as palindromic.
Julian F. G. W.
999 x 1000 = 000999000.
However, I have thought of four more:
0 x 1 = 0
1 x 0 = 0
0 x 1i = 0
1i x 0 = 0
I would not think that a negative product would be palindromic due to the minus sign so there would be no other solution involving imaginary numbers.
I trust that base 10 should be assumed in the absence of controrary information.
Infinity is a concept, not a number, although the product would be symmetrical as well as palindromic.
Julian F. G. W.

 Lemon Quarter
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Re: Consecutive
jfgw wrote:One of my answers is not valid as one of the numbers is not below 1000:
999 x 1000 = 000999000.
However, I have thought of four more:
0 x 1 = 0
1 x 0 = 0
0 x 1i = 0
1i x 0 = 0
I would not think that a negative product would be palindromic due to the minus sign so there would be no other solution involving imaginary numbers.
I trust that base 10 should be assumed in the absence of controrary information.
Infinity is a concept, not a number, although the product would be symmetrical as well as palindromic.
Julian F. G. W.
The complex numbers are unordered, that is to say there no ordering that has the analytic properties that are expected. Both i < 1000 and i > 1000 are nonsense (i = 1000 is at least false, which is better than nonsense). Under the axiom of choice, the complex numbers are well ordered, but you are not going to like that order as it will not have the ordinary analytic properties.
If you're going to expand the problem you might try the rationals or the reals, which at least have well defined subsets < 1000.

 Lemon Slice
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Re: Consecutive
UncleEbenezer wrote:Rules clarification please.
Does 10x11 = 0110 qualify?
Or indeed the trivial 1x2 = 2?
0110, etc do not qualify. 2 and 6 do, trivially. Negative numbers do not  you can't just ignore the minus sign.
Cinelli

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Re: Consecutive
9873210 wrote: If you're going to expand the problem you might try the rationals or the reals, which at least have well defined subsets < 1000.
Ok, so what counts as a consecutive number for those two sets, if there's no limit to the number of decimal places?

 Lemon Quarter
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Re: Consecutive
It may be worth a reminder that this is the puzzles board not the maths board! i.e. a puzzle posted here will likely constitute an actual puzzle with associated fairly obvious assumptions, not a thorough investigation of the properties of numbers!
GS
GS

 Lemon Quarter
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Re: Consecutive
GoSeigen wrote:It may be worth a reminder that this is the puzzles board not the maths board! i.e. a puzzle posted here will likely constitute an actual puzzle with associated fairly obvious assumptions, not a thorough investigation of the properties of numbers!
Hey, we're having fun here
Julian F. G. W.

 Lemon Slice
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Re: Consecutive
jfgw wrote:Even if leading zeros are not allowed, I suspect that the number of solutions is infinite if you allow negative numbers.
Julian F. G. W.
Have you given any thought to a proof of this? I would have thought that as numbers get larger and larger, the chance of hitting upon a palindrome gets smaller and smaller. I wonder what the next one past 289982 is.
Cinelli

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Re: Consecutive
cinelli wrote:I wonder what the next one past 289982 is.
Cinelli
2629262.
Extending it a bit (told you I used brute force ):
1 x 2 = 2
2 x 3 = 6
16 x 17 = 272
77 x 78 = 6006
538 x 539 = 289982
1621 x 1622 = 2629262
2457 x 2458 = 6039306
5291 x 5292 = 27999972
5313 x 5314 = 28233282
52008 x 52009 = 2704884072
142401 x 142402 = 20278187202
143498 x 143499 = 20591819502
1610151 x 1610152 = 2592587852952
1713543 x 1713544 = 2936231326392
4670028 x 4670029 = 21809166190812
5218983 x 5218984 = 27237788773272
15137566 x 15137567 = 229145919541922
15282411 x 15282412 = 233552101255332
15814148 x 15814149 = 250087292780052
47370058 x 47370059 = 2243922442293422
50702751 x 50702752 = 2570769009670752
142594226 x 142594227 = 20333113431133302
166691108 x 166691109 = 27785925652958772

 Lemon Slice
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Re: Consecutive
That is a very impressive list from UncleEbenezer. It is interesting how those ending in 6 peter out quite early. Only 6 occurrences out of 23.
Cinelli
Cinelli

 Lemon Slice
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Re: Consecutive
cinelli wrote:Only 6 occurrences out of 23.
Cinelli
D'oh! Can't count. That should be just 3 out of 23.
Cinelli

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Re: Consecutive
cinelli wrote: This puzzle is to find all occurrences of two consecutive numbers below 1000 which produce a palindromic product.
Cinelli
The puzzle doesn't specify which base is used for the numbers, although I expect you have an unwritten assumption of base 10. But wearing my pedant's hat for a moment and trying it out in binary integers, suppressing leading zeros, it appears to offer an easy solution as all the answers are even. Even numbers in binary all end in zero so there's only one solution:
0 x 1 = 0
That works for the nonnegatives anyway, when it gets to representing negative binary numbers I'm a bit out of my depth. But on the basis that two negatives multiplied make a positive, it looks like there's one more solution:
1 x 0 = 0
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