Donate to Remove ads

Got a credit card? use our Credit Card & Finance Calculators

Thanks to kiloran,88V8,Ravomas,SalvorHardin,Blagdon, for Donating to support the site

Consecutive

cinelli
Lemon Slice
Posts: 596
Joined: November 9th, 2016, 11:33 am
Has thanked: 268 times
Been thanked: 176 times

Consecutive

#682349

Postby cinelli » September 3rd, 2024, 11:03 am

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

Cinelli

UncleEbenezer
The full Lemon
Posts: 11373
Joined: November 4th, 2016, 8:17 pm
Has thanked: 1548 times
Been thanked: 3165 times

Re: Consecutive

#682351

Postby UncleEbenezer » September 3rd, 2024, 11:11 am

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?

UncleEbenezer
The full Lemon
Posts: 11373
Joined: November 4th, 2016, 8:17 pm
Has thanked: 1548 times
Been thanked: 3165 times

Re: Consecutive

#682357

Postby UncleEbenezer » September 3rd, 2024, 12:36 pm

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

jfgw
Lemon Quarter
Posts: 2633
Joined: November 4th, 2016, 3:36 pm
Has thanked: 1157 times
Been thanked: 1203 times

Re: Consecutive

#682400

Postby jfgw » September 3rd, 2024, 5:42 pm

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.

jfgw
Lemon Quarter
Posts: 2633
Joined: November 4th, 2016, 3:36 pm
Has thanked: 1157 times
Been thanked: 1203 times

Re: Consecutive

#682448

Postby jfgw » September 4th, 2024, 7:42 am

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.

9873210
Lemon Quarter
Posts: 1048
Joined: December 9th, 2016, 6:44 am
Has thanked: 242 times
Been thanked: 325 times

Re: Consecutive

#682493

Postby 9873210 » September 4th, 2024, 1:11 pm

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.

cinelli
Lemon Slice
Posts: 596
Joined: November 9th, 2016, 11:33 am
Has thanked: 268 times
Been thanked: 176 times

Re: Consecutive

#682510

Postby cinelli » September 4th, 2024, 2:25 pm

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

Harry23
2 Lemon pips
Posts: 142
Joined: February 18th, 2023, 2:31 pm
Has thanked: 308 times
Been thanked: 52 times

Re: Consecutive

#682530

Postby Harry23 » September 4th, 2024, 4:45 pm

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?

GoSeigen
Lemon Quarter
Posts: 4672
Joined: November 8th, 2016, 11:14 pm
Has thanked: 1715 times
Been thanked: 1715 times

Re: Consecutive

#682657

Postby GoSeigen » September 5th, 2024, 9:36 am

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

jfgw
Lemon Quarter
Posts: 2633
Joined: November 4th, 2016, 3:36 pm
Has thanked: 1157 times
Been thanked: 1203 times

Re: Consecutive

#682669

Postby jfgw » September 5th, 2024, 10:55 am

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 :D


Julian F. G. W.

cinelli
Lemon Slice
Posts: 596
Joined: November 9th, 2016, 11:33 am
Has thanked: 268 times
Been thanked: 176 times

Re: Consecutive

#682670

Postby cinelli » September 5th, 2024, 11:13 am

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

UncleEbenezer
The full Lemon
Posts: 11373
Joined: November 4th, 2016, 8:17 pm
Has thanked: 1548 times
Been thanked: 3165 times

Re: Consecutive

#682691

Postby UncleEbenezer » September 5th, 2024, 1:17 pm

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

cinelli
Lemon Slice
Posts: 596
Joined: November 9th, 2016, 11:33 am
Has thanked: 268 times
Been thanked: 176 times

Re: Consecutive

#682721

Postby cinelli » September 5th, 2024, 3:58 pm

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
Posts: 596
Joined: November 9th, 2016, 11:33 am
Has thanked: 268 times
Been thanked: 176 times

Re: Consecutive

#682775

Postby cinelli » September 5th, 2024, 8:15 pm

cinelli wrote:Only 6 occurrences out of 23.

Cinelli

D'oh! Can't count. That should be just 3 out of 23.

Cinelli

Harry23
2 Lemon pips
Posts: 142
Joined: February 18th, 2023, 2:31 pm
Has thanked: 308 times
Been thanked: 52 times

Re: Consecutive

#682807

Postby Harry23 » September 5th, 2024, 11:27 pm

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 non-negatives 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


Return to “Games, Puzzles and Riddles”

Who is online

Users browsing this forum: No registered users and 1 guest