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Hi Nasir;

I got exactly the same as you did (including the two given questions), and, like you, didn't include numbers with repeated digits (in the spirit of the puzzle).

I initially considered omitting questions whose incorrect copy was also one of the original questions - which would reduce the number of questions (including the two given questions) to 14 - but decided against that because Hungry Horace wouldn't have noticed the 'repeats' as he hadn't recorded the originals.

Here's a link to the puzzle: Sides Reversed Is Puzzle

Hi Bob;

Thanks, that works perfectly well...and nice proof!

The formula solves AB for a range of BMs (can be non-integer) without using the 3:4:5 triangle rule. LK is always 3.

Hi Bob;

bob bundy wrote:

I'm confused about this:

I drew parallelogram LKCM (a fixed-dimension shape).

Sorry, but for brevity's sake I omitted some steps and comments...which are:

1. I drew LM, which from the 3:4:5 triangle rule on ΔLKM = 5.

2. Then I drew KC = 5 (from being parallel to LM, and KL being parallel to BM).

3. CM = 3 (from the 3:4:5 triangle rule on ΔCMK).

4. As the opposite sides of LKCM are parallel, LKCM is a parallelogram.

5. Also, LKCM's 4 angles and side lengths are constant because of the right angles at K and M, and so LKCM retains its 'fixed-dimension shape' while it rotates around its hinge point L...a critical feature enabling me to draw the conclusion I did about AB and BC changing in length by equal amounts.

I'll have to look at your proof tomorrow coz I'm off to bed now, but thanks for your work!

phrontister wrote:

Luckily for me, BM is 5, or I'd be well and truly stuck!

I wondered if I could find something...and I think I might've.

Using Geogebra, I drew parallelogram LKCM (a fixed-dimension shape).

Rotating LKCM around hinge point L changed AB and BC by equal amounts, and I spotted the following formula that finds AB for a range of BMs:

AB = KL + BM

This formula is simply from observation, and I don't know any other means of arriving at it.

Here are images of solutions for 3 sample BMs:

And here is a Geogebra animation of solutions for the range BM = 5.9 to 3.9 (accuracy to 1 decimal place).

Hi Bob;

The Escher Waterfall drawing (see image) uses certain visual techniques to show water flowing uphill from the wheel to the upper aqueduct...which water can't do.

Likewise, the Wiki drawing uses incorrect visual sizes of A's intersection angles and BC's intersection lengths, and shows AOD as a crooked line...which it can't be.

Conclusion: Escher's and Wiki's drawings are illusions, each representing something that is untrue as being true.

Conversely, my drawing in post #12 shows that LC is actually a straight line, with M lying on it.

I've renotated post #6's image to try and show more clearly how my solution works, including that M must lie on LC.

Luckily for me, BM is 5, or I'd be well and truly stuck!

I wonder what maths level this puzzle is aimed at.

Hi Bob;

Wow...that's quite some proof!

However, the trig's too advanced for me to understand it...but I'm happy to wait for a simplified solve if you can find one.

I'm glad, though, that your proof confirms that my solution is correct.

bob bundy wrote:

The properties like M on LC, DM = 4 and AKM a straight line, then follow.

Btw, I don't think that AKM is a straight line (see my Geogebra image below).

Hi ninjaman & Bob;

There's a similar thread here that might help: Measure length without bend?

And here are some explanatory drawings:

Hope that helps!

Hi Bob;

The only dimensions given by the OP are those of 3:4:5 triangles, and therefore I strongly suspected that the solution would involve such triangles. (barring red herrings, that is!)

So I drew the LABC rectangle, which led to the image below (Geogebra) and my post #2 solution.

That's probably not a proper proof as such, but I've failed to find one of those things, though I've spent some time trying.

And I can't find any other solution...

Hi Bob;

Hi Bob;

I received my Excel file ok, thx.

Only had time to have a quick look at it, and all I did was check for duplications in the columns by using the UNIQUE function that's part of the new Excel Dynamic Arrays family.

Example: Entering '=COUNTA(UNIQUE(B2:B47))' in B48 returns '46' there, confirming that your 46 column B entries are unique. Similarly with Column G.

I'll have to take a closer look at earlier posts to try and understand this thread before examining further. Previously I've only skimmed through the posts.

Might take me a bit of time, though!

Hi tony123;

Hi Bob,

bob bundy (post #12) wrote:

I put the two lists side by side and checked carefully. Definitely no repeats from my program and all possibilities are 'legit'...

Anyway here they are:

1. QR RB RB RB 2. QR RB RB WR extra

3. QR RB RB TE 4. QR RB RB PK

5. QR RB RB TM 6. QR RB RB WR extra

Aren't 2 & 6 the same? Or are my cataracts playing up?

I thought it wouldn't hurt to have someone else check carefully, too.

Humble apologies. But I refer you to the third line of my signature

I'm onto you there!

I used Mitch's spreadsheet layout from post #4.

Couldn't get the right answer with MIF's calculator. Tried quite a few options that made sense to me, but not to the calc!

I got a bit further with Mathematica, but need Bobby to tidy up some things for me cos I just don't know enough!

Hi Mitch,

I've been casually following this thread and so haven't understood all that much.

However, I thought I'd check some of your workings and Bob's, and found what may be an error in post #9:

UCanCallMeMitch (post #9) wrote:

QB - 1

RB - 2 (Max of 4)

WR - 2 (Max of 4)

TE - 1(Max of 2)

PK -1

TM - 1

Flex Players - 2 (Either RBs, WRs, or TEs); there are six combinations:1 QB, 2 RBs, 2 WRs, 3 TEs, 1 PK, 1 TM

1 QB, 2 RBs, 3 WRs, 2 TEs, 1 PK, 1 TM

1 QB, 2 RBs, 4 WRs, 1 TE, 1 PK, 1 TM

1 QB, 3 RBs, 2 WRs, 2 TEs, 1 PK, 1 TM

1 QB, 3 RBs, 3 WRs, 1, TE, 1 PK, 1 TM

1 QB, 4 RBs, 2 WRs, 1 TE, 1 PK, 1 TM

Isn't the first combo invalid, as it exceeds the max number of TEs? If so, that would reduce the number of valid combinations to 5.

Or aren't Flex Players included in the Max count?

Hi Bob and Hannibal;

bob bundy wrote:

I used Microsoft Excel to do a few years' calculations.

That would be "a few months' calculations", Bob.

bob bundy wrote:

You should be able to see that the 'Principle' is gradually going down but not as fast as you thought.

I try not to drop my principles, but this $10,000 principal fairly plummeted! Test spotted, Bob!

Hannibal lecter (post #8) wrote:

balance = 10,000 + 0.5% = 10050

now 10050 - $500 = 9950

now add 9950 + 0.5% = 9597.75

after that again sub 9597.75 - $500

The algorithm is right, but there are two errors in the figures. They should read:

balance = $10,000 + 0.5% = $10050

now $10050 - $500 = $9550

now add $9550 + 0.5% = $9597.75

after that again sub $9597.75 - $500

Hannibal lecter (post #2) wrote:

Every month, $500 is withdrawn to meet college expenses.

The problem doesn't say at which end of the month the money is withdrawn, but the timing makes a difference to the amount of interest earned, as shown in my image below by comparing the column D and E figures.

This interest difference can affect the month in which a zero balance is reached: eg, for withdrawals of $501 instead of $500, the column D algorithm reaches a zero balance in the month prior to column E's zero balance.

Btw, Excel has a handy function that calculates the number of periods (months, in this case) for the bank balance to reach zero. It gives the same answers as Hannibal's and Bob's algorithm (the same as my algorithm B) and also my algorithm A...which is comforting to know.

Enter the function into one cell for algorithm A and another cell for algorithm B. The respective solution for each algorithm will be displayed in whole + part months.

Here's a link to a very helpful MIF page: Compound Interest.

Hi ganesh;

There's an error in your -8x+3x calc...

Hi;

I spent a bit of time trying find a solution that used the OP's requirements, but I failed and went back to what I had tried before with TEXTJOIN because I felt I'd got close then (sorry, Wood).

This time I think I succeeded (see image). The method returns column D results that are distinct, comma-delimited, and with each cell's contents sorted into descending order...as per post #1's requirements.

Some comments:

1. I used Excel's 'Sort Z to A' feature to sort each column D cell into descending order (don't know how to do it via formula).

2. I haven't met post #10's requirements (don't know how).

3. I populated column C via LARGE, with distinct values in descending order.

I'm sorry, Wood, but I know nearly nothing about arrays and {INDEX,MATCH,LARGE} formulas, and unfortunately I don't have enough time to research and learn this area.

I hope someone else here can help you (and maybe Bob knows how), but if not it may be best for you to try an Excel forum.

Hi Wood,

Could you please show us the best formulas you have tried.

Thanks.

Hi Bob & Wood,

Here's an image of my attempt, with formulas displayed in the lower section. I used some different values in column A than in your posts, just for testing purposes.

B1 contains the search value (5, in this case), which with my spreadsheet meant using an extra column (E). However, it enables changing the search value (if required) easily.

Column C returns all values <= the value in B1 ('5').

Column D gives the list of all unique Column C values (excluding blanks), in descending order.

The single cell E1 contains the comma-delimited output (in descending order) of the range D1:D4. Unfortunately I couldn't think of an automatic way to use TEXTJOIN that doesn't require having to select the cell range.

============================================================================================

Hi Bob,

With my version of Excel, 'CONCATENATE' joins numbers and returns the result as a text string, so there's no need for column C's conversion to text.

You may already know this: CONCATENATE has an alternative 'ampersand operator' option, used like so:

=C1&","&C2&","&C3&","&C4&","&C5

Hi Allerious,

The puzzle can be solved by logic, without any guessing.

Row 1 solves logically, as Bob said, and leads to multiple elimination of duplicates in the lower rows.

From there, eliminating the 2 from R3C2 ('R' = row, 'C' = column) leaves {3,4} in both R3C2 and R3C3, which enables us to eliminate the 3 from R3C1 and R3C4 because the 3 must be in either R3C2 or R3C3.

...etc, etc.

Hi Anthony,

A computer program I ran found that 1331 ⇒ 3544453 is the next smallest palindromic pairing of *n* ⇒ n(2n+1) after your 66 ⇒ 8778.

It gave the following list as the first 12 such pairings:

0 ⇒ 0

1 ⇒ 3

5 ⇒ 55

9 ⇒ 171

66 ⇒ 8778

1331 ⇒ 3544453

123321 ⇒ 30416261403

1332222331 ⇒ 3549632679762369453

13322222331 ⇒ 354963215686512369453

123322223321 ⇒ 30416741529792514761403

133222222331 ⇒ 35496321045754012369453

1233222223321 ⇒ 3041674104186814014761403

Hi Anthony & Jaspers;

I found these:

bob bundy wrote:

Just a bobbym search with no keyword specified gives me this:

HTTP 500 error

I get the same error if I'm logged in, but when logged out I get access to 192 pages of 'Topics with posts by bobbym'!