You are not logged in.
Aarrgh!!
Equally painful! I wonder what the 'correct' answer is.
835 solutions! Getting close to "infinitely many"!
That's the formula in post #14 by Relentless.
Congratz!
But how did you manage to win with an incorrect formula?
plus all digits by 4 times (-D1 + D2 - D3+ D4 - D5 + D6 - D7 + D8)..
its like this??
That is very different from what I did, and gives a very wrong answer.
My method uses several sets, because, although they're not exactly spelt out to the letter, I understood the rules to be these:
(a) only full sets can be used in the calculation;
(b) the numbers must retain their original sequence;
(c) no out-of-sequence numbers can be used (even though that number may be one that appears in the set);
(d) no new numbers (ie, numbers not in the set) can be introduced.
In your format, my formula looks like this:
4(D1+D2+D3+D4+D5+D6+D7+D8)-(D1-D2+D3-D4+D5-D6+D7-D8)
The second set looks different from that in post #12 because of the effect of parentheses on the minus signs.
I use the set more than once - which is not allowed - and so Relentless's method in post #14, in which the set appears only once, suits better.
if calculate one by one using normal calculator (not copy and paste)...it still get 210.
is there any other way?
This one, which is the one in my post, yields 162: 3+4+5+6+7+8+9+0+3+4+5+6+7+8+9+0+3+4+5+6+7+8+9+0+3+4+5+6+7+8+9+0-3+4-5+6-7+8-9+0
This one yields 210: 3+4+5+6+7+8+9+0+3+4+5+6+7+8+9+0+3+4+5+6+7+8+9+0+3+4+5+6+7+8+9+0+3+4+5+6+7+8+9+0
Did you use all "+" signs, as in the second one, and overlook the four "-" signs that are in the first one?
Hi Relentless;
Yes, I can see that.
Your previous post's formula works perfectly well.
I usually don't mind puzzles like this and have a reasonable success rate with them, but trying to find a solution can sometimes be as exasperating as determining the value of the next term in a series. You can either see it or not, and the greatest klutz can surprise.
I tend to call it a day after my brain has switched off and I've headbutted the keyboard...but then, after a good night's sleep, sometimes light dawns at dawn's light...or in the middle of the night.
I may try this puzzle once more, again.....or not.
Hi oranghijau;
Answer for number 3 is 210 if using that formula..
210 is incorrect, so I suppose you made an input error.
There are a number of easy ways to check the answers of the sums in my post that avoid input errors. Just copy the whole line before the "=" sign and paste it into a software or online calculator on your computer that accepts such input. Many do.
eg, Mathsisfun: here, or WolframAlpha here.
You can even use the address bar in Google Chrome and Microsoft Edge...just don't press Enter, otherwise you'll get an error message because of the large input size.
You'll find that the answers I posted are correct.
Hi guys;
I have a solution based on parts of posts #2 and #8, but it involves what I'd say is trickery.
I wonder if the minimum number of "4"s is 16. But I won't look any further.
Now it is more picturesque than before.
I used Excel's 'Conditional Formatting' on the whole 32 x 32 grid to automatically apply the different colours according to which number I entered into a cell. Very quick and easy. It helps me keep track visually of what I'm doing, and makes the end result easy to understand.
The colouring also helped confirm that the last two grids were error-free...which I hadn't done properly at first, hence my earlier errors.
Well, the rain has finally stopped, but the builders next door have started...and so has their noise! So I may as well go outside and see if my noise can beat theirs: my 9" angle grinder and petrol-driven posthole digger versus their generator, pneumatic nailing guns and rap 'music'! It's war!!
Definitely a team effort!
I'd still be poking away at it in the dark if it wasn't for the insight of the other team members.
Hi;
I found another answer, but still for X = 4.
It uses 12 less "4"s than in my previous solution.
I tried for X = 3 again, but can't find a solution for it.
Raining outside...can't dig holes in the mud!
This is the best I've got so far:
X = 4.
Sorry, thickhead, but I have to rush off now for a few hours and can't spend time on your post until this evening.
I'd logged on again to edit my post #34, which I finished just a few minutes ago.
Catch you later...
Quick edit before I go: Just had a thought and checked my images. There's a critical error! Grumble! Back to the drawing board!!
Hi Anna;
You should compare columns with columns and rows with rows, not columns with rows. The 'x' diagonal only goes in one direction (NW to SE), which throws the expected balance out, but there is another way of looking at this that shows we have the balance we need.
Here are two coloured images from my spreadsheet to help demonstrate. All green cells contain "2" and all blue cells "3".
In this version, if you draw a line through the centre to two cells opposite to, and equidistant from, each other, you'll find that one of them will contain a "2" and the other a "3". Always. In the following examples of this, the first letter denotes the row and the second the column.
Ae ("2") is opposite fB ("3").
OR ("3") is opposite RO ("2")
And so forth, and so fifth...etc.
There are an equal number of each colour/number, distributed equally in a balanced pattern about the 'x' diagonal.
This version comprises two diagonally-cut halves (with the NE sector rotated 180°) alongside each other for comparison. You'll see that all "2"s and "3"s in the right-way-up image align exactly with their respective "3"s and "2"s counterparts in the 180°-rotated image. That pattern applies throughout, which could not happen without 100% balance.
Brimming with balance!
Just had a thought and checked my images. There's a critical error! Grumble! Back to the drawing board!! Maybe you're right. Have to go out now for a few hours, and will look at this again this evening.
The table is perfectly symmetrical about the 'x' diagonal.
eg,
Row "A" has 1 @ "x", then 30 @ "2", then 1 @ "3", and row "f" has 1 @ "2", then 30 @ "3", then 1 @ "x";
Column "A" has 1 @ "x", then 31 @ "2", and column "f" has 31 @ "3" then 1 @ "x".
Also, starting with column A = 62, the A.P. of +1 requires the following handshake totals:
A=62, B=63, C=64...d=91, e=92, f=93.
And the total for column "f" can only be 93 by 31*3 and 1 @ "x".
Also, a Countif search of the table finds 32 @ "x", 496 @ "2" and 496 @ "3".
Ah...yes! Exactly, thickhead.
X = 3.
And thanks for spotting the flaws in my table.
"e" will have 2*2+29*3=91
"f" will have 2+30*3=92
I've got the following for those:
e = 2 + 30*3 = 92;
f = 31*3 = 93
Because f is the 31st term from A, and A = 62, then f = 62 + 31 = 93...and e is 1 less.
Hi;
I think I agree with thickhead's answer in post #6.
Hi Relentless;
Both correct!
My HP 32SII (below), working 'perfectly well' again! I must've walked off with it, and the refund! I'll have to 'pay' the shop a visit...
You're welcome!
At the end of my previous post I'd included a link to a similar puzzle that I posted in 2009, but then deleted it because when I read the thread I thought I'd made a critical error in one of the posts.
However, I checked it again just now and all seems to be ok, so here's the link again if you'd like to have a look at it: The missing dollar (& the 2 extra dollars).
Now you owe your mum $49 and your Dad $49 so 49+49=98 plus the $1 with u is $99.
Yes, that is $1 less than $100...which is the initial amount of Mum and Dad's loan.
However, the above quote should read: Now you owe your mum $49 and your Dad $49 so 49+49=98 less the $1 with u is $97...which is the cost of the shirt.
Adding, instead of subtracting, the $1 only seems to be going in the right direction with it because of the close proximity of $99 to $100. If the $100 were some other figure - like $1,000,000 - it would be obvious that something was up, even though one figure is just as (ir)relevant as the other.
The correct sum for arriving at $100 is this:
$49 + $49 + the $2 already returned to Mum and Dad = $100...which is the initial amount of Mum and Dad's loan.
There is a list of BBCodes here...or, when you are logged in, you can get there by clicking on the BBCode button near the bottom of the thread page, on the left-hand side. The code tags are near the bottom of the list.
Also, again when you are logged in, you can click on the 'Quote' button in the bottom right-hand corner of my previous post to see how I used it.
The chart looks warped because of single and double digit numbers.
Here's the same chart in 'code' tags.
If you want to align the single-digit numbers with the right digit of the double-digit numbers, you'd have to include spaces where needed...but this is the simple solution.
Those equations work, but they also work for smaller values of r.
Is there a notation that could be incorporated in the equation to indicate that maximum r is required, or should that be worded somehow with the answer, or don't we worry about putting it in the answer because it's in the question?
Also, is there some way of arriving at a solution without testing several values of r to find out which is the maximum? I suppose you could solve for r where the rhs = 500, and round the answer down to the nearest even integer. Btw, my little HP 32SII calculator's Solve function gives the answer r = 18.1617121098.
In my previous post I had this equation:
That came from this:
I should have included the inequality:
Ok, thanks. I think I have the picture now.