Jeg er forsigtigt i gang med at afprøve HelloFresh. Det lyder bare rigtig godt, at man får en kasse med præcis det, der er brug for, hverken mere eller mindre. Ikke noget med et helt hvidløg her, nej nej. Hvis der står 2 fed i opskriften, så får du 2 fed.
Men jeg er ved at blive fortvivlet over alt, hvad der har med betaling at gøre. Se nu fx, da jeg ville have overblik over prisen på næste kasse.
Billedet til venstre fortæller mig, at en kasse normalt koster 430. Plus 50 i fragt. Billedet i midten fortæller til gengæld, at jeg fĂĄr 130 i rabat. Og sĂĄ skal jeg ogsĂĄ huske billedet til højre, der lige nĂĄr at fĂĄ med, at der skal lægges 2 x 40 oveni. AltsĂĄ 430 + 50 – 130 + 2*40 = 430. SĂĄdan i runde tal.
Svaret på det er (gætter jeg på), at jeg får den her udregning som mail. Det var i hvert fald noget i den retning, der skete ved en tidligere lejlighed. Der havde jeg bestilt en kasse til levering om onsdagen. Søndag fik jeg det her.
Der har jeg altså fået udregningen første gang. (I mærkelig formatering, men skidt pyt.) Det er søndag, 20.52. Næste skridt kommer mandag, 9.02.
Jeg har i god tid sikret mig, at der ville være penge nok på kontoen om onsdagen. Men det er altså ikke nok. Og da jeg ikke reagerer hurtigt nok, så kommer næste besked 6 timer senere.
(Måltidksasse?) Bare ærgerligt.
Et par dage senere prøvede jeg at spørge via chat, om jeg på forhånd vidste, at pengene skulle være på kontoen om mandagen. Havde det stået i en mail? Eller stod det måske et sted på hjemmesiden? Det kunne jeg ikke rigtig få et klart svar på.
Og lidt mere.
Der er lidt af hvert, jeg er i tvivl om her. SĂĄsom: Overholder det her loven?
I just read a really good book about the game Minesweeper. And one of the interesting topics was: What came before? Let’s take a trip back in time. And read all the way to the end for a surprise.
A very tiny Alice (…) is racing across a 2-by-2 chessboard, as shown below, where each smaller square has a side length of 1 cm. Alice starts at the bottom-left corner of the bottom-left black square and is trying to reach the top-right corner of the top-right black square.
There’s just one catch. Alice moves faster on the white squares than she does on the black squares. Her speed on the white squares is 1 cm per minute (again, she’s very small), while her speed on the black squares is 0.9 cm per minute.
What’s the least amount of time it will take her to reach the finish?
Actually, that link will also take you to a solution to the extra credit puzzle:
Instead of a 2-by-2 chessboard, Alice now wants to traverse an 8-by-8 chessboard, as shown below. Again, she’s going from the bottom-left corner of the bottom-left black square to the top-right corner of the top-right black square.
As before, her speed on the white squares is 1 cm per minute, while her speed on the black squares is 0.9 cm per minute.
What’s the least amount of time it will take her to reach the finish?
There are some excellent youtube channels out there, like Numberphile, if you want more math fun in your life. The concept is something like professors talking about all sorts of math, both classic theorems and new discoveries. Like this one about sudoku:
(1) What is the next number after 95 whose digits add to a multiple of 7?
It can’t be 96-99, because I only add 1-4 to the sum here, and can’t get all the way to a new multiple.
It can’t be 100, the sum is 1, not a multiple of 7.
If we add 6 to 100, we should get there. Answer: 106.
(2) Can you find a pair of consecutive numbers whose digits both add to multiples of 7?
My first guess is x9 and y0, where x+1 = y. Then it could be 69 and 70. The digit sum of 70 is 7, but the digit sum of 69 is wrong. When I go from 69 to 70, I add 1 to the sum and subtract 9, -8 in all. Of course I can’t get from one multiple of 7 to another that way.
How about 699 and 700? When I move from one to the other, I add 1 to the sum and subtract 2*9, -17 in all. 17 isn’t a multiple of 7, this won’t work.
So, when I go from 69…9 to 70…0, I add 1 and subtract n*9 from the digit sum. What I want is: 7 | 1-n*9. A quick check reveals, n = 4 works. 69999 and 70000 is a solution. Answer: Yes.
(I use | as “divides”.)
(3) What about a pair of numbers under 100 whose digits both add to multiples of 7, but that are two apart rather than consecutive?
Inspired by what I already did, 68 and 70 works. So, the answer is yes or 68, 70.
And let’s look at a way to solve it. Hopefully you can follow my symbols. I have steps, where each time I add all the new information I can, based on what I already know. At one point I make a guess, which turns out to be wrong, so I go back and try again.
Thankfully I reached the same solution as Enigmatic Code, even though I did it differently.