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33

how about this? this is allowed right?


17

We have the following COCA +COLA ----- SODA Next, notice something similar in the Since we have a 4 digit number as the result, we know that But: Thus, Also, we know Thus, the solution is;


12

Loophole that I discovered: (1) Pretty much based on micsthepick's answer... (2) self-innovated: (3) self-innovated: (4) self-innovated


8

Since we know that Therefore $A$ Hundreds value must carry since $O \neq 0$ Therefore Therefore $O$ We now get And since $S<9$ Then there are many possibilities... any relations I missed out?


7

I think that in the first case the total sum is Reasoning For the second case


7

Full Solution Notation: Deducing $e$ therefore, Deducing $c, f, h$ regarding B, C: only pair available for distance of 4 is: Therefore, Sum of 25 rule renders Deducing the rest regarding A, E: only pair available for distance of 2 is: Therefore, Sum of 25 renders


6

The answer is Reason: Example:


5

TL;DR Preliminary deductions Left column: Bottom row: In particular, the numbers Option 1 Let's assume Top row: Right column: So we have The remaining numbers are Option 2 Let's assume Top row: Right column: So we have one of the following two possibilities: The remaining numbers are respectively $2,6,8$ or $4,6,8$, so the complete grid is ...


5

Here's some other words that should get you free ice cream: But why? The answer lies within Words with no two: Words with a two: Now the only question I have is what kind of training the cashiers go through to figure these out.


5

I'm not sure if this is allowed, but


4

I believe A, B, P are Some deductions I came to before brute forcing


4

Answer Explanation


4

I have a bad feeling that the answer you’re after is Because This, of course,


3

I'm going to assume that if the fourth card is drawn, and a multiple of 3 is never reached, the game is considered drawn and this is not a win condition for Jack. Let's start with the easy part: Now going through other possibilities: If Jack draws 1: If Jack draws 2: If Jack draws 4: Ultimately, this gives him odds of: EDIT: Adjusting for the fact ...


3

Here's one that I found: Here is one similar to micsthepick's answer: And here is a hybrid of the two:


3

It works with We can rewrite the equation as follows: Clearly this works if which yields the solution stated.


3

Here is one solution. Some observations


2


2

I’m thinking this is wrong, because there are some words that break the pattern, but could it be that So a word that could get you ice cream is I have no idea if this has anything to do with anything, but So could I have a scoop of


2

Let a be Mark and b be Mindy. Then So


1


1

Assuming that when the final card of the deck is drawn, the running total remains and the deck is reshuffled. The game will continue at most 3 times through the deck. The count begins at zero (0 modulo 3), after the first pass through the deck the sum is ten (1 modulo 3), and after the second it will be twenty (2 modulo 3), and then thirty (0 modulo 3). ...


1

Wrote an R script: install.packages("combinat") require(combinat) #Creates all permutations for drawing four cards data<-permn(1:4) table<-matrix(nrow=length(data),ncol=4) table<-as.data.frame(table) for(i in 1:length(data)){ for(j in 1:4){ table[i,j]<-data[[i]][j] } } #Don't care about Jill's second pull table$V4<-NULL table$sum<-...


1

Maybe I'm missing something but... Reason:


1

The sequence is: Explanation:


1

191, 426, 931, ???, ???, 646, 971, ???, ??? After looking for few minute I found pattern as 1.Sum of digit in series as 11,12,13, ???, ???,16,17,???, ??? 2.the first digit: square of series 1,4,9,1(6),2(5),3(6),4(9),6(4),8(1) 3.the last digit is flipping between 1& 6 So using 1&2&3 pattern :626,591,486,991


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