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;
13
Based on Omega Krypton's answer,
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
5
Complete answer.
because
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
I have a bad feeling that the answer you’re after is
Because
This, of course,
4
Using the Vinculum:
we have:
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
Here is a suggestion for 86:
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
ma is the age of Mark
my ist the age of Mindy
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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