Never mind. Fixed it.
solution
| A | I | |||
| E | C | F | D | |
| B | G | J | H |
A through E is easy. Did you have to trial and error the F, then? I did that and happened to pick the right one and it was easy after that, too. But the trial and error felt like cheating.
yep, I messed that up
You guys are good at these. I donāt know that I could have gotten the HAWK to DOVE one.
1/11/21 - Security Door
āThe security door passcode,ā said Miranda, āis a seven-digit number whose digits total 35. The fourth digit is three more than the first digit, the fifth digit is four more than the second digit, the sixth digit is one less than the fourth digit, the last digit is one less than twice the second digit, and the sum of the first and third digits is one more than the fourth digit.ā Miranda continued, āHowever, the passcode has no repeated digits.ā What is the security doorās passcode?
1/11/21 - this one I was able to figure out, albeit with some guess and check.
Answer
Well, I thought I had the answer until I checked. I got 3426857. I messed up on the last clue āthe sum of the first and third digits is one more than the fourth digit.ā In my equation I did one less instead of one more.
Hereās the posted solution:
5248673
If A is the first digit and B is the second digit, then we can use the information given to create expressions in terms of a and b for the seven digits in order: (a, b, 4, a+3, b+4, a+2, 2b-1). The third digit must be 4 since the sum of the first and third digits is one more than the fourth digit. Since the sum of all of the digits is 35, we have the equation 3a+4b+12=35, or 3a+4b=23. In this equation, b must be less than 6 or the left side would total more than 23. Make a chart. Only two pairs of numbers work: (1, 5) and (5, 2). If we choose a=1 and b=5, then the passcode would be 1544939. This has repeated digits, which are not allowed. So instead, a=5 and b=2, and the passcode is 5248673.
nm
My logic for solving
They took a different approach (probably simpler) than I did.
I first wrote out all the clues like this:
seven digits total 35
four = one +3
fifth = two + 4
sixth = four - 1
seventh = second*2 -1
first + third = fourth + 1
no repeated digits
Then I subbed in the four to that last equation to get that the third equaled 4.
I also noted that 35 means you have to subtract 10 from the sum of 1-9, so we are missing a pair that adds up to 10.
From there I wrote all the possibilities for each number over the top of seven squares (for example, the first one has to be between 1 and 6 since the fourth canāt exceed 9 and so on). I iterated through a little bit accounting for no duplicates and found that 1 is not in the puzzle, so removed 9 as well. This then lets everything fall into place without having to plug and try numbers.
I donāt know if logic-ing something out is the same as trial and error, but itās how I solve a lot of kakuro puzzles, I know what numbers are going to sum, but Iām not sure of placement until I look at/figure out a few (or more) of their crossways pieces.
In this instance, the f could only go one of two places, and in the lower spot, the h, i, and j (and I guess the g too, in some orderings) would fail the non-touching condition.
You guys are quick at these. Hereās the next one.
1/12/21
Three of these objects have six-letter names and three have four-letter names. From each longer word, subtract the letters of one of the shorter words to get two leftover letters. For example, WICKET minus KITE leaves CW. Rearrange the six leftover letters to make a final word. What is the final word?
Solution
- powder (I had the hardest time with this one, only got it because of the rope, I was stuck between bottle and shaker before that :shake:) - leaving WD
- star
- pipe
- zipper (remove pipe leaves ZR)
- rope
- tiaras (remove star leaves IA)
WIZARD
Nice!
Summary
I thought it was salt or shaker for #1. I had #4 and #3 first, and was trying to figure out what I was doing wrong on #6. Didnāt think to make tiara plural 
Not sure how that happened, Iāve unlocked it.
placing a table here is just a matter of using vertical bars ( | | ) with the first row being a ācolumn headersā, a second row of just a dash (or two or three) for each column (this indicates that the prior row is a header row), and then so on . . .
|A|B|C|D|E|
|F|G|H|I|J|
The above could be the body of your table, but add TWO of the following to the beginning:
|-|-|-|-|-|
And you get:
| - | - | - | - | - |
|---|---|---|---|---|
| A | B | C | D | E |
| F | G | H | I | J |
You missed that trainā¦ by a lot. Weāre already to January 13th (which Iāll post on Monday) for goodness sake!
MA already told me I can just copy/paste from Excel, so thatās what Iāve been doing.
You act all surprised by this. Have you never seen a game that Iāve heckled?
1 Like
lol, I suppose I shouldnāt be surprised.
1/13/21
Place the numbers 1 to 10, one per circle, so that each side of this pentagon totals 14. Weāve started you off with the 1 and 8. Can you finish the pentagon?
1/13/21
Solution
I started with the side with an 8. The two blanks on that side had to equal 6, and since the 1 is already taken they had to be 2 and 4. Just tried the different possibilities until I went around the circle and everything worked.
1/14/21
In this numberless crossword, the clues for the Across and Down words have been intermingled into a single clue list. However, within that list, the Across clues appear in order from top to bottom, and the Down clues appear in order from left to right. Can you complete the grid?Clues: train unit, daddy, plotting groups, glass squares, humiliated, lawlessness, somewhat, recess, bring back, individuals, tatter, thumbs-up, tinted
Summary
|||||||
| P | A | P | A | |||
|---|---|---|---|---|---|---|
| C | A | B | A | L | S | |
| A | N | A | R | C | H | Y |
| R | E | S | T | O | R | E |
| S | E | L | V | E | S | |
| D | Y | E | D |
Summary
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Are they supposed to take less than a minute? Just woke up, so didnāt see other solutions.
1/13/21 solution
Clockwise from top:
(4, 9, [1), 10, (3], 6, [5), 7, (2], 8, 4)
Just looked at your solution. I started with the 1, seeing as the 10 has to go somewhere. If the 10 goes up and to the left of the 1, the top is 3, then the edge with the 8 has another 3, so, the 10 goes down and the left of the 1. Then, put the 9 up and to the right. Solves itself from there.