Daily Puzzles (15/23 Solved+1Q) [+5M]
Replies
I was having the same problem with puzzle 5 yesterday!! I was thinking, could it be possible that the liars only lie about their own age, and not their age in relation to that of others? (ie: "i am 51" is a lie, but " is older than i am" is not?)
Will sell soul for :')
edit nvm. i took a wrong step. rip.
inb4 someones age is 40
Ok, with the PMed correction, the ages should be:
- 38
- 39
- 68
- 29
- 26
- 61
- CROSSHINT Happiness Ejects ? Riotous Impetus Confess Antics Logistics ? Pious
arts by the amazing
Ok, with the PMed correction, the ages should be:</p>
<ol>
<li>38</li>
<li>39</li>
<li>68</li>
<li>29</li>
<li>26</li>
<li>61
</ol>
<p> - CROSSHINT
Happiness
Ejects ?
Riotous
Impetus
Confess
Antics
Logistics ?
Pious
One word wrong. Can give 9/10 credit. Remaining 1/10 to anyone who guessed the last word right. Yup, trades~
Which one did I miss?
arts by the amazing
Which one did I miss?
It's a group activity, not a singular activity. It is also voluntary (active) as opposed to passive (done to you).
Exodus?
Make a trade lot~
Added another puzzle.
I suck at explaining things, but I'm going to try!
So, for puzzle , it is not congruent. I tried drawing this out in my head, and the conclusion I came to is this: Dice blocks and match. Dice blocks and match. You can't really tell what's on the left hand side of the dice blocks, which leaves as a maybe. However, Dice block contradicts blocks and , as the U is on the opposite side of L.
I may have just pulled all of that out of my butt, but after staring at it for hours this is the solution I came to.
~ ~ Wishlist!
<p>I suck at explaining things, but I&;m going to try!</p>
<p>So, for puzzle , it is not congruent. I tried drawing this out in my head, and the conclusion I came to is this:
Dice blocks and match.
Dice blocks and match.
You can&;t really tell what&;s on the left hand side of the dice blocks, which leaves as a maybe.
However, Dice block contradicts blocks and , as the U is on the opposite side of L.</p>
<p>I may have just pulled all of that out of my butt, but after staring at it for hours this is the solution I came to.
40% right. You are making one assumption wrong about the nature of the blocks. hint. Sides of Cubes. fill
Puzzle 14
Base 9
-
8*7 = 56 base 10 = 54 + 2 = 62 base 9 1 is telling the truth 9 is the only correct base where this is true.
-
18 base 9 = 17 base 10 is prime 41 base 9 = 37 base 10 is prime 2 is telling the truth
-
35 base 9 = 27+5 = 32 base 10 is not prime So 3 is lying
-
63 base 9 = 57 base 10 is not divisible by 4 So 4 is lying This is false in every base, since we must have that 6 times the base is odd
9 is the optimal reasonably small base since it's the only one that satisfies 1, and the best possible without 1 is to satisfy 2 and 3. But these can't both be satisfied since either x+8 or 3x + 5 is even (and so not prime)
Puzzle 11
Dice 1 and 4 are the same die rotated, as are dice 3 and 5, so we'll cross out the latter for ease.
Notably, we must have a U, P, N/Z, L, and E. This requires five sides.
Dice 1 and 6 have two different orientations of an N/Z with respect to P, so we must have two of these symbols.
But dice 2 and 3 have U's with respectively an L and Z above them, so we must have two U's.
So there must be 7 sides.
Assuming that 5/6 dice are consistent together, either 2 or 6 is inconsistent.
From examination and constraint solving, 1, 2, 3, 4, and 5 are consistent together, so 6 is wrong (it puts an N next to a P in an incompatible way)
"The interesting hyperbolic structures on a pair of pants are easily classified" -Wikipedia, Pair of pants
I could be completely wrong but 12
A
Won 3-0 against B, won 2-1 against D, won 1-0 against C
B Drew 1-1 with D, won 1-0 to C, lost 3-0 against A
C Lost 1-0 against B, lost 1-0 against A, won 1-0 against D
D Drew 1-1 with B, 2-1 loss to A, lost 2-0 to C
Please make trade lots you two.
puzzle 14 correct. puzzle 11 40% right for initial pairs of consistent views(1&4, 3&5) 60% right for math for correct conclusion(6 is inconsistent), correct observation(multiple Us)
<p><details>
<summary>Puzzle 14</summary>
Base 9</p>
<ol>
<li>
<p>8*7 = 56 base 10 = 54 + 2 = 62 base 9
1 is telling the truth
9 is the only correct base where this is true.</p>
</li>
<li>
<p>18 base 9 = 17 base 10 is prime
41 base 9 = 37 base 10 is prime
2 is telling the truth</p>
</li>
<li>
<p>35 base 9 = 27+5 = 32 base 10 is not prime
So 3 is lying</p>
</li>
<li>
<p>63 base 9 = 57 base 10 is not divisible by 4
So 4 is lying
This is false in every base, since we must have that 6 times the base is odd</p>
</li>
</ol>
<p>9 is the optimal reasonably small base since it&;s the only one that satisfies 1, and the best possible without 1 is to satisfy 2 and 3, which isn&;t satisfied by any reasonably small bases.</p>
<p>
</details></p>
<p><details>
<summary>Puzzle 11</summary>
Dice 1 and 4 are the same die rotated, as are dice 3 and 5, so we&;ll cross out the latter for ease.
Notably, we must have a U, P, N/Z, L, and E. This requires five sides.
Dice 1 and 6 have two different orientations of an N/Z with respect to P, so we must have two of these symbols.
But dice 2 and 3 have U&;s with respectively an L and Z above them, so we must have two U&;s.
So there must be 7 sides.
Assuming that 5/6 dice are consistent together, either 2 or 6 is inconsistent.
From examination and constraint solving, 1, 2, 3, 4, and 5 are consistent together, so 6 is wrong (it puts an N next to a P in an incompatible way)
</details>
<p>I suck at explaining things, but I&;m going to try!</p>
<p>So, for puzzle , it is not congruent. I tried drawing this out in my head, and the conclusion I came to is this:
Dice blocks and match.
Dice blocks and match.
You can&;t really tell what&;s on the left hand side of the dice blocks, which leaves as a maybe.
However, Dice block contradicts blocks and , as the U is on the opposite side of L.</p>
<p>I may have just pulled all of that out of my butt, but after staring at it for hours this is the solution I came to.
You have one incoherent statement in your answer.
Puzzle 16
1 dollar = 8 pencils
1 dollar = 1 pen
10 dollars = marker
let n rep. number of groups of pencils let p rep. number of pens let m rep. number of markers 8n + p + m = 100 n + p + 10m = 100 8n + p + m = n + p + 10m 7n - 9m = 0 7n = 9m since n,m,p must be whole numbers, 7n and 9m must be a common multiple of 7 and 9 under 100 common multiples of 7 and 9: 0, 63 if 7n = 0 7n = 0 n = 0 0 = 9m 0 = 9m 0 = m 0 + p + 10(0) = 100 p = 100 8(0) + p + 0 = 100 p = 100 m,n = 0; p=100 if 7n = 63 7n = 63 n = 9 7(9) = 9m 63 = 9m 7 = m 9 + p + 10(7) = 100 79 + p = 100 p = 21 8(9) + p + 7 = 100 79 + p = 100 p = 21 n = 9; p = 21; m = 7 therefore the student either bought 0 pencils, 100 pens, and 0 markers or the student bought 9 pencils, 21 pens, and 7 markers
Oops, that should be
12
C won 2-0 against D
Your systematic equations are wrong. This problem does not require algebra, but it can be used if the initial modeled equations are right. Your conclusions violate either or of the requirements of the puzzle also. , correct - please make a trade , yes the last equation should sub in to = # of supplies to get 72 pencils, 21 pens and 7 markers. please make a trade lot.
<details>
<summary>Puzzle 16</summary>
1 dollar = 8 pencils
1 dollar = 1 pen
10 dollars = marker</p>
<p>let n rep. number of groups of pencils
let p rep. number of pens
let m rep. number of markers
8n + p + m = 100
n + p + 10m = 100
8n + p + m = n + p + 10m
7n - 9m = 0
7n = 9m
since n,m,p must be whole numbers, 7n and 9m must be a common multiple of 7 and 9 under 100
common multiples of 7 and 9:
0, 63
if 7n = 0
7n = 0
n = 0
0 = 9m
0 = 9m
0 = m
0 + p + 10(0) = 100
p = 100
8(0) + p + 0 = 100
p = 100
m,n = 0; p=100
if 7n = 63
7n = 63
n = 9
7(9) = 9m
63 = 9m
7 = m
9 + p + 10(7) = 100
79 + p = 100
p = 21
8(9) + p + 7 = 100
79 + p = 100
p = 21
n = 9; p = 21; m = 7
therefore the student either bought 0 pencils, 100 pens, and 0 markers
or the student bought 9 pencils, 21 pens, and 7 markers
</details>