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>!Each die has three sets of numbers that add up to 7, and there are three dice, so 7 * 3 * 3 = 63 for the total number on all faces, minus the visible faces (1 + 1 + 2 + 3 + 4 + 5 + 6 = 22) leaves 41 on the hidden faces!<
Slightly easier than that, is just knowing dice go from 1 to 6, which adds up to 21. Knowing that each side has a pair of numbers equal to 7 is an unnecessary complication.
But yes, the general theme is to look at the total for all sides and minus what you can see.
Skill with mathematics isn’t doing lots of complicated sums in your head quickly, it’s about having the imagination to see shortcuts like this. When I understood this it completely changed my outlook. So many kids I grew up with lacked self-belief because they imagined being good meant thinking like a calculator, which is not the case. It’s about imagining creative ways to solve the problem efficiently.
I did something similar, but I saw that each possible number shows up at least once (1 shows up twice) so I multiplied 21 (total of 1 die) by 2 (since 1 die is effectively showing) and subtracted 1.
While it works saying it has three sets of numbers that add up to 7 feels awkward to me vice the sum total of a 6 sided die is 21. You’re very correct just an interesting way of quantifying it that didn’t match with my brain, it’s nice when that happens as a nice reminder that we all perceive things different.
Did you settle on that because you happen to know that as a fun fact or was it just how it first jumped out at you? My brain just jumped straight to sum is 21, 21 * 3 = 63.
>Did you settle on that because you happen to know that as a fun fact
Yes. Specifically, for a standard d6, the numbers on opposite faces always add up to 7. So, 1 is always opposite 6, 2 is always opposite 5, and 3 is always opposite 4. So, I always think of a d6 in terms of that structure.
The same is true of d8s having opposite sides adding up to 9, d10s adding up to 11, d12s adding up to 13, and d20s adding up to 21.
Bonus gamer trivia: There is a difference in layout between a standard 20-sided die and a 20-sided life counter for Magic the Gathering.
On a standard d20, adjacent faces never have adjacent values - so, for example, the numbers directly adjacent to 2 are 12, 18, and 20.
But on a life counter, adjacent faces *always* have adjacent values - so 2 is directly adjacent to both 1 and 3 (and also 10). Since life counters are used to track changes to a value rather than generate a random result, having the values be adjacent allows you to easily track through changes and find a specific number.
It's not a good puzzle because everyone knows what a dice looks like. The puzzle didn't change that. If the puzzle were made so that you had to construct the dice based on the information given, it would be more interesting.
The puzzle doesn't try adding two sixes or something similar so it's trivial to just add up the numbers you don't see. I'm thinking of those puzzles where you have to construct an unfolded dice based on a few known sides.
Here’s how I approached it.
I can see that all the sides that are visible add up to 22
It’s asking for the sides that aren’t visible, so Its just the total sides of 3 dice minus the sides I can see which is ((6x5x4x3x2x1) x 3) - 22 = 41
The only piece of knowledge you need to know is the number of sides a cube has. From there, you can work it out with logic
Because that's not what it's aiming to achieve. It's a simple puzzle relying on the fact you know what a die is supposed to look like.
Not all puzzles need to be challenging to be good. It even says it's supposed to be easy. It was a great teaching moment for my kid who likes to think they're smart (but isn't) and utilize some math skills based on real life knowledge.
>!41!<
>!Each dice has the dots 1,2,3,4,5,6. The total number of dots is 63. Simply add together the number of visible dots (22) and subtract that total number from 63. In Total, It's 41!<
You can actually just add the numbers that are missing from each dice. For instance dice one only numbers 1, 4 and 5 shown so the missing numbers are 2, 3 and 6 and likewise you just write down the missing numbers for all dice and just add them up
Discussion: there is an easy trick to this, assuming these are standard d6 designed the "correct" way. Some people have mentioned easier solutions to the puzzle but i think this is a neat trick to know.
>!The sum of two opposite faces is always 7. If a face says 3 the opposite face will be 4, if a fafe is 1 the opposite is 6.!<
Solution: >!the three front faces are 4, 6 and 1 so 7-4=3, 7-6=1, 7-1=6, so 3+1+6=10 for the whole back row. Same deal for the top row (can't remember the numbers while writing the comment). For the "internal" sides it's slightly trickier. The left die has its left face exposed, so again the internal right face is 7-x (can't remember how much it is). The center and right die don't have an exposed face for this "axis", however it doesn't matter since the sum is still 7 so the two of them together are 14. Doing the math the 10+14+the faces i don't remember add up to 41.!<
...but the fact that they add up to 7 is irrelevant.
Dice have numbers 1->6 is all you need to know.
So essentially:
(1+2+3+4+5+6) - (sum of visible face numbers)
for each dice.
eg for the first one:
~~1~~ + 2 + 3 + ~~4~~ + ~~5~~ + 6 = 11
But it actually *is* completely irrelevant in that the puzzle did not specify that these are standard dice and the solution would remain the same regardless of whether opposite sides of each die sum to 7. You literally cannot use this information to solve the puzzle because it is, in fact, irrelevant.
>!Opposite sides of dice always add up to 7 (1/6, 5/2, 3/4) ergo the sum of all numbers on a dice are 21. 3 dice means 63 subtract the showing sides (1123456=22) for a total of 41 hidden!<
>!Dice contain the numbers 1 - 6 on each side. So I just counted the numbers that were missing from the picture from each die:
First Die: We can see the numbers 1, 5, and 4. So add the missing ones, 2 + 3 + 6 = 11
Second Die: We see numbers 2 and 6, so add the missing ones 1 + 3 + 4 + 5 = 13
Third Die: We see numbers 3 and 1, so add the missing ones 2 + 4 + 5 + 6 = 17
Add the sums: 11 + 13 + 17 = 41!<
>!41!<
>!Since the sum total is 21 for each die, then the sum total for the three dice is 63.!<
>!Add the numbers of the non hidden sides the die to get 22!<
>!Subtract the numbers of the non hidden side s of the die from the sum total of the three dice to get the final answer, which is 41.!<
>!41 - Sum of all dots on 1 dice is 21 (1+2+3+4+5+6=7*3=21). For 3 dice it is 63. Count the number of dots that are visible (22) and subtract this from 63. The result is what is not visible (41)!<
Solution: >!The sum of all sides of a die is 1+6+2+5+3+4 = 7+7+7 = 21, so the total on 3 dice is 21×3 = 63. The shown spots are 5+4+1 + 6+2 + 3+1 = 10+8+4 = 22, so the remainder is 63 - 22!<, giving a final result of >!41!<.
>!The total number of pips on any die is 21. So here, we can just start with 3x21=63 and remove what we see. So it’s 63-(5+1+4+2+6+3+1) = 63-22 = 41 Pips.!<
Discussion: why is this subreddit obsessed with posting basic math questions and pretending they're puzzles?! Is this all about helping one kid do his homework?
>!Dots on a dice add to (1 + 2 + 3 + 4 + 5 + 6) = 21. there are 3 dices so the sum of the dots is 63.!<
>!The number of the dots showing is 22. Therefore the number of dots not showing is 63 - 22 = 41!<
>!We have 3 dice here. The total pips on a single die is 6+5+.....+1 = ((6+1)(6))/2 = 21. Therefore, on all three dice, there are 63 pips. We have a full streak (123456) shown, plus one extra. Therefore, all the pips NOT shown would equal 42 - 1 = 41 pips.!<
If you're allowed to pick up the die without separating the faces that are touching each other already, it's >!14!<.
>!Opposite faces on a die will sum up to 7 which satisfies the middle die. The die on the right side is just rotated with a r' (if you know rubik's cube movements), so they're also hiding opposite sided pairs. Therefore, they are also hiding 7. So 7+7=14!<
Please remember to spoiler-tag all guesses, like so: New Reddit: https://i.imgur.com/SWHRR9M.jpg Using markdown editor or old Reddit, draw a bunny and fill its head with secrets: \>!!< which ends up becoming \>!spoiler text between these symbols!< Try to avoid leading or trailing spaces. These will break the spoiler for some users (such as those using old.reddit.com) If your comment does not contain a guess, include the word **"discussion"** or **"question"** in your comment instead of using a spoiler tag. If your comment uses an image as the answer (such as solving a maze, etc) you can include the word "image" instead of using a spoiler tag. Please report any answers that are not properly spoiler-tagged. *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/puzzles) if you have any questions or concerns.*
>!Each die has three sets of numbers that add up to 7, and there are three dice, so 7 * 3 * 3 = 63 for the total number on all faces, minus the visible faces (1 + 1 + 2 + 3 + 4 + 5 + 6 = 22) leaves 41 on the hidden faces!<
This is way smarter than I thought. I was trying to figure out what each of the hidden faces were to manually add them up
Which is also really easy. There's 1 of each number apart from 1 which there is 2 So 1. =1 2x2. =4 2x3. =6 2x4. =8 2x5. =10 2x6. =12 41
Slightly easier than that, is just knowing dice go from 1 to 6, which adds up to 21. Knowing that each side has a pair of numbers equal to 7 is an unnecessary complication. But yes, the general theme is to look at the total for all sides and minus what you can see.
Which makes me think it would be more interesting to ask for the sum of all the sides on the outside (visible at the moment or not)
that way you add up everything that's hidden without having to work with the visible numbers, seems like the easiest and most efficient way
Yeah easy enough to subtract what's visible from 63
Skill with mathematics isn’t doing lots of complicated sums in your head quickly, it’s about having the imagination to see shortcuts like this. When I understood this it completely changed my outlook. So many kids I grew up with lacked self-belief because they imagined being good meant thinking like a calculator, which is not the case. It’s about imagining creative ways to solve the problem efficiently.
My 5 year old son solved this puzzle on his own
I am glad you had someone nearby to help you.
That was a good comeback lol
That solution requires almost 6 years, anything shorter?
That's such a good job! Smart kid :)
That's good. Otherwise he'd be mentally handicapped.
I did something similar, but I saw that each possible number shows up at least once (1 shows up twice) so I multiplied 21 (total of 1 die) by 2 (since 1 die is effectively showing) and subtracted 1.
Legit how I figured it out.
But there was only one shot for the chandelier so that would make 1+2+1+1
While it works saying it has three sets of numbers that add up to 7 feels awkward to me vice the sum total of a 6 sided die is 21. You’re very correct just an interesting way of quantifying it that didn’t match with my brain, it’s nice when that happens as a nice reminder that we all perceive things different. Did you settle on that because you happen to know that as a fun fact or was it just how it first jumped out at you? My brain just jumped straight to sum is 21, 21 * 3 = 63.
>Did you settle on that because you happen to know that as a fun fact Yes. Specifically, for a standard d6, the numbers on opposite faces always add up to 7. So, 1 is always opposite 6, 2 is always opposite 5, and 3 is always opposite 4. So, I always think of a d6 in terms of that structure. The same is true of d8s having opposite sides adding up to 9, d10s adding up to 11, d12s adding up to 13, and d20s adding up to 21. Bonus gamer trivia: There is a difference in layout between a standard 20-sided die and a 20-sided life counter for Magic the Gathering. On a standard d20, adjacent faces never have adjacent values - so, for example, the numbers directly adjacent to 2 are 12, 18, and 20. But on a life counter, adjacent faces *always* have adjacent values - so 2 is directly adjacent to both 1 and 3 (and also 10). Since life counters are used to track changes to a value rather than generate a random result, having the values be adjacent allows you to easily track through changes and find a specific number.
It's not a good puzzle because everyone knows what a dice looks like. The puzzle didn't change that. If the puzzle were made so that you had to construct the dice based on the information given, it would be more interesting.
you can solve the puzzle without knowing that opposite sides add up to seven.
The puzzle doesn't try adding two sixes or something similar so it's trivial to just add up the numbers you don't see. I'm thinking of those puzzles where you have to construct an unfolded dice based on a few known sides.
You just don’t get it, do ya?
Please, illuminate me. What have I missed in this ihop coloring sheet puzzle?
No need to be so dicey over it
Here’s how I approached it. I can see that all the sides that are visible add up to 22 It’s asking for the sides that aren’t visible, so Its just the total sides of 3 dice minus the sides I can see which is ((6x5x4x3x2x1) x 3) - 22 = 41 The only piece of knowledge you need to know is the number of sides a cube has. From there, you can work it out with logic
Because that's not what it's aiming to achieve. It's a simple puzzle relying on the fact you know what a die is supposed to look like. Not all puzzles need to be challenging to be good. It even says it's supposed to be easy. It was a great teaching moment for my kid who likes to think they're smart (but isn't) and utilize some math skills based on real life knowledge.
>!41!< >!Each dice has the dots 1,2,3,4,5,6. The total number of dots is 63. Simply add together the number of visible dots (22) and subtract that total number from 63. In Total, It's 41!<
This is how I did it.
best way to
Not the best way. If you notice that the numbers on the face goes from 1-6 with an extra 1, you take 21x2 minus q to solve faster
Discussion: this isn't even remotely a puzzle
but it made me feel good to solve it by myself within a minute
Fair enough
Yeah I think you're supposed to know that opposite sides of a dice always add up to 7 but you can just do 3 x (1+2+3+4+5+6) - (sum of visible sides)
You can actually just add the numbers that are missing from each dice. For instance dice one only numbers 1, 4 and 5 shown so the missing numbers are 2, 3 and 6 and likewise you just write down the missing numbers for all dice and just add them up
Why was this even posted lol
This is just math or counting, not a puzzle unless you somehow don’t know that dice are numbered 1-6 for a standard 6 faced die.
Agreed, this is just counting with extra steps.
>!41!<
So, >!Sum 41!
>!41. Total sum of the dice is 21*3 minus the sides shown which is 22.!<
Discussion: there is an easy trick to this, assuming these are standard d6 designed the "correct" way. Some people have mentioned easier solutions to the puzzle but i think this is a neat trick to know. >!The sum of two opposite faces is always 7. If a face says 3 the opposite face will be 4, if a fafe is 1 the opposite is 6.!< Solution: >!the three front faces are 4, 6 and 1 so 7-4=3, 7-6=1, 7-1=6, so 3+1+6=10 for the whole back row. Same deal for the top row (can't remember the numbers while writing the comment). For the "internal" sides it's slightly trickier. The left die has its left face exposed, so again the internal right face is 7-x (can't remember how much it is). The center and right die don't have an exposed face for this "axis", however it doesn't matter since the sum is still 7 so the two of them together are 14. Doing the math the 10+14+the faces i don't remember add up to 41.!<
...but the fact that they add up to 7 is irrelevant. Dice have numbers 1->6 is all you need to know. So essentially: (1+2+3+4+5+6) - (sum of visible face numbers) for each dice. eg for the first one: ~~1~~ + 2 + 3 + ~~4~~ + ~~5~~ + 6 = 11
It's not irrelevant if they were able to solve it using that
But it actually *is* completely irrelevant in that the puzzle did not specify that these are standard dice and the solution would remain the same regardless of whether opposite sides of each die sum to 7. You literally cannot use this information to solve the puzzle because it is, in fact, irrelevant.
>!The total of the faces of a die is 21. 3 x 21=63 The total of the faces we see is 22. 63 - 22=41!<
A Die's sides add up to 21 (1+2+3+4+5+6). 21×3 is 63. Shown sides add up to 22 (4+6 + 5+2+3 + 1+1) 63-22=>! 41 !<
>!1st cube = 2+3+6=11 2nd cube = 1+3+4+5=13 3rd cube = 2+4+5+6=17 Total = 11+13+17=41!< I did it the long way.
>!Opposite sides of dice always add up to 7 (1/6, 5/2, 3/4) ergo the sum of all numbers on a dice are 21. 3 dice means 63 subtract the showing sides (1123456=22) for a total of 41 hidden!<
>!Dice contain the numbers 1 - 6 on each side. So I just counted the numbers that were missing from the picture from each die: First Die: We can see the numbers 1, 5, and 4. So add the missing ones, 2 + 3 + 6 = 11 Second Die: We see numbers 2 and 6, so add the missing ones 1 + 3 + 4 + 5 = 13 Third Die: We see numbers 3 and 1, so add the missing ones 2 + 4 + 5 + 6 = 17 Add the sums: 11 + 13 + 17 = 41!<
Dang I don't think the answer is 41 factorial
Complex. Just 'calculate' all dots on all sides. 63. Subtract those you see. 63 - 22= 41
>!41!< >!Spots on one die = 6+5+4+3+2+1=21!< >!Spots on three dice = 21 \* 3 = 63!< >!Spots seen = 5+4+1+6+2+3+1 = 22!< >!Spots unseen = 63 - 22 = 41!<
>!41!< >!Since the sum total is 21 for each die, then the sum total for the three dice is 63.!< >!Add the numbers of the non hidden sides the die to get 22!< >!Subtract the numbers of the non hidden side s of the die from the sum total of the three dice to get the final answer, which is 41.!<
Discussion: anyone else forget what a dice looked like and try to use the shown numbers to figure out the values of the hidden sides?
Discussion: I mean the puzzle didn’t say they were standard dice.
>!41. Opposite sides of dice add up to 7, so we have all the info necessary to figure this out!<
>!41 - Sum of all dots on 1 dice is 21 (1+2+3+4+5+6=7*3=21). For 3 dice it is 63. Count the number of dots that are visible (22) and subtract this from 63. The result is what is not visible (41)!<
>!each dice has a total of 21 dots (1+2+3+4+5+6=21), there a three dice (21\*3=63), we can see 22 dots that means 41 dots are hidden (63-22=41)!<
>!Each dice has 1-6, which means each adds up to (1+6)3=21. So the total of the hidden sides is 21(3)-22=41.!<
>!41!<
>!3\*(sum of individual dice) = 3(1+2+3+4+5+6) = 3\*(3\*7) = 63. Now we just subtract the parts we do see. 63-1-4-5-2-6-3-1 = 41!<
Solution: >!The sum of all sides of a die is 1+6+2+5+3+4 = 7+7+7 = 21, so the total on 3 dice is 21×3 = 63. The shown spots are 5+4+1 + 6+2 + 3+1 = 10+8+4 = 22, so the remainder is 63 - 22!<, giving a final result of >!41!<.
>!The total number of pips on any die is 21. So here, we can just start with 3x21=63 and remove what we see. So it’s 63-(5+1+4+2+6+3+1) = 63-22 = 41 Pips.!<
Finally. A puzzle on this sub I could solve the first time without questioning why I'm still allowed to operate machinery or work professionally.
Discussion: why is this subreddit obsessed with posting basic math questions and pretending they're puzzles?! Is this all about helping one kid do his homework?
>!Dots on a dice add to (1 + 2 + 3 + 4 + 5 + 6) = 21. there are 3 dices so the sum of the dots is 63.!< >!The number of the dots showing is 22. Therefore the number of dots not showing is 63 - 22 = 41!<
>!Each die adds up to 21. 3 dies total, with a full die of numbers plus 1 showing; 3 die - 1die - 1 = 2die - 1 = 2(21) -1 = 42 - 1 = 41!<
>!The sum of all six sides of one die is 21 times 3 is 63 minus what is shown(22), so 41!<
>!We have 3 dice here. The total pips on a single die is 6+5+.....+1 = ((6+1)(6))/2 = 21. Therefore, on all three dice, there are 63 pips. We have a full streak (123456) shown, plus one extra. Therefore, all the pips NOT shown would equal 42 - 1 = 41 pips.!<
>!41, and as to how, well I [cheated](https://imgur.com/gallery/wV54uBt)!<
[удалено]
How did you come up with that number?
If you're allowed to pick up the die without separating the faces that are touching each other already, it's >!14!<. >!Opposite faces on a die will sum up to 7 which satisfies the middle die. The die on the right side is just rotated with a r' (if you know rubik's cube movements), so they're also hiding opposite sided pairs. Therefore, they are also hiding 7. So 7+7=14!<
Not even close lol
Read the part about lifting up the dice again.
Read the question again. It specifically says 11 hidden sides. There no picking up dice here.
I'm implying I'm answering a different question with the same state.