I was doing this by myself and FINISHED IT, but then I thought there was a fault in the problem because I got two answers. BUT. When Brian said they only got one of two... I felt like a genius.
I actual like your sponsor portions of your videos because they are always new and interesting. (I also love math games. I may not be good at them, but I like them.)
I wish they would have said that there were two correct answers at the START of the video. I paused it and wasted a whole lot of time trying to figure out why I couldn't find a unique solution.
This was indeed not cool. Even a 'there might be more than one answer' or 'find all possible answers for a beer' but some way to let us know or suspect that there is more than a unique answer would be nice. But then again Brian also got in the way of the crew working the puzzle a number of times in his excitement. Not a great episode. But thank you for your comment, because i scrolled down to see if someone discussed this exact issue.
If a puzzle has multiple correct answers, that you get in a very obvious way... not a good puzzle. Or, to put it another way, this wasn't a puzzle. Just a basic equation you'd see in middle school. I'm glad your stuff got better over time ^_^ can actually see you evolve from a "meh" channel to something that puts everyone else to shame. Kudos.
I'm always checking previous videos and current ones for stuff I haven't watched yet. I love scam school. I think Brian should collaborate with other cons and magicians like he does with Diamond Jim. That guy is great too.
I know I'm bragging here, but I found both answers in less than two minutes after watching just the intro. Aside from the bragging, I would like to say that consistently watching Scam School is one of the reasons I was able to do this.
paused the video and did the puzzle myself and ended up scratching my head because I had two possible solutions. Watched the rest of the video and found out they are both correct and now I feel like a genius :D.
I was getting mad becasue i thought there was only one answer and I couldn't logically figure out why both set of numbers couldn't work...then he said there were 2 answers lol
If yo take the remaining numbers that don't get used in either solution you get 2, 4, and 9. if you turn those into letters with a simple number-to letter cipher you get B, D, and I. If you rearrange the letters it can spell "Bid"! So if you want have fun with in you can present it with: "Solve this puzzle and you will see; what remains will be the key; a proper welcome from one to thee." And with a glorious red herring in the 123 pun at the end.
I was mad and thought I got it wrong because there were two answers. It was messing with me and I didn't wanna spoil it. Glad I finally gave up to see I was right!
before solution: o must be 5-9 (Brian's hint) e must be 1 (three digit number plus three digit number can't be more than 2000) d must be 5 (third digit in EVEN must be one, so dd + dd must leave a 1 in the tens place) n must be 0 (55 + 55) if o is 5, 6, 7, 8, or 9, v will be 1, 3, 5, 7, or 9 respectively after number can't repeat rule: 655 + 655 = 1310 or 855 + 855 = 1710 after solution: lit, got it right
E has to be 1, O has to be grater than 5. D has to be 5 which forces N to be 0. I'm pretty sure those are the only requirements. For example: 555 + 555 = 1110; 655 + 655 = 1310; 755 + 755 = 1510; 855 + 855 = 1710; 955 + 955 = 1910.
Since ODD+ODD is 5 digits, O>4 and E = 1 Since DD + DD = 1N or 11N, D = 5 and so N=0. 2(O55) = 1V10, O can be 5,6,7,8, or 9, and V will be (2O-10)+1 = 2O-9. If numbers can't be repeated, then O isn't 5, it's also not 9 since then V=9, and it's not 7 since that makes V=5. Thus, O = 6 or 8, and V = 3 or 7, respectively. ODD+ODD=EVEN 655+655=1310 855+855=1710
As an algebra lover, this was painful to watch. One does not simply try out the numbers, it's much simpler and more professional to just write down equations (e.g. 2D=N and 2D=E or 2D=N+10 and 2D+1=E) etc.. Eventually solving em all. By the way for anyone interested there's an app called cryptogram that gives a few of these problems, I think they're really fun.
Paused the video before the two minute mark and in about a minute got the values. If you haven't figured it out, don't keep reading, but O = 6, D = 5, E = 1, V = 3, and N = 0. Educated guesses FTW!
Dude, I paused for a while to try and figure it out, I was stumped because I got down to 2 valid answers (655 and 855) and I played the video to find a mistake, but yall confirmed I was right!
1. E must be 1 because two 3-digit numbers added together to create a 4-digit number must be at least 1000, but _cannot_ be 2000. 2. (Since that makes the 3rd digit of EVEN also 1), DD + DD must be 55 + 55 because it is the only combination of two 2-digit same-digit numbers (11, 22, 33, 44, 55, 66, 77, 88, 99) added together that results in a 1 in the tens column. 3. Therefore, N must be 0, since 55 + 55 = 110. 4. O + O + 1 (carried from DD + DD) is at least 10 (as EVEN is 4 digits), so O must be at least 5. There is not enough information to solve, but possible solutions that do not repeat any digits from other solutions are 6 and 8. (655 + 655 = 1310, 855 + 855 = 1710). 555, 755, and 955 do not work because they repeat digits: 555 + 555 = 1110 (that obviously repeats a few), 755 + 755 = 1510 (D and V would both be 5), and 955 + 955 = 1910 (O and V are both 9).
One thing he didn't mention in the video is that these equations are called "Cryptarithmetics". I see some of you in the comments found this easy. if you want a challange, try this one: SODAS - AIM = BEER @Brian This video went excelently! I'm so glad you got the sudoku aspect of the puzzle! This is how you can tell some tru puzzles apart from other... it's also why Minesweeper can be a brilliant scam of a puzzle.
You can also solve it graphically, although its a bit mathy... First do some algebra and rewrite the problem: 2(ODD)=EVEN 2(100*o + 10d+d)=1000e+100v+10e+n 200o+22d=1010e+100v+n Next we know that the one's place of d+d has to equal n because it is the last letter, we can write this algebraically as follows: mod(2d,10)=n We also know that e has to be 1 based off of the logic explained in the video So plugging these into the above equation we get: 200o+22d+1010 + 100v + mod(2d,10) We now have three variables but we know that o can only be 5,6,7,8,9 and so we can write an equation for each of these values Now we have 5 equations which we can graph as a function of v For simplicity I have made the following substitutions: v=y, d=x This gives us: y = (200*o +22d - 1,010)/100 Graphing this, as I did here: www.desmos.com/calculator/dreddhgqse allows us to visually see that the only integer solutions to the equation required x to be 5 when x (which is d) is 5 in each of our six equations we see that y (which is v) must be 1,3,5,7,9 when o is 5,6,7,8,9 respectively We now know that these are the only possible answers and they give us the following 555 + 555 = 1110 655 + 655 = 1310 755 + 755 = 1510 855 + 855 = 1710 955 + 955 = 1910 Finally, we can sort through these to see that they satisfy o=/d=/e=/v=/n and we are left with 655 + 655 = 1310 855 + 855 = 1710 And, well, this is exactly what they found in the video... but at least we have a proof for it now :p
hey Brian! is there any chance you could do a compilation of tricks with pens? I thought about this because I'm in high school and once in class I was thinking "wouldn't it be cool if I were just vanishing a pen repeatedly in the back of the class until someone just turned around and said :how the hell are you doing that?!"
1:44 Oh, cool, like Sideways Arithmetic from Wayside School! (In fact, they probably used this one, but I've since forgotten it.) Okay, let's figure this out... Two digits added together will never be greater than 18, so if you carry anything, it will be a 1. ..ODD +ODD _____ EVEN With that said, there was nothing in the thousands column, but we ended up with E. So we know E is 1: ..ODD +ODD _____ 1V1N Now, looking at the tens column, we know D+D should be even, but there's a 1 under that sum. So there must be a 1 carried over from the ones column, and D+D ends in 0. But the units column, which we just said must carry a 1, is D+D. Since D+D both ends in 0 and carries a 1, it must be 5+5=10 (which also means that N must be 0). ..O55 +O55 _____ 1V10 Now we know there's a 1 carried over to the 100s column, (and that the result of the sum "carries" over to the 1000s column), so O+O+1=V+10. Rearranging a little, 2*O=V+9. Now if V were even, V+9 would be odd, but it's obvious that 2*O is even, so V must be odd. (And so 2*O=V+9 an ODD no. plus an ODD no. equalling an EVEN no., reminiscent of the problem itself!) I assume the digits are different (other comments seem to confirm that he eventually says this), so V can't be 1 or 5. That leaves 3, 7, and 9. We'll have to test these through brute force. Solving for O, we know O=(V+9)/2. That should give us a good starting point. And the key to eliminating possibilities is that all letters are different digits, so if one letter (in particular, O) is the same as another, we have a contradiction and our premise must be wrong. If V=3, O=(3+9)/2=6 If V=7, O=(7+9)/2=8. That could be. If V=9, O=(9+9)/2=9. But then O would be the same as V, so that can't be. So I can't work it out right now, but I've only started the video. Maybe there's a rule that none of the letters are in the names of the digits they represent? Then it would have to be ..855 +855 _____ 1710 but I haven't heard all the rules yet
Saw that E had to be one, figured D was probably five making N zero. Then just used six as my first option for O to try it out and it was right. Took about 30 seconds to get the solution, but I didn't think to check for any other solutions.
If we force rule that different letters correspond to different digits, then we are left only with: 655 + 655 = 1310, 855 + 855 = 1710 It's much easier to solve if you write it like it's long addition: _ODD _ODD EVEN you instantly get E = 1, so it is _ODD _ODD 1V1N and here we have not much left for letter D: two of them have to sum up to odd number 1, so there has to be carry out from previous position, and it was D as well, so D is at least 5. Also we have to have exactly 1 so without carry it has to be 0, so D is either 0 or 5, but previously we deduced that it is at least 5 therefore it's 5, so now we have: _o55 _o55 1V1N after that we get that N = 0 and for o we've got only 4 different values we need to check, 6, 7, 8 and 9, from which only 6 and 8 satisfy condition that all digits have to be different. Lol, this is pretty much majority of things which children do at 4th-5th grade in my country and this is quiet useful for developing logic and ability to deduce things, I don't understand why it took adults so much time, it's very easy.
The equation for this problem is E times the square root of 3 plus V times the square root of 3 plus N times O = 2 times D plus 2 times D. This means only 855+855=1710 can be the answer. The other, 655 is a false positive.
For those interested in a mathematical proof of the solutions, I came up with this (feel free to correct me): Let's find the digits O,D,E,V,N making ODD + ODD = EVEN Because the two same 3-digits numbers sum up to a 4-digits number, we know that O ≥ 5. We also know that two 3-digits numbers can't sum up to more than 1998 (in general x-digits number summing up to a (x+1)-digit number will be 1-something). Which means that E = 1. Let's suppose that D < 5. Then D + D = N D + D = E But that's impossible, because all digits are different. Which means that D ≥ 5. This means that D + D = 10 + N (N is the last digit of a number between 10 and 19) E = (N + 1)%10 (that's D+D plus the 10 of the previous one) EV = O + O + 1 So far we have (solving the equations above) E = 1 N = 0 D = 5 This leaves us with O and V. We know that O ∈ {6,7,8,9}. If we add 1 to the double of each of these numbers, we get V. So V ∈ {3,5,7,9}. But each digit is unique, which leaves us with (O, V) ∈ {(6, 3), (8, 7)} Which leaves us with two only possible solutions 655 + 655 = 1310 855 + 855 = 1710 Please do correct!
Problem: Find all combinations of digits that satisfy this pattern, each number must be different: ODD + ODD = EVEN O=6,8 D=5,5 E=1,1 V=3,7 N=0,0 Or: 655 + 655 = 1310 855 + 855 = 1710 I believe these are all the solutions. Logic: First e must equal 1, because it's two three digit numbers, and there is no way for three two digit numbers to be added to produce a number greater than or equal to 2000. (999+999 = 1998, the highest we can go). Therefore, the target number is bounded between 1000 and 1998. For O since the answer is bounded between 1000 and 1998, we know O must be greater than or equal to 5, and less than or equal to 9. That's all we know for now, we'll come back. For D, we need to produce the 1v1n pattern, so the 10s place must have a 1 in it. The selected 0 doesn't affect the 10s place, it only affects the 100s and1000s place (we can convince ourselves by this line of reasoning: ODD + ODD = 2*ODD = 2*(O*100 + D*10+D) = (200*O + 20*D+2*D), so we can clearly see for the 10s place and 1s place, we see that O will be multiplied by 0. and will have no effect, but D will. . That means DD*2 must have a 1 in the 10s place. We can try all 10 numbers if we want (00,11,22,33,44,55,66,77,88,99) to convince ourselves that 110 is the only number that will produce this, so D *must* equal 5, since 55*55 = 110. Okay, now that we've got that sorted, we know N has to be 0. Then V depends on our selected O. Again, we try all 5 values of O (5,6,7,8,9) and we find that the only ones that have a digit in the V place that isn't already taken are 6 and 8. Edit: I just saw their solution part of the video, just to explain why O=9, O=7, and O=5 don't work (for the sake of completeness of my own solution), it is because the digits are already used. D is already 5, so O can't be that, O=7 doesn't work, because it gives 1510, but then V=D=5, which isn't allowed, and O=9 doesn't work because it gives 1910, so then O=V=9, which isn't allowed because they say you must have a different number of each letter. :)
1:52 Alright, right off the bat...ODD+ODD=EVEN. ODD is a 3-digit number and EVEN is a 4-digit number. Because ODD is 3 digits, EVEN has to be between 1000 and 1998 (2 * odd# = even#). This means that E = 1. N MUST be an even number because it's 2 * D. Since EN has to be 1X where X is even, the options for D are limited...if D is 1, for instance, 1+1=2=N and 1=E=1+1=2, which is obviously wrong. Therefore the sum of D+D>9 because it must affect the outcome of E. If D=5 then 5+5=10 meaning N=0, carry the 1, 5+5=10 add the remainder 1 to make 11, carry the 1, you have DD+DD=110 and EN=10. In fact, D has to equal 5 as it's the only number which makes E=1 in that scenario. So E=1, N=0, D=5. O>4 because ODD*2 has to be >999. O=/=5 because 5=D. If O=7 then you have 755+755=1510, meaning that D=V. If O=9 then 955+955=1910 and O=V. This leaves 2 solutions of: 655+655=1310 and 855+855=1710.
I honestly just stumbled across the right answer when I was going through the options for D. Assuming each letter had to be a different value, and the fact 5 worked to solve for E, I just moved O to 6.. o.o
Says, "algebra question", then has lets represent digits with implied concatenation? Seems to me, it should be multiplication if you wanted to claim it is an algebraic statement.
posting before answer is shown but i got it figured out O=Any # > 4, D=5, E=1 N=0 ,and V is a little tricky but first i will create a new variable x which is for solving for v, x = 9 - O, And v = O - X
So for those doing this at home and are reading the comments before watching the full video, two important facts that weren't mentioned initially: 1) Each letter represents a _unique_ integer. 2) There are two solutions. (If the letters don't represent unique integers, there are four solutions).
he doesn't have to state that each letter is a unique integer as that is very obvious else there would've been hundreds of solutions such as 666+666= 1332, 667+667= 1334 etc
Using the puzzle as presented at the start (can have duplicate numbers) 3digt+3digt=4digit means O>4 and E=1. xDD+xDD=1x1x means D=5 because that's the only way to get 1 in the tens column of the answer. This leaves 555+555=1110 655+655=1310 755+755=1510 855+855=1710 955+955=1910 and I can't see any way to improve that. Because you can't have duplicate numbers (as Brian added later) 655+655=1310 and 855+855=1710 so I don't know how you solve it. Oh, it turns out there are TWO solutions, Well fuck that.
I just looked at the d and said ok what double digit will set me up like this at the end. idk I'm good at math but I actually did 55+55 first and I figured out there was actually multiple answers before I even finished the puzzle but I chose 655.
I worked out that e has to be 1 and d has to be 5 but from what i can see from my own math is that odd can be either 655, 855, or 955 and idk how to narrow it further because all results show 1_1_ with 2 unique numbers in the spaces.
There are plenty of more interesting puzzles of this sort. I've seen addition and subtraction at the same time, multiplication, long division, puzzles where you are only given the number of digits and where one digit is (so something like -x+-x=x--, which has multiple solutions, but there are some with unique solutions) ODD + ODD = EVEN is just the tip of the iceberg Another well known one is SEND + MORE = MONEY Which has the unique solution 9567+1085=10652
He's needs to include concatenation signs, or else they're all factors, it would be O times D times D. 2(ODD)=EVEN is how it could be solved. Then ODD equals ODD when divided by two. So it's not even really a problem. Just fill in numbers.
I had it before they started working it out but then you said to focus on odd and even so i started thinking it was wrong because 6 and 8 weren't odd...i still got it in about a minute and a half so I'm happy anyway xD
I was doing this by myself and FINISHED IT, but then I thought there was a fault in the problem because I got two answers. BUT. When Brian said they only got one of two... I felt like a genius.
"I'm not helping you guys anymore!"
"Let me just give you one more piece of advice..."
Lol Brian.
Once you have eliminated the impossible whatever remains, however improbable, must be the truth.
"The V can't be the same as the D"
- Brian Brushwood 2017
Try telling that to an activist
@@spicytrashpanda damn I forgot I said this 😭😭😭 I'm queer now 🏳️⚧️🏳️🌈
Before Brian said that they were all different numbers, I though this was a really bad puzzle... 000+000=0000
Tommy Dancer Zero isn't odd.
Kyle Kern Did he state that the number had to be odd? (Also zero's not even either) (Never mind I got to the point in the video he said it)
Zero is even. It fits all the criteria.
And that's numberwang!
Tommy Dancer 59+41=100
That bald signal is just the moon.
there's actually a whole book with stuff like this, Sideways Arithmetic from Wayside School by Louis Sachar
Those domain.com ads are so good I think I'll buy a domain anyway even though I don't need one
655 + 655 = 1310
I just literally started by plugging in a 5 for D, saw it fit, plugged in a 6 for O, saw that fit, and suddenly was done. Oops.
I actual like your sponsor portions of your videos because they are always new and interesting. (I also love math games. I may not be good at them, but I like them.)
I wish they would have said that there were two correct answers at the START of the video. I paused it and wasted a whole lot of time trying to figure out why I couldn't find a unique solution.
This was indeed not cool. Even a 'there might be more than one answer' or 'find all possible answers for a beer' but some way to let us know or suspect that there is more than a unique answer would be nice. But then again Brian also got in the way of the crew working the puzzle a number of times in his excitement. Not a great episode.
But thank you for your comment, because i scrolled down to see if someone discussed this exact issue.
Glad it wasnt just me!
If a puzzle has multiple correct answers, that you get in a very obvious way... not a good puzzle.
Or, to put it another way, this wasn't a puzzle. Just a basic equation you'd see in middle school.
I'm glad your stuff got better over time ^_^ can actually see you evolve from a "meh" channel to something that puts everyone else to shame. Kudos.
655 and 855 work, but 7 and 9 repeat. Oh, you got it as I wrote this. As you were!
I'm always checking previous videos and current ones for stuff I haven't watched yet. I love scam school. I think Brian should collaborate with other cons and magicians like he does with Diamond Jim. That guy is great too.
Did it without using pen/paper. Many of your puzzles are too easy man, some are insanely awesome.
I spent about 10-15 minutes on this when he gave the prompt. Was super happy with myself when I got it.
Oh. I missed something that would have made it easier and faster. Eh. Still got it.
I know I'm bragging here, but I found both answers in less than two minutes after watching just the intro. Aside from the bragging, I would like to say that consistently watching Scam School is one of the reasons I was able to do this.
I figured out the answer insanely quick on this one. I love that type of puzzle and I've done quite a few of them. I even made them at one point.
paused the video and did the puzzle myself and ended up scratching my head because I had two possible solutions. Watched the rest of the video and found out they are both correct and now I feel like a genius :D.
Wow! first time I won ahead of the others. I just went at it differently but also came out with the same answer option. Loved it!
SPOILER
My best guesses are
655+655 = 1310
855+855 = 1710
I was getting mad becasue i thought there was only one answer and I couldn't logically figure out why both set of numbers couldn't work...then he said there were 2 answers lol
If yo take the remaining numbers that don't get used in either solution you get 2, 4, and 9. if you turn those into letters with a simple number-to letter cipher you get B, D, and I. If you rearrange the letters it can spell "Bid"! So if you want have fun with in you can present it with: "Solve this puzzle and you will see; what remains will be the key; a proper welcome from one to thee." And with a glorious red herring in the 123 pun at the end.
King Kasper very clever
I have to admit. That ad in the video was so smooth.
Really good puzzlers used here. Great vid.
'You saw the bald signal'
...That's the moon, Brian.
Gave up after finding the right answers because I thought there could only be one answer. Gdi Brian.
I was mad and thought I got it wrong because there were two answers. It was messing with me and I didn't wanna spoil it. Glad I finally gave up to see I was right!
"You saw the bald signal..."
The bald signal? You mean just a regular signal light? 😂😂😂
before solution:
o must be 5-9 (Brian's hint)
e must be 1 (three digit number plus three digit number can't be more than 2000)
d must be 5 (third digit in EVEN must be one, so dd + dd must leave a 1 in the tens place)
n must be 0 (55 + 55)
if o is 5, 6, 7, 8, or 9, v will be 1, 3, 5, 7, or 9 respectively
after number can't repeat rule:
655 + 655 = 1310 or 855 + 855 = 1710
after solution:
lit, got it right
Loved the ad read
As soon as Schwood said algebra I thought: (O*D*D) + (O*D*D) = E*V*E*N. Glad it wasn't that complicated.
like a $0.25 paperback mystery novel, you changed the rules at the very end.
E has to be 1, O has to be grater than 5. D has to be 5 which forces N to be 0. I'm pretty sure those are the only requirements. For example: 555 + 555 = 1110; 655 + 655 = 1310; 755 + 755 = 1510; 855 + 855 = 1710; 955 + 955 = 1910.
Since ODD+ODD is 5 digits, O>4 and E = 1 Since DD + DD = 1N or 11N, D = 5 and so N=0. 2(O55) = 1V10, O can be 5,6,7,8, or 9, and V will be (2O-10)+1 = 2O-9. If numbers can't be repeated, then O isn't 5, it's also not 9 since then V=9, and it's not 7 since that makes V=5. Thus, O = 6 or 8, and V = 3 or 7, respectively.
ODD+ODD=EVEN
655+655=1310
855+855=1710
I sat there a good fifteen minutes thinking that their couldn't be two right answers. Then gave up and watched the rest, guess I was right!
As an algebra lover, this was painful to watch.
One does not simply try out the numbers, it's much simpler and more professional to just write down equations (e.g. 2D=N and 2D=E or 2D=N+10 and 2D+1=E) etc.. Eventually solving em all.
By the way for anyone interested there's an app called cryptogram that gives a few of these problems, I think they're really fun.
Trevor is finally in an ad! \o/
Trevor is great!
More than that, he's finally back
bald guy with glasses is the mvp of this group.
Paused the video before the two minute mark and in about a minute got the values. If you haven't figured it out, don't keep reading, but O = 6, D = 5, E = 1, V = 3, and N = 0. Educated guesses FTW!
Dude, I paused for a while to try and figure it out, I was stumped because I got down to 2 valid answers (655 and 855) and I played the video to find a mistake, but yall confirmed I was right!
Do the 12 balls, 3 weighing logic puzzles. It would probably be the hardest puzzle you have done on this channel.
1. E must be 1 because two 3-digit numbers added together to create a 4-digit number must be at least 1000, but _cannot_ be 2000.
2. (Since that makes the 3rd digit of EVEN also 1), DD + DD must be 55 + 55 because it is the only combination of two 2-digit same-digit numbers (11, 22, 33, 44, 55, 66, 77, 88, 99) added together that results in a 1 in the tens column.
3. Therefore, N must be 0, since 55 + 55 = 110.
4. O + O + 1 (carried from DD + DD) is at least 10 (as EVEN is 4 digits), so O must be at least 5. There is not enough information to solve, but possible solutions that do not repeat any digits from other solutions are 6 and 8. (655 + 655 = 1310, 855 + 855 = 1710). 555, 755, and 955 do not work because they repeat digits: 555 + 555 = 1110 (that obviously repeats a few), 755 + 755 = 1510 (D and V would both be 5), and 955 + 955 = 1910 (O and V are both 9).
One thing he didn't mention in the video is that these equations are called "Cryptarithmetics". I see some of you in the comments found this easy. if you want a challange, try this one: SODAS - AIM = BEER
@Brian This video went excelently! I'm so glad you got the sudoku aspect of the puzzle! This is how you can tell some tru puzzles apart from other... it's also why Minesweeper can be a brilliant scam of a puzzle.
here are all 48 answers to the soda beer
#1
sodas: 2450
aim : 563
beer : 1887
~~~~~~~~~~~~~~~
#2
sodas: 2450
aim : 567
beer : 1883
~~~~~~~~~~~~~~~
#3
sodas: 3450
aim : 561
beer : 2889
~~~~~~~~~~~~~~~
#4
sodas: 3450
aim : 569
beer : 2881
~~~~~~~~~~~~~~~
#5
sodas: 3460
aim : 681
beer : 2779
~~~~~~~~~~~~~~~
#6
sodas: 3460
aim : 689
beer : 2771
~~~~~~~~~~~~~~~
#7
sodas: 3560
aim : 671
beer : 2889
~~~~~~~~~~~~~~~
#8
sodas: 3560
aim : 679
beer : 2881
~~~~~~~~~~~~~~~
#9
sodas: 4250
aim : 581
beer : 3669
~~~~~~~~~~~~~~~
#10
sodas: 4250
aim : 589
beer : 3661
~~~~~~~~~~~~~~~
#11
sodas: 4560
aim : 671
beer : 3889
~~~~~~~~~~~~~~~
#12
sodas: 4560
aim : 679
beer : 3881
~~~~~~~~~~~~~~~
#13
sodas: 4690
aim : 912
beer : 3778
~~~~~~~~~~~~~~~
#14
sodas: 4690
aim : 918
beer : 3772
~~~~~~~~~~~~~~~
#15
sodas: 5190
aim : 963
beer : 4227
~~~~~~~~~~~~~~~
#16
sodas: 5190
aim : 967
beer : 4223
~~~~~~~~~~~~~~~
#17
sodas: 5690
aim : 912
beer : 4778
~~~~~~~~~~~~~~~
#18
sodas: 5690
aim : 918
beer : 4772
~~~~~~~~~~~~~~~
#19
sodas: 6230
aim : 341
beer : 5889
~~~~~~~~~~~~~~~
#20
sodas: 6230
aim : 349
beer : 5881
~~~~~~~~~~~~~~~
#21
sodas: 6280
aim : 831
beer : 5449
~~~~~~~~~~~~~~~
#22
sodas: 6280
aim : 839
beer : 5441
~~~~~~~~~~~~~~~
#23
sodas: 7230
aim : 341
beer : 6889
~~~~~~~~~~~~~~~
#24
sodas: 7230
aim : 349
beer : 6881
~~~~~~~~~~~~~~~
#25
sodas: 7340
aim : 451
beer : 6889
~~~~~~~~~~~~~~~
#26
sodas: 7340
aim : 459
beer : 6881
~~~~~~~~~~~~~~~
#27
sodas: 7280
aim : 831
beer : 6449
~~~~~~~~~~~~~~~
#28
sodas: 7280
aim : 839
beer : 6441
~~~~~~~~~~~~~~~
#29
sodas: 7380
aim : 821
beer : 6559
~~~~~~~~~~~~~~~
#30
sodas: 7380
aim : 829
beer : 6551
~~~~~~~~~~~~~~~
#31
sodas: 7490
aim : 932
beer : 6558
~~~~~~~~~~~~~~~
#32
sodas: 7490
aim : 938
beer : 6552
~~~~~~~~~~~~~~~
#33
sodas: 8290
aim : 954
beer : 7336
~~~~~~~~~~~~~~~
#34
sodas: 8290
aim : 956
beer : 7334
~~~~~~~~~~~~~~~
#35
sodas: 9130
aim : 354
beer : 8776
~~~~~~~~~~~~~~~
#36
sodas: 9130
aim : 356
beer : 8774
~~~~~~~~~~~~~~~
#37
sodas: 9270
aim : 714
beer : 8556
~~~~~~~~~~~~~~~
#38
sodas: 9270
aim : 716
beer : 8554
~~~~~~~~~~~~~~~
#39
sodas: 10231
aim : 345
beer : 9886
~~~~~~~~~~~~~~~
#40
sodas: 10231
aim : 346
beer : 9885
~~~~~~~~~~~~~~~
#41
sodas: 10241
aim : 463
beer : 9778
~~~~~~~~~~~~~~~
#42
sodas: 10241
aim : 468
beer : 9773
~~~~~~~~~~~~~~~
#43
sodas: 10251
aim : 584
beer : 9667
~~~~~~~~~~~~~~~
#44
sodas: 10251
aim : 587
beer : 9664
~~~~~~~~~~~~~~~
#45
sodas: 10281
aim : 835
beer : 9446
~~~~~~~~~~~~~~~
#46
sodas: 10281
aim : 836
beer : 9445
~~~~~~~~~~~~~~~
#47
sodas: 10381
aim : 824
beer : 9557
~~~~~~~~~~~~~~~
#48
sodas: 10381
aim : 827
beer : 9554
~~~~~~~~~~~~~~~
this was insanely easy I knew the answer right away
Solved it in my head in a couple of minutes. Once I realized E was 1, it was quick to get D was 5 and then it was easy.
Paused and got it in 6 minutes.
But I was sober, so I guess I didn't do any better than Team Melon Head.
Finally all of the math I learned in High school and collage paid off :p
You can also solve it graphically, although its a bit mathy...
First do some algebra and rewrite the problem:
2(ODD)=EVEN
2(100*o + 10d+d)=1000e+100v+10e+n
200o+22d=1010e+100v+n
Next we know that the one's place of d+d has to equal n because it is the last letter, we can write this algebraically as follows:
mod(2d,10)=n
We also know that e has to be 1 based off of the logic explained in the video
So plugging these into the above equation we get:
200o+22d+1010 + 100v + mod(2d,10)
We now have three variables but we know that o can only be 5,6,7,8,9
and so we can write an equation for each of these values
Now we have 5 equations which we can graph as a function of v
For simplicity I have made the following substitutions:
v=y, d=x
This gives us:
y = (200*o +22d - 1,010)/100
Graphing this, as I did here: www.desmos.com/calculator/dreddhgqse
allows us to visually see that the only integer solutions to the equation required x to be 5
when x (which is d) is 5 in each of our six equations we see that y (which is v) must be 1,3,5,7,9 when o is 5,6,7,8,9 respectively
We now know that these are the only possible answers and they give us the following
555 + 555 = 1110
655 + 655 = 1310
755 + 755 = 1510
855 + 855 = 1710
955 + 955 = 1910
Finally, we can sort through these to see that they satisfy o=/d=/e=/v=/n
and we are left with
655 + 655 = 1310
855 + 855 = 1710
And, well, this is exactly what they found in the video... but at least we have a proof for it now :p
Brian was a proud parent in this episode.
I solved this in less than a minute... and I kept yelling THIS IS IMPOSSIBLE!!!! because there wasn't just one correct answer. Gah.
Not really algebra, more like a simple cypher, if it was algebra you would simplify to 2O(D^2)=(E^2)VN
But it's actually 200×O + 22×D = 1010×E + 100×V + N
hey Brian! is there any chance you could do a compilation of tricks with pens? I thought about this because I'm in high school and once in class I was thinking "wouldn't it be cool if I were just vanishing a pen repeatedly in the back of the class until someone just turned around and said :how the hell are you doing that?!"
Whoever the video needs a raise!
Hah - "Hope y'all got degrees in Russian finance" - Yeah, sage advice it seems ;)
I'm glad it didn't turn out to be "The square root of 28491 times 2/59 to the power of 8 PLUS ...... = ......"
1:44
Oh, cool, like Sideways Arithmetic from Wayside School! (In fact, they probably used this one, but I've since forgotten it.) Okay, let's figure this out...
Two digits added together will never be greater than 18, so if you carry anything, it will be a 1.
..ODD
+ODD
_____
EVEN
With that said, there was nothing in the thousands column, but we ended up with E. So we know E is 1:
..ODD
+ODD
_____
1V1N
Now, looking at the tens column, we know D+D should be even, but there's a 1 under that sum. So there must be a 1 carried over from the ones column, and D+D ends in 0.
But the units column, which we just said must carry a 1, is D+D. Since D+D both ends in 0 and carries a 1, it must be 5+5=10 (which also means that N must be 0).
..O55
+O55
_____
1V10
Now we know there's a 1 carried over to the 100s column, (and that the result of the sum "carries" over to the 1000s column), so O+O+1=V+10. Rearranging a little, 2*O=V+9. Now if V were even, V+9 would be odd, but it's obvious that 2*O is even, so V must be odd. (And so 2*O=V+9 an ODD no. plus an ODD no. equalling an EVEN no., reminiscent of the problem itself!)
I assume the digits are different (other comments seem to confirm that he eventually says this), so V can't be 1 or 5. That leaves 3, 7, and 9. We'll have to test these through brute force.
Solving for O, we know O=(V+9)/2. That should give us a good starting point. And the key to eliminating possibilities is that all letters are different digits, so if one letter (in particular, O) is the same as another, we have a contradiction and our premise must be wrong.
If V=3, O=(3+9)/2=6
If V=7, O=(7+9)/2=8. That could be.
If V=9, O=(9+9)/2=9. But then O would be the same as V, so that can't be.
So I can't work it out right now, but I've only started the video. Maybe there's a rule that none of the letters are in the names of the digits they represent? Then it would have to be
..855
+855
_____
1710
but I haven't heard all the rules yet
I solved this in like five mins. this is one of your easier puzzles thank you.
Saw that E had to be one, figured D was probably five making N zero. Then just used six as my first option for O to try it out and it was right. Took about 30 seconds to get the solution, but I didn't think to check for any other solutions.
If we force rule that different letters correspond to different digits, then we are left only with:
655 + 655 = 1310, 855 + 855 = 1710
It's much easier to solve if you write it like it's long addition:
_ODD
_ODD
EVEN
you instantly get E = 1, so it is
_ODD
_ODD
1V1N
and here we have not much left for letter D: two of them have to sum up to odd number 1, so there has to be carry out from previous position, and it was D as well, so D is at least 5. Also we have to have exactly 1 so without carry it has to be 0, so D is either 0 or 5, but previously we deduced that it is at least 5 therefore it's 5, so now we have:
_o55
_o55
1V1N
after that we get that N = 0 and for o we've got only 4 different values we need to check, 6, 7, 8 and 9, from which only 6 and 8 satisfy condition that all digits have to be different.
Lol, this is pretty much majority of things which children do at 4th-5th grade in my country and this is quiet useful for developing logic and ability to deduce things, I don't understand why it took adults so much time, it's very easy.
"The V cannot be the same as the D." Remind Bud Light drinkers this everydaym
The equation for this problem is E times the square root of 3 plus V times the square root of 3 plus N times O = 2 times D plus 2 times D. This means only 855+855=1710 can be the answer. The other, 655 is a false positive.
Trevor with the STL Cardinals hat 🙌🏻
If they were allowed to do doubles, which Brian said wouldn't matter (2:40), then 755 + 755 = 1510 and 955 + 955 = 1910 are both legit answers
For those interested in a mathematical proof of the solutions, I came up with this (feel free to correct me):
Let's find the digits O,D,E,V,N making
ODD + ODD = EVEN
Because the two same 3-digits numbers sum up to a 4-digits number, we know that O ≥ 5. We also know that two 3-digits numbers can't sum up to more than 1998 (in general x-digits number summing up to a (x+1)-digit number will be 1-something). Which means that E = 1.
Let's suppose that D < 5. Then
D + D = N
D + D = E
But that's impossible, because all digits are different. Which means that D ≥ 5. This means that
D + D = 10 + N (N is the last digit of a number between 10 and 19)
E = (N + 1)%10 (that's D+D plus the 10 of the previous one)
EV = O + O + 1
So far we have (solving the equations above)
E = 1
N = 0
D = 5
This leaves us with O and V. We know that O ∈ {6,7,8,9}. If we add 1 to the double of each of these numbers, we get V. So V ∈ {3,5,7,9}. But each digit is unique, which leaves us with
(O, V) ∈ {(6, 3), (8, 7)}
Which leaves us with two only possible solutions
655 + 655 = 1310
855 + 855 = 1710
Please do correct!
Problem: Find all combinations of digits that satisfy this pattern, each number must be different:
ODD + ODD = EVEN
O=6,8
D=5,5
E=1,1
V=3,7
N=0,0
Or:
655 + 655 = 1310
855 + 855 = 1710
I believe these are all the solutions.
Logic:
First e must equal 1, because it's two three digit numbers, and there is no way for three two digit numbers to be added to produce a number greater than or equal to 2000. (999+999 = 1998, the highest we can go). Therefore, the target number is bounded between 1000 and 1998.
For O since the answer is bounded between 1000 and 1998, we know O must be greater than or equal to 5, and less than or equal to 9. That's all we know for now, we'll come back.
For D, we need to produce the 1v1n pattern, so the 10s place must have a 1 in it. The selected 0 doesn't affect the 10s place, it only affects the 100s and1000s place (we can convince ourselves by this line of reasoning:
ODD + ODD = 2*ODD = 2*(O*100 + D*10+D) = (200*O + 20*D+2*D), so we can clearly see for the 10s place and 1s place, we see that O will be multiplied by 0. and will have no effect, but D will.
. That means DD*2 must have a 1 in the 10s place. We can try all 10 numbers if we want (00,11,22,33,44,55,66,77,88,99) to convince ourselves that 110 is the only number that will produce this, so D *must* equal 5, since 55*55 = 110.
Okay, now that we've got that sorted, we know N has to be 0.
Then V depends on our selected O. Again, we try all 5 values of O (5,6,7,8,9) and we find that the only ones that have a digit in the V place that isn't already taken are 6 and 8.
Edit: I just saw their solution part of the video, just to explain why O=9, O=7, and O=5 don't work (for the sake of completeness of my own solution), it is because the digits are already used. D is already 5, so O can't be that, O=7 doesn't work, because it gives 1510, but then V=D=5, which isn't allowed, and O=9 doesn't work because it gives 1910, so then O=V=9, which isn't allowed because they say you must have a different number of each letter. :)
1:52
Alright, right off the bat...ODD+ODD=EVEN. ODD is a 3-digit number and EVEN is a 4-digit number. Because ODD is 3 digits, EVEN has to be between 1000 and 1998 (2 * odd# = even#). This means that E = 1. N MUST be an even number because it's 2 * D. Since EN has to be 1X where X is even, the options for D are limited...if D is 1, for instance, 1+1=2=N and 1=E=1+1=2, which is obviously wrong. Therefore the sum of D+D>9 because it must affect the outcome of E. If D=5 then 5+5=10 meaning N=0, carry the 1, 5+5=10 add the remainder 1 to make 11, carry the 1, you have DD+DD=110 and EN=10. In fact, D has to equal 5 as it's the only number which makes E=1 in that scenario.
So E=1, N=0, D=5. O>4 because ODD*2 has to be >999. O=/=5 because 5=D. If O=7 then you have 755+755=1510, meaning that D=V. If O=9 then 955+955=1910 and O=V. This leaves 2 solutions of: 655+655=1310 and 855+855=1710.
I solved this really fast. The numbers I used were 0, 6, 5, 1, and 3. They are in no particular order, but those are one solution.
Brian, where/when do you announce the winners of the throwing cards giveaway thingy.....
O=6, D=2, E=4, V=3, N=1 was my first thought, then they went on about each letter representing a part of the number instead of multiplication
I could see this trick being effective with drunk people.
Nobody saw the letters on the refrigerator door that said "hope yall don't degress Russian finance"
Either I'm getting to smart or the puzzles are getting to simple for me to think there actually are puzzles
Brian got scammed on this so bad. He loses a beer either way. Choose your words wisely!!
I honestly just stumbled across the right answer when I was going through the options for D. Assuming each letter had to be a different value, and the fact 5 worked to solve for E, I just moved O to 6.. o.o
Says, "algebra question", then has lets represent digits with implied concatenation?
Seems to me, it should be multiplication if you wanted to claim it is an algebraic statement.
I got to where they got with 655 or 855. I didn't realize having two answers was acceptable so I finished watching. Now I am slightly disappointed lol
posting before answer is shown but i got it figured out O=Any # > 4, D=5, E=1 N=0 ,and V is a little tricky but first i will create a new variable x which is for solving for v, x = 9 - O, And v = O - X
love this show
So for those doing this at home and are reading the comments before watching the full video, two important facts that weren't mentioned initially:
1) Each letter represents a _unique_ integer.
2) There are two solutions.
(If the letters don't represent unique integers, there are four solutions).
he doesn't have to state that each letter is a unique integer as that is very obvious else there would've been hundreds of solutions such as 666+666= 1332, 667+667= 1334 etc
I love how this was uploaded 10 minutes ago and already has about 450 views
Spoiler
Any number from 655 to 955 that ends in *55 will work
For 4 minutes I couldn't decide why I kept getting 2 answers
Almost got it. In the end I did not realize 955+955=1910 is not a valid answer.
I'm confused.. is there a c in schwood?
Martin Sobotka Of course. It comes from his last name, Bruschwood.
ah ah ah! Brushwood. there is no "c"
Dafuq? They worked it out and Brian talked them out of it.
It's an evenly odd number.
watching this as a tornado is Goin over my house
Live in NC?
ohio
Ohio... Definitely Ohio.
Ohio is a shit show right now
I slept through it man
Brian Brushwood Whats the difference between russian finance and american finance, wouldnt they be the same aside from curancy conversions
Love the vids Bri Guy!
Awesome to see Fraser Cain!
Using the puzzle as presented at the start (can have duplicate numbers)
3digt+3digt=4digit means O>4 and E=1. xDD+xDD=1x1x means D=5 because that's the only way to get 1 in the tens column of the answer. This leaves
555+555=1110
655+655=1310
755+755=1510
855+855=1710
955+955=1910
and I can't see any way to improve that.
Because you can't have duplicate numbers (as Brian added later) 655+655=1310 and 855+855=1710 so I don't know how you solve it.
Oh, it turns out there are TWO solutions, Well fuck that.
I just looked at the d and said ok what double digit will set me up like this at the end. idk I'm good at math but I actually did 55+55 first and I figured out there was actually multiple answers before I even finished the puzzle but I chose 655.
Figured it out in about a minute...
I worked out that e has to be 1 and d has to be 5 but from what i can see from my own math is that odd can be either 655, 855, or 955 and idk how to narrow it further because all results show 1_1_ with 2 unique numbers in the spaces.
Nvm -_- damn "odd+odd=even" being a legit rule
and nvm again.....i got 3 correct lolFUCK
2 -_-
There are plenty of more interesting puzzles of this sort.
I've seen addition and subtraction at the same time, multiplication, long division, puzzles where you are only given the number of digits and where one digit is (so something like -x+-x=x--, which has multiple solutions, but there are some with unique solutions)
ODD + ODD = EVEN is just the tip of the iceberg
Another well known one is
SEND + MORE = MONEY
Which has the unique solution 9567+1085=10652
He's needs to include concatenation signs, or else they're all factors, it would be O times D times D. 2(ODD)=EVEN is how it could be solved. Then ODD equals ODD when divided by two. So it's not even really a problem. Just fill in numbers.
I got the 655+655=1310 Answer in a random way of logic. I am surprised I got it actually.
problem with this is that the letters are added together if this where a real math problem. example O*D*D + O*D*D = E*V*E*N.
I wrote up a 28 line program that solved it in 0.11389565467834473 seconds.
I had it before they started working it out but then you said to focus on odd and even so i started thinking it was wrong because 6 and 8 weren't odd...i still got it in about a minute and a half so I'm happy anyway xD