I Will Predict Your Sum. Trick Explained

[@ObsidianParis]

The trick doesn't amaze me as much as the fact that I randomly picked 7-10-1-16 at first…

[@Bradleybus]

Dude thats crazy. I picked 1-7-10-16. Different order but same numbers

[@madanmatcha7484]

Same here! 10 - 7 - 1 - 16 for me.

[@reedeek1473]

Same

[@RyanG-ks9ev]

Me too. Different order, same numbers!

[@SacoSilva]

10-7-4-13 for me, still got spooked by the first 2

[@Yehan-xt7cw]

Just sum the average of each row. Meaning: row 1: 2.5 row 2: 6.5 row 3: 10.5 row 4: 14.5 ---------------- + total: 34 Since each column is used once, each deviation (-1.5, -0.5, +0.5, +1.5) of the average is used once, which means they cancel out, leaving only the sum of averages.

[@deerh2o]

I like your approach. I did something similar. I started with the average of the whole shebang (the numbers from 1 to 16): 8.5. Now: 8.5 x 4 = 34. For the 5 x 5 grid, the average of the first 25 numbers is 13: 13 x 5 = 65.

[@osliverpool]

I went for... highest column total = 40 but then you have to choose -1, -2 and -3 regardless of how you switch to other columns. Or lowest column total = 28 and then you have to choose +1, +2 and +3. (Wrote that before I watched the solution, so let's see...)((aha!))

[@craigfjay]

General formula is: (N^3 + N) / 2 Grid size is NxN. Divide every number in each row by N (and round down), the numbers in each row are 0,1,2,3,4…(n-1). You have one number from each row, so sum these up. It’s the triangular number for n-1: (n^2 -n)/2. Multiply by N, since the digit in any row is the digit above plus N: (N^3 + N^2) / 2 Now add the remainders up for the digits when divided by N. These will be 1,2,3,4…N. The triangular number for N: (N^2 + N) / 2. Add these together and the N^2 terms cancel out, leaving the general case formula: (N^3 + N) / 2

[@kennyalbano1922]

It even technically works for the base cases of a 1X1 and 2X2 grid

[@hobbesnonumbers]

you could have generalized it even more to an NxN grid, replacing ri and ci with just x where the new function becomes: SUM[X=1->N](N*X+X-N) which then simplifies to (N^3 +N)/2

[@cmuller1441]

Each line is 1 2 3 4 But +0 for the 1st, +4 for the 2nd, +8 for the 3rd and +12 for the last. You choose one number per column. So you basically choose 1+2+3+4 plus +0+4+8+12 ie 10+24=34

[@Micboss1000]

This is the same method I found.

[@shambhav9534]

Similar solution to mine. Divide the rows into four parts. (1, 2, 3, 4) (5, 6, 7, 8) (9, 10, 11, 12) (13, 14, 15, 16). You have to pick one from each row. So least sum, taking the least numbers from each of the groups is 1 + 5 + 9 + 13 = 28. But, you also have to pick one from each column. Think of column number as deviation from the first number. That is, 1 is 1 + 0, 2 is 1 + 1, 3 is 1 + 2, and 4 is 1 + 3. The numbers to be added are 0, 1, 2, 3. Recall that we have to use each column at least once, so each of the four numbers has t be used at leat once. So, let's add them. 0 + 1 + 2 + 3 = 6. Finally, 28 + 6 = 34.

[@djdoc06]

It’s easier to conceptualize if you use a 10x10 matrix. Each row is the 10s place. Each column is the 1s place. If you pick all different rows & columns, your sum has to simply be 0+10+20+30+ …90 + 1+2+3+…9+0 since every column and every row is represented exactly once.

[@nawdrawg8436]

Just want to say I've followed you for a few years now, and your channel, your knowledge, your presentation, and your posiivity are all a delight! Thank you for all your amazing work! Keep on keeping on with many thanks 🙏🙂

[@JLvatron]

Here's a number magic trick: -Pick an integer between 1 and 3, but not including 1 and 3 -multiply by 17 -subtract 3, then take the square root -Watch your 2nd favorite (available) sitcom -Read your 4th email from the top I can guess your original number was, ...2. Kinda takes yer breath away, don't it?!

[@majbach1968]

That's about what I got out of it

[@muskyoxes]

I can't get to the sitcom because i'm not done with the square root. The decimal doesn't seem to run out...

[@MostLovedGod]

Such is a way of a rotten monkey... copying the form without function.

[@proboiz_50]

It takes my life away ,bro. How did you do that? There was 1 total possibility and still you guessed it right👏👏 btw what's sitcom?

[@JLvatron]

@@proboiz_50 haha, I have a talent for guessing 1-possibility numbers! In US television, comedies are usually called 'Situation Comedy' = Sitcom. Usually a 22 minute show, 30 minutes including commercials.

[@forcelifeforce]

*@ MindYourDecisions* -- The special sums are the same as the Magic Constants as the Magic Squares for each appropriate square of numbers.

[@glennsampson5945]

I've seen this trick before. Martin Gardner wrote about it. A much simpler explanation is that the grid is simply an addition table. Write to the left of each successive row, the numbers 0, 4, 8, and 12. Write above each successive column, the numbers 1, 2, 3, and 4. Each entry in the table is the sum of the corresponding row number and column number. Picking an element in the table corresponds to picking its row number and column number. Since each element picked is in a different row and column, you will pick each row number once and each column number once. So the total will be the sum of the row and column numbers which in this case is 0 + 4 + 8 + 12 + 1 + 2 + 3 + 4 = 34. This generalizes quite easily. Pick any numbers you like for each row and column, fill in the addition table, and the sum will be the sum of the numbers you picked for the rows and columns.

[@muskyoxes]

Then that would totes work for a multiplication table. Build it from 1,2,3,4 and their four numbers will always multiply to 576

[@glennsampson5945]

@@muskyoxes Nice observation.

[@jakubdraws395]

I picked 1, 7, 10, and 16! What black magic is this?!

[@darksidegaming9806]

Wow! But there is always math behind these magic looking tricks :)

[@markdagley4213]

For odd numbered grids, multiply the center number by number of rows; For even numbered grids multiply average of center diagonals time the number of rows.

[@z000ey]

the formula for any number of rows and same number of columns (the number being "n") is: n*(n^2+1)/2 this comes from the arithmetic sum of all numbers in the whole table where you've N=n^2, but is divided by the number of rows or columns (not both at the same time) as that is how many same sums there are in the sum of all numbers. Thus you've N*(N+1)/2 (geometric sum of all numbers in table), insert n^2 instead of N, divide by n and you've got it. If you want to look it through the biggest number in the square table (the N), just calculate directly through N*(N+1)/2*sqrt(N), as N will always be a square number you always get an integer, but it's easier through the number of rows and columns

[@n00bxl71]

Before watching the video, I thought of it this way. Each row has a constant, and so does each column. Every next column adds 1 to the total, starting with 1. Every next row adds 4 to the total, starting with zero. So, r1=0, r2=4, r3=8, r4=12 And c1=1, c2=2, c3=3, c4=4. So any given position, such as row 2, column 3, can be calculated by adding r2 and c3, 4+3=7. The important thing to realise is that we are being made to choose exactly 1 from every row and from every column. So the total will be 1+2+3+4 (columns) + 0+4+8+12 (rows) = 34

[@wmpowell8]

Every number is 1 + a + 4b, where a and b are 0, 1, 2, or 3 depending on the row or column. Every row and column is used once, so the sum of all 4 is 1 + 1 + 1 + 1 + 0 + 1 + 2 + 3 + 0 + 4 + 8 + 12 = 4 + 6 + 4*6 = 34. The 4 in the simplified sum is the side length and 6 is the triangle number (N-1)(N)/2 = (N^2-N)/2 corresponding to side length N-1 where N is the square side length. One factor of (N-1)(N) is always even and the terms in N^2-N are either both even or both odd, which is why they are always even. So the general solution for the magic number for side length N, as others have pointed out, is N + (N^2-N)/2 + N*(N^2-N)/2 = N + (N*N^2+N^2-N*N-N)/2 = (N^3-N)/2 + N = (N^3 + N)/2.

[@ಠಎಠ]

Every cell value can be thought of as a difference, (X-N). A cell's X value is determined by its row, and is equal to the right-most number in that row - so X will be 4, 8, 12 or 16. A cell's N value is determined by its column - going left to right, the N values for the columns are 3, 2, 1, 0. You're asked to pick four cells, ensuring that none of them share the same row or same column. So, regardless of which cells you pick, all four X values and all four N values are represented exactly once. No matter how the X & N values are paired, the sum remains the same: (4 + 8 + 12 + 16 - 0 - 1 - 2 - 3) = 34

[@MrNostril]

My solution was you're ultimately asking the pariticipant to choose 4 numbers that follow the form 4×0+a, 4×1+b, 4x2+c, 4×3+d where a, b, c, and d are 1, 2, 3, and 4, but not necessarily respectively, the particpant's gets to decide which variable goes with which value, but all 4 values, must be used. Because it doesn't matter what order you sum numbers a+b+c+d=10 regardless of how the values are assigned. So, 4x0+a+4×1+b+4x2+c+4×3+c = 4×6+(a+b+c+d) = 24+10 = 34

[@brightwriter]

Choose one cell from each row and one from each column, as per instructed, and note the sum. Start over and repeat, such that no previously selected cell is chosen. Repeat twice, again never selecting a previously chosen cell. You now have four totals. Add up the number in all the cells 1 + 2 + ... + 16, which is ½(16)(17) or 136. But this is four times the average number of the 4-cell selction totals. Therefore, divide 136 by 4, which is the number of groups of 4-cell selections. Answer 34. This method makes the undefended assumption that the four cells selected per the rule always add up to the same number, but if it is accepted then my method will work.

[@rbid]

Remembers me what is called a 'diabolic square', when you reorder the 16 numbers in this way: 1 15 14 4 12 6 7 9 8 10 11 5 13 3 2 16 Here all rows sum 34, all cold sum 34, the diagonals sum 32, even all 4 numbers that are symetrically placed Greetings from the Galilee.

[@lucky_loot]

1:57 you scared me for the whole minutes thinking that you read my mind, you are so go that your example IS my number sequence, am i that predictable

[@JMMC1005]

Haha I had the same thing happen. I went in a different order but picked the exact same numbers. He's good...

[@vcvartak7111]

Easiest proof of for getting sum (n^3+n)/2 is start counting diagonal from 1st row last column to last row 1st column. So sum is n+(2n-1)+(3n-2)+...+(n*n-(n-1)) n(1+2+3...n)-(1+2+3+(n-1) (n^2)(n+1)/2)-n(n-1)/2 which will result (n^3+n)/2 ( I feel it's easiest)

[@cmuller1441]

Better generalization: Put any numbers in the 1st row. Then add a constant to make the 2nd row Then another one for the next row Etc... Total is sum of 1st line + sum of constants...

[@onradioactivewaves]

He's an easier and simpler way. There is 4²×3²×2²×1² = 576 possible orders. Now we just write each of 576 cases and see they all have the same sum. Easy! ( no calculators allowed, abacus is ok)

[@glennsampson5945]

The table is an addition table. Each entry is simply the sum of numbers assigned to the corresponding row and column.

[@popogast]

34 is equal to the sum of 1 to 16 divided by 4 (rows, columns) for a 4x4-grid. 65 is equal to the sum of 1 to 25 divided by 5 (rows, columns) for a 5x5-grid. Another way to express the rule behind the "magic".

[@felipelopes3171]

If you're numering columns and rows from 1 to 4, the number at the cell will be 4*(row-1)+column With your restrictions, every combination of four numbers will have rows and columns as a permutation of the list (1,2,3,4). So their sum will be 1+2+3+4=10 in the column term, and 0+4+8+12=24 in the row term. Summing them you get 24+10=34

[@wormopolis9802]

my mind was boggled when you picked the actual 4 numbers I picked....

[@hikari1690]

I know how this works but I'll gladly fall for it everytime~

[@franklhota5019]

Another quick way to solve this: first subtract 1 from each cell, so that the numbers run from 0 to 15. Now write each number in base 4: 00 01 02 03 10 11 12 13 20 21 22 23 30 31 32 33

[@shashwatgupta4613]

Common S=Ax1+ Ay2 Az3 +Aa4 Now x y z a can be A Permutation of 1 to 4 Map S so that xyza is replaced by 1234 in summ. This is ok since it does not change sum. (This mapping gives sum of diagonal elements) Villa you proved that this is true for any l matrix. Also true if rows are AP with same d. Eg a possible matrix can be A transpose. Rows form an AP with same d. To generalise. This is true for any matrix such that the value is the sum across any diagonal (leading or off)

[@drelijahmikail3916]

they will always be in the same positions relative to each other no matter the transposition. and since they are fixed from the start positions from the start relative to increment of 1 (column-wise) and +4 (row-wise) and +5 (diagonal SE) and +3 (diagonal SW), their fate of summing to a fix constant is sealed.

[@francomiranda706]

My first thought was "The numbers must spread each row, so the sum is at least 0+4+8+12 = 24. Then, the numbers spread each column, so the total is 24+1+2+3+4 = 34"

[@shambhav9534]

I did that too, but your way of saying it is quite succinct and even clearer.

[@platypi_otbs]

So I then you can modify it by changing the increments. say 2 for columns and 3 for rows 3 5 7 9 6 8 10 12 9 11 13 15 12 14 16 18 The answer should be 42. I didn't check all combinations but I checked a few

[@DaaaarK147]

start with the number "n" and start scrolling up or down resulting in a sum of n + (n+-x) + (n+-y) + (n+-z), n will always be (34 +- x +- y +- z)/4

Your channel is where every video is a new story and discovery. Keep on surprising and inspiring us!🌿🤘👹

[@Vienticus]

I saw the trick once you mentioned about getting the fourth number. What's funny is you picked the same first three numbers I did, although I did 10, 1, 7.

[@kiranvootori8101]

We can further generalize for grid of size n to [n(n+1)^2/2]-n^2

[@pbenikovszky1]

it would have been also amazing if you use the fact that the determinant of this matrix is 0 :)

[@mandolinic]

I have a better trick. Fill a grid with all "1". Now pick ANY four squares from the grid with NO restrictions. You could have picked any squares from the grid, but whatever you pick their numbers will always add to 4. Thank you - I'm here all week. Matinee performance on Saturday.

[@Aama1807]

1:53 Him: There's no way I could have known what your first number was, your second number was, your third numbers was or your fourth number. Also him: Chose the exact numbers I picked for demonstrating (The order was different tho)

[@JohnDlugosz]

At 2:59 the original challenge did not specify that the numbers had to be in the cells in order. I can arrange them in many different orders, including randomly, making a Magic Square, and spiral. Most such orders will not work. So, it's no wonder people got it wrong!

[@evolsdog126]

That was scary - we picked the same numbers

[@nutsymcgregor]

For any sequentially numbered grid of size MxM is there a grid NxN (N

[@kristofcserpes1502]

I actually picked 10, 7, 1,16. It freaked me out a lot😂

[@GunSpyEnthusiast]

" Pick 4 numbers. these numbers must each be in a different collum and row. Do that, and I will predict the total." " But all of the options would add up to 34? " " *Pick. 4. Numbers.* "

[@ΚΑΝΑΒΟΣΜΑΝΩΛΗΣ]

It works for any table AxA. For examble 6x6 gives Sum of 111

[@Shreyam-q6b]

This Video was made by you only several years back. Approx 9 years ago.

[@vindi167]

as soon as i saw how there being three rows & columns taken means 1 is left, i though "huh, this trick probably has something to do with this"

[@Pineapple-bs1jo]

BRO I CHOSE THE SAME NUMBERS AS YOU WHAT 🎉 THERES ACTUALLY NO WAY

[@badams52]

I bet that 1 out of every 24 people who watch this video will choose the same numbers as you.

[@brahimajaani]

​@@badams52 How we can calculate this probability 🤔

[@kicorse]

​​@@brahimajaani 16*9*4*1/4! = 24. That's a 1 in 24 chance, so not surprising that it happened for someone almost immediately. If our Pineapple friend picked the same numbers in the same order, that's a 1 in 576 chance (16*9*4*1), which is more notable, but still, more than 576 people are going to watch this video and pick numbers.

[@brahimajaani]

@@kicorse hmm i see, thank you for your explanation ♥️

[@xinpingdonohoe3978]

There's no way? It's hardly that rare of an occurrence, especially given the viewer sample size is currently 20,000.

[@slimeboi314]

"I will not guess your 1st, 2nd, 3rd, and 4th number" he says after getting all my numbers and mixing up my 2nd and 3rd choice

[@therealjoshtodd]

yeah i was interested for sure, then you said pick the 2nd number. i raised an eyebrow. at the 3rd number, i was like "ok, this is getting a bit narrow scoped" then with the 4th number I just rolled my eyes. this is not impressive

[@alejandrofigueroa6958]

If the square has an odd number of rows/columns just sum the middle column

[@itstrivikram9619]

You know something I also choose 1,7,10,16

[@maikrentsch6392]

It is the same sum for the magic squares (n^3+n)/2

[@coritosalegrespentecostes]

is it normal tha i picked 3 6 9 4. sum is 22

[@wschmrdr]

With a 5x5 grid, it's always 65.

[@f5673-t1h]

Easy: each row adds 4, each column adds 1.

[@xeuszzz]

The funniest thing is that I selected 10, 7, 1, and 16 in this very order. 😂

[@mlkwy]

Pick in a different column and row from the others: Too tight restriction that it seems obvious that you will always get the same sum. You can't call this a trick.

[@daddykhalil909]

2:44 very nice

[@sparshsharma5270]

You are just 9 years late in explaining the magic square puzzle! (Those who don't know, he made video on I will guess your number with the same magic square (size 4) 9 years ago!)

[@akinamegu9896]

i like this one !

[@banibratamanna5446]

I am not amazed because the problem is inherently an Assignment problem in Operations Research where the minimum cost is 34 for all 4! assignments..... alternatively the problem can be framed as Travelling Salesman Problem where the minimum cost will be 34 for any 4! path....

[@CutTiesWithYourCaptor]

(√n×(n+1))/2 n = largest number

[@miguelribeiro9396]

I solved this in 10 seconds of seeing the what I needed to do. Yes I'm amazing no need to say it. it wilkl always sum 34

[@MrWhyWhyWhy]

Super cool😮😮😮!!!

[@azrobbins01]

Now I have crayon on my monitor...

[@apoorva.720]

15+12+5+1= 33

[@markbothum4338]

Ugh. Thinking hard. Video game easy. What to do rest of afternoon...

[@shamilbabayev8405]

(1+2+3+...+16)/4=34

[@narayanmandal8078]

I want a problem from JEE ADV on logarithm ❤❤❤

[@tomahawk3645]

Dammit the first number was 10

[@Shirasagi77]

1:04 i actually chose 10..

[@iliyagharibi7506]

16+11+6+1=34

[@illiil9052]

I just watched 10s of this video and already know answer is 34.

[@NussbaumBros]

1:55 I picked 1 7 10 and 16 omg

[@lukasjetu9776]

4, 14, 9, 7

[@lukasjetu9776]

34

[@brandorrrrgaming1544]

12-7-2-13

[@Gizmo9707]

No way! I picked exactly the same numbers you did and in the same order 10 - 7 - 1 - 16

[@PrimordialOracleOfManyWorlds]

will it work any n by n matrix?

[@xinpingdonohoe3978]

Why wouldn't it?

[@atifarshad7624]

It will work if N is an even number.

[@inthemidwest3514]

wrong

[@mortoopz]

Obvious pattern is obvious.... I suspect I share this feeling with most of your audience.

[@Fensmiler]

He already posted about this kzbin.info/www/bejne/nGjVg5xmerp7n68

[@TheSheep1]

Not first 😊

[@UnknownUnrecognized]

first again

[@Pineapple-bs1jo]

Bro you 9?

[@Mohammed_gothacked]

First