Can You Solve this Logic Puzzle? | Fill the Purple Boxes!

Learn how to solve this purple box puzzle. Fast and easy explanation by PreMath.com

[@aakashkarajgikar3935]

This was fun!

[@PreMath]

So nice of you dear Thanks Aakash for the feedback. You are awesome 👍 Take care dear and stay blessed😃

[@aakashkarajgikar3935]

@@PreMath You too, You're Welcome.

[@study-dot_vinit]

@@PreMath kzbin.info/door/D_E-R0ro6dwZqp7C9WQG7g

[@asdsad7476]

A*b=21 A+c=8 C-d=5 B*d=9 B=9/d A+b=c+d

[@study-dot_vinit]

kzbin.info/door/D_E-R0ro6dwZqp7C9WQG7g

[@Bronzescorpion]

I was actually a bit bothered that it turned out to not being whole numbers. I still solved it, but I kind of feel robbed aesthetically. There is nothing wrong with fractions or decimal numbers, but these kind of puzzles are just more beautiful with whole numbers, at least in my opinion.

[@colinyu9517]

multiply the numbers 21,5,9,8 by 10 and you have integer answers

[@Andrey_Semenov582]

Yes. There is no "zest". They shouldn't post this on KZbin.

[@timetraveller497]

@@colinyu9517 not true. 21 × 100 not = 210, 100 × 9 not = 90. The equations no longer hold true.

[@pawnoir]

@@DanielKalanja your solution does not work

[@Andrey_Semenov582]

@@DanielKalanja Something is wrong. Try algebra.

[@jas2819]

Very clever, but most people would have been working on the assumption of whole number values. 🤔😕

[@pricelessgm8292]

Useless questions very basic

[@mess9755]

Hahahahah I will just put randomly number and see if that number are correct or not

[@eksortso]

They cannot assume that though. Nothing says they have to be integers.

[@georgesedeno7891]

Exactly. Freaking little mind game.

[@norwaydbp]

There's nothing really clever about an assumption here. The only way to solve this is via simple algebraic substitution.

[@warrior10ize]

At 2:38 add the first two lines, ab+bd=21+9 b(a+d)=30 b=10 because a+d=3, then just insert the value of b in ab=21 and bd=9

[@PreMath]

Excellent! You are the best Warrior. Stay blessed. Love and prayers from the USA!

[@serdalhocailematematik1930]

My math puzzle kzbin.info/www/bejne/g37Od2eKgJWoi6s

[@dxnich]

Top left is "a". Top right is "21/a". Bottom right is "9a/21". Bottom left is "9a/21+5". So, "a+9a/21+5=8" => "30a/21=3" => "10a=21" => "a=2.1" => all other.

[@makwieli]

How is bottom left 9a/21 + 5 ??

[@3KLANGMUSIC]

@@makwieli if d = 9a/21 and c - d = c - 9a/21 = 5 then you can solve the equation to c and get c = 5 + 9a/21

[@davidseed2939]

yes i did something similar. i started filling in with abcd : 7,3,1,3 and then found second row /=5. so i then replaced the grid with 7/x , 3x , 8-7/x, 3/x Now I have the equation 8-10/x = 5

[@josegonsalves5248]

A+9a/21+5=8 -> 30a/21=3 -> 10a=21 -> a=21 -> all other

[@What_In_This_World]

A=2.1 B=10 C= 5.9 D=0.9

[@TheCarpenterUnion]

This isn't solving logically. This is solving algebraicly.

[@benchapple1583]

Algebra is logical.

[@TheCarpenterUnion]

@@benchapple1583 only if you've never been to college

[@brucecarter8296]

the logic is turning it into an algebra problem. it ends there

[@GeoffCanyon]

At 2:17 when you know a + d = 3, you can add the first two equations to get ab + bd = 21 + 9. Then factor out b and get b(a + d) = 30. We know a + d = 3, so 3b = 30, b = 10. Then a = 2.1, d = 0.9 follows from the given equations, and finally c = 5.9 from that.

[@MichaelPolymhxanos]

I thought straight away the same thing. Way simpler ;)

[@abbynormal4740]

That's what I thought too. Once you could factor out b then it's an easy substitution of 3 for a + d, then you're done in a flash. 💡😉

[@tablamurugesan]

Very nice. Thanks.

[@alandunstan5485]

Silly me. Here I was thinking we would be working with integers.

[@gary9793]

I bet you’re not alone

[@tonybates7870]

You're not. Alone, that is. 😁

[@ryukoros]

same here.

[@ignxtius1683]

Same

[@patraic5241]

Ditto

[@srf2112]

I love this channel. Reasoning and intelligence are not common on Y/T.

[@ivanbajo474]

When i saw "logic puzzle" in the title i thought that this would be posible to do without needing equations

[@liamsmith4017]

Hm. I did the same to find that a+d=3... But then noticed that if ab=21 and bd=9, then ab+bd=30, i.e. b(a+d)=30, then straight to b=10. From there the rest fell into place.

[@wayneblackburn9645]

Yep, much easier than what's shown.

[@hanslanda1227]

Easy is for everyone

[@clairer9551]

I did it this way too, much simpler

[@Sam_on_YouTube]

ab=21 bd=9 ab+bd=9+21=30 b(a+d)=30 a+c=8 c-d=5 a+d=3 b×3=30 b=10. Everything else follows trivially. Basically, add those first two equations. It makes the algebra MUCH easier.

[@briantruelove9647]

Smiling as the algebra learned decades ago unlocks in the brain. Thanks for existing brotha. Have a great day!!!

[@geraldillo]

Nice puzzle! Just one small point of criticism; converting the fractions into decimal doesn't help to see the logic. In my opinion it would be better to make the denominators the same. 🙏🏻

[@ChrisM541]

Excellent point.

[@watewmark]

Well.. technically he converted it into a 1 denominator.

[@7curiogeo]

I say whole numbers only, as gthe answer were all 4 whole numbers. The use of fractions destroyed the symmetry.

[@serdalhocailematematik1930]

My new math puzzle kzbin.info/www/bejne/p2akhJtuqpt_sJY

[@martinbennett2228]

It is good that you present problems of a range of difficulties: an astute strategy on your part.

[@serdalhocailematematik1930]

My math puzzle kzbin.info/www/bejne/g37Od2eKgJWoi6s

[@MiniLemmy]

Easy peasy! The way you explained it too made it a walk in the park….

[@AngelA-wf1sg]

Thank you. I haven't done algebraic equations in decades. Good to exercise the old brain now and again🤓

[@Skank_and_Gutterboy]

Playing around with some early trial-and-error, it became apparent very quickly that integers would not work. With the only integer factors for 9 being 3 and 3, and 3 and 7 for 21, there was just no way to get there with integers. Also, with two of the equations being ab=21 and bd=9, you weren't just solving a system linear equations. It took some creative substitution to get there.

[@tintinfan007]

Exactly....I got stuck In the same way

[@jonathanjose6531]

Had fun doing this. Got it. Thanks.

[@serdalhocailematematik1930]

My math puzzle kzbin.info/www/bejne/g37Od2eKgJWoi6s

[@iqmathsciencelogicalreason2770]

Excellent skill of teaching

[@TimKyn]

Hi. Nice puzzle … but seems a lot more simple to just solve it with one variable. Set bottom right be x. So bottom left is 5+x. So top left is 3-x. So top right is 21/(3-x). Then top right x bottom right is 9. So 21x/(3-x)=9. Solve to get x =9/10. 😀

[@DennepeerRelaxation]

Really like this one 👍👍👍 Didn't see it like this myself. 👏

[@MofoMan2000]

I got these same answers using a substitution method (solving for one variable in terms of another), whereas you've mostly used elimination in this video (subtracting equations). I thought my answers seemed strange but they match up!

[@Mirolao]

Il got the result quicker just doing b(a+d)=30, combined with a+d=3

[@chrisj5443]

I also used substitution, and arrived at the correct answers faster and with fewer steps than the presenter.

[@serdalhocailematematik1930]

My new math puzzle kzbin.info/www/bejne/p2akhJtuqpt_sJY

[@charlsreyn]

I'm now wondering whether I never learning adding/subtracting two equations or I just forgot you can do that

[@owlsmath]

Nice!!! love these fill in the box problems :)

[@johnopalko5223]

I don't know why you called it a logic puzzle. It's just a system of simultaneous equations. Basic algebra.

[@Gabsboy123]

The box puzzle thingy is a great way to teach algebra to kids. It helps them understand the concept of variables as being placeholders, visualize systems of equations and to check their solutions. I won't be surprised if somebody already used systems of equations to solve Sudoku puzzles

[@Firefly256]

The first algebra question you had was in 1st grade 2 + ? = 5

[@benchapple1583]

I was thinking of doing exactly that, it's just not interesting. 9 equations, 9 unknowns. It must solve generally. I don't think any maths geeks should do it though. It would ruin the quite legitimate fun people have in solving these (utterly boring to me) problems.

[@serdalhocailematematik1930]

My new math puzzle kzbin.info/www/bejne/p2akhJtuqpt_sJY

[@serdalhocailematematik1930]

My math puzzle kzbin.info/www/bejne/g37Od2eKgJWoi6s

[@akunog3665]

@@benchapple1583 sudoku is a much more complicated system than 9 equations and 9 unknows. There are 9 rows, 9 columns, and 9 boxes, and they must all have every integer, which is a much different condition than an equation. If you just set each row, column, and box equal to 1+2+3+4+5+6+7+8+9 this will lead to non integer solutions, and repeat numbers. Also, it would be 27 equations and (81-givens) unknowns, which would be unsolvable without other constraints (like only integers, and no repeats) The only time I programmed a sudoku solver it used a bit of a guess and check method to get there. Probably not the most elegant solution, but it got me credit in my class, so it was good enough :D

[@bhalsak_tebaghar]

i'm "dumb as a rock" when it comes to math, but because of you good sir i can honestly say i've graduated to "dumb as a potato" - thank you.

[@ericjohnson8001]

Good to know -- if I ever get a job doing these kinds of puzzles.

[@mosesmanaka8109]

I thought you had to color in the boxes.

[@SparkyNJ]

This was great. One question though. What allowed you to just combine the first two C and D equations to get your equation #1? It seemed like you just did it. What principle allowed you to do that?

[@hansschotterradler3772]

At about 2:00 he subtracts one equation from the other, or added them after multiplying the bottom one by -1.

[@nol2521]

This is actually the golden rule of algebra- what you do on one side you do to the other. Since c-d = 5, you're subtracting the same quantity from both sides. By subtracting c-d from a+c, you must also subtract c-d from 8. But, c-d = 5 so instead of writing 8-(c-d) you can substitute c-d with 5, so you write it as 8-5. Therefore, you can work with the rule, "if a = b and c = d then a+c = b+d because c = d" Hopefully this clears things up

[@hansschotterradler3772]

@@nol2521 yep, when 2 equations are true, then the addition of them ( or subtracting one from the other) must be true. a + b = e c + d = f a + b + c + d = e + f Also a + b + f = c + d + e

[@fredr4258]

Check out his algorythm. It's dead on. Rules of algebra are what allowed the combination. Don't lose focus of what you're trying to solve and that should give you that answer.

[@aparnagattupalli7071]

Simultaneous equations method

[@xrty9y874]

GG @PreMath, you just advanced to level 1!

[@hadassahsoddsandends]

I solved using two variables in two equations. I started with xy=21 and solved for x. x=21/y. Then I substituted for x in the equation: (8-x)-(9/y)=5. I came up with (8-21/y)-(9/y)=5 and went on from there. I came up with all the same answers for the boxes, but I do not think I could have finished the problem by myself using four variables. And, I agree: this was fun! Thank-you!

[@tintinfan007]

Awesomely done....saatdick saved the day

[@bobybull]

Excellent brain-training mate! 👍

[@davendraroyeppen4676]

Great example, beautifully solved. Thank you , Sir

[@serdalhocailematematik1930]

My math puzzle kzbin.info/www/bejne/g37Od2eKgJWoi6s

[@thesoutherngentleman1600]

I had my first course in Algebra when I was in the 7th grade over 62 years ago and didn't like it but made a B in that grade in the last half of the 7th grade. From there in college when I got out of the Navy in 67 had to take a pre lim math course and had a 21 or 22 year old teacher who was at LSUNO back then and became good friends who later invited me to his Masters exam which he passed in flying colors. But since then last contact with him and when I see things like this remember times from long ago. Nicely done...

[@serdalhocailematematik1930]

My math puzzle kzbin.info/www/bejne/g37Od2eKgJWoi6s

[@MrAdamNTProtester]

Write out all the equations with w, x, y & z as the unknown values [variables] Then select the three easiest of the equations & solve as a matrix... then you sub for the last equation to solve final variable

[@highlyeducatedtrucker]

Actually had to pull out a pen and paper for this one.

[@PreMath]

Keep it up my friend Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃

[@knotwilg3596]

label the boxes a,b,c,d as you do combine a+c=8 and c-d=5 into a+d=3, as you do multiply by b: ba+bd=3b => 21+9=3b => 30=3b => b=10 resolve for a,c,d

[@brianegendorf2023]

I solved it, but thought I'd made a mistake. It would be helpful if in the title you at least let people know that the answers don't have to be whole numbers.

[@calculusfan1]

I solved ab=21 to get a=21/b and bd=9 to get d=9/b. Then I substituted into a+d=3.

[@hichamroka9428]

Thank you. Excellent

[@daldi5211]

Looking forward for complex number version of this puzzle Please info if someone had already made it

[@serdalhocailematematik1930]

My math puzzle kzbin.info/www/bejne/g37Od2eKgJWoi6s

[@spartacus4698]

Thank you so much! İt's beatiful questions.

[@lougay9669]

wow, I haven't done algebra & trig since 73, thanks

[@davidfromstow]

Another excellent video Another excellent video!

[@wesleysuen4140]

- 2:30: Just divide line 1 by line 2, and you’ll have a/d = 21/9; i.e., 3a = 7d!! - Alternative Solution (quick). Express a, b and c in terms of d: c = 5 + d; a = 8 - c = 3 - d; b = 9/d. Then (3 - d)*9/d = 21 gives the value of d readily, and hence those of the other unknowns.

[@PreMath]

Thanks Wesley for the feedback. You are awesome

[@tdwebste]

I solved the problem myself. Didn't watch the video. Or read comments. And Wesley's approach is most obvious.

[@mauriciorodrigues9991]

great job!

[@sumanthkris777]

Cool ! learning once again.. LOL

[@deepsaha6187]

I was able to do it in less than 2 minutes. I used algebra, but went for relative values with 1 variable. I took the first box value as 'X' and by relation i deduced the third box as "8-X". Similarly then i deduced fourth box as " 3-X" and again deduced the second box as "9/3-X". Now i took the first box and second box relation "X × 9/3-X = 21" which gave me "X=2.1". After that it was all about putting the values of "X" in other boxes expressions.

[@serdalhocailematematik1930]

My math puzzle kzbin.info/www/bejne/g37Od2eKgJWoi6s

[@OneWithManyThings]

I solved differently after equation 1. If you add the top two equations together you get ab + bd = 30 b(a+d) = 30 Using equation 1: 3b = 30, this b=10 :)

[@242math]

got it, thanks for the challenge bro

[@PreMath]

Any time! Thanks my dear friend for the feedback. You are awesome 👍 Take care dear and stay blessed😃 Love and prayers from the USA! 😃

[@eyjeong9658]

오~ 명쾌한 설명, 정확한 답!! 재미있고 속 시원합니다 👍

[@rathikamuthukumar413]

Its fun. We always expect whole number so its really fun

[@JPTaquari]

Very nice, herr professor!!! I nomened A , B , C , D Let's go to solution: A + B = 21 D * B = 0 Entonce, B ( A + D ) = 30 I know that A + D = 3 + A + C - 8 - C + D = - 5 , them, I have that A + D + 3 B * 3 = 30 B = 10 A * B = 21 A = 2,1 A + C = 8 C = 5,9 B * D = 9 D = 0,9 C - D = 5

[@lindafromcalifornia1155]

I enjoy these boxes challenges. They are so fun to do.

[@PreMath]

Glad you like them! Thanks Linda dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃

[@Gabsboy123]

This is an excellent way to teach basic algebra to middle-school kids. Helps them visualize systems of equations and come up with a systematic procedure to solve them.

[@mustafizrahman2822]

I have seen 1st stage that you let the boxes a,b,c or d . When you let them, I paused the video. Though I have solved but I have needed to see the 1st stage. Otherwise, I couldn't solve it. However, thanks for giving this logical questions. You are great.

[@PreMath]

Well done Rahman dear! Keep it up Love and prayers from the USA! 😃

[@mustafizrahman2822]

Sir, have you heard the name of Suborno Isac Bari?

[@mohamedshafeeq7265]

Sir, let me suggest one more approach Put first box as x, then 2nd box wll be 21/x and hence last box will be 9x/21.Also third box is 8-x Now, from the relation of last two boxes, we have 8-x - 9x/21 =5 Solve for x, to get x=2.1 Now we can fill all boxes..

[@mustafizrahman2822]

@@mohamedshafeeq7265 Wow! Excellent! Thanks for sharing. Great you are.

[@khatharrmalkavian3306]

Labeling the boxes in reading order: A,B,C,D: Given C-D=5 ; C=5+D Given A+C=8 ; A+(5+D)=8 ; A+D=3 Given AB=21 and BD=9 ; AB+BD=21+9 ; B(A+D)=30; B(3)=30 ; B=10 AB=21 ; 10A=21 ; A=21/10 BD=9 ; 10D=9 ; D=9/10 C=5+D ; C=5+9/10 ; C=59/10 Running them back through the original equations, they check out.

[@frednurks8528]

A good example of using simultaneous equations to solve. Four unknowns and four equations.

[@serdalhocailematematik1930]

My math puzzle kzbin.info/www/bejne/g37Od2eKgJWoi6s

[@justgame5508]

Not really a logic puzzle, just a straight algebra question presented in a different form

[@Listenandchillzz]

In my mind I was pertaining the answer should be a whole number 😂

[@bygonehope6158]

This all made sense but why did you multiply equation one by three. It was just assumed we would know. But I’m not sure

[@anthonypasnin2270]

Excellent wow im amazed

[@archnof0]

I don't look at it as a logic puzzle, but 4 equations in 4 unknowns: axb=21,a+c=8, bxd=9, c-d=5. Solving them gets a=21/10, b=10, c=59/10, d=9/10

[@111111111Tiger]

great job... I just revisited high school

[@icantranch9398]

You have very satisfying handwriting

[@emiliospinhouse3262]

Wow... long time since I've had this much fun with math logic

[@serdalhocailematematik1930]

My math puzzle kzbin.info/www/bejne/g37Od2eKgJWoi6s

[@firebreathers7459]

Incredible..

[@ASChambers]

Marvellous. Great fun.

[@YuvrajSingh-wf9jn]

I solved in my own way using substitution and I got A= -21/4, B= -4, C= 53/4 and D= -9/4. Although the answers are different, the equations still hold true. What a wonder!!

[@thewolfdoctor761]

But C-D isn't 5

[@takshilajola234]

In a stage where you create 1 equation from 2.. a+c=8 and c-d= 5 when you inverted the signs.. why its not -a in the ans

[@wayyudao8907]

Easy, 4 variables 4 equations, but nice thoughts. Well thought

[@simplyib9942]

You may consider shortening the method by adding firs2 eq to get b(a+d) = 30 and you have already foung=d a + d - 3, Thus you get b = 10 thereafter just calculate backward. Thanks

[@AmarYadav-kh2ze]

I also did it n that also in less than 5 minutes using simple algebra by using x and y as variables n then I formed the equation lyk 8-x as the values

[@piochechicha4038]

More easy . Can you do that with nein boxes ?

[@tonymunn]

Awesome. My brain needs exercise.

[@noduh629]

I haven’t watched the video or tried the problem, but isn’t it just 4 equations 4 unknowns? You can just set up a matrix

[@noduh629]

Nah wait it’s not linear, can’t do a matrix but it is 4 eq 4 unknowns which can be brute forced

[@direwolf6234]

at first one is led to believe we'd be using whole numbers not fractions...

[@ryateo1]

I solved it faster than I could click on the thumbnail. I solved it with familiarity and not arithmetic. They're all groups of numbers that we have been working with since we were five.

[@archIDch]

From a+d=3, then ab+bd=21+9. It can be b(a+d)=30. B(3)=30. B=10.

[@aashsyed1277]

thanks so much dear you are awsome

[@williamwingo8952]

No peeking, no calculators: Starting at the upper left square and moving clockwise, label the boxes x, y, w, and z. Then: xy = 21; x + z = 8; z - w = 5; and yw = 9. Hmmm.... If yw = 9, then w = 9/y. If z - w = 5, then w = z - 5. So: w = 9/y = z - 5; but z = 8 - x; so: 9/y = 8 - x - 5 = 3 - x; but x = 21/y, so: 9/y = 3 - 21/y; add 21/y to each side: 9/y + 21/y = 3; collect terms on the left: 30/y = 3; multiply through by y: 30 = 3y; so y = 10. Voila! Since xy = 21, 10x = 21, and x = 21/10 = 2.1; since x + z = 8, z = 8 - x = 8 - 2.1 = 5.9; and since wy = 9, w = 9/y = 9/10 = 0.9. At this point, I ran the video, and it looks like we agree. Your a, b, c, and d are my x, y, z, and w. Good thing we didn't make the unconscious assumption that everything was supposed to be integers. Cheers. 🤠

[@jeetendrachaurasia446]

From 2:24, if u solve for b in equation 1, that will be more easy. In equation 1, put a= 21/b & d=9/b. So all value will come in form of b & u can calculate the value of b. Then u can get value of a &d. Then find c. It will be so simple.

[@louaiello1493]

Agree. As a former algebra teacher, this is really the only true way to quickly solve the problem. The steps he went through work, but require a process that is not truly generic enough to solve other similar problems. Your (my) way took several fewer steps.

[@evescent7312]

I solved for B first, and it was the only integer. That was beautiful xD

[@serdalhocailematematik1930]

My math puzzle kzbin.info/www/bejne/g37Od2eKgJWoi6s

[@michaelyano6094]

This is just 4 equations and 4 unknowns. No lateral thinking required here. Just straight arithmetic

[@user-pw9my3yg3l]

Thank you It was very good

[@echchakhaouikhalifa7468]

Easy if we notice that a+d=3. After that ab+bd=21+9=30 then b=10. The other fall like this a=21/10=2.1 d=9/b=0.9 Finally c=5+d=5.9

[@moth36]

It's really nice problem 👍

[@argonwheatbelly637]

Delightful, this!

[@manojv5091]

Upload more videos like this sir

[@mnazar3601]

Because the answser is not straight foward n i need to find the answer. Mind blown question. Challenged our mind.

[@oldyoung9379]

A good moment. Thank's.

[@anvarmuqimov4540]

Extremely cool puzzle

[@gwg605]

The task description is not clear enough. I opened this video to just see how it will be solved with integers.

[@djr093r5]

Hah! Worked this one out. Slightly different route, but essentially the same method...

[@audreyDsouza]

I really love to do this kind of puzzle I don’t know why🤷🏻‍♂️..my brain is always excited to count or solve puzzle or reading🤔mayb i’m sick 🤕

[@USMC-cx2nj]

My head hurts, lol, but it was pretty nice!

[@MusicLover-gp6dl]

I got the same answer but my method was different, I wrote the given first ab=21 bd=9 a+c=8 c-d=5 Then I subtracted a+c=8 and c-d=5 a+d=3 Then I find the value of a and d a=21/b d=9/b then I add those two 21/b+9/b=3 30/b=3 30=3b b=10 then I simply put that value on bd 10d=9 d=9/10 or 0.9 Then I put that value on c-d=5 c-0.9=5 c=5.9 Then lastly I put that value on a+c=8 a+5.9=8 a=2.1 And I got all the values a=2.1 b=10 c=5.9 d=0.9

[@Lord_Volkner]

That took about 30 seconds, but it was still fun.

[@olivieru]

a+c = 8 and c-d = 5 ---> d=3-a ----> 9 = b.d = b.(3 - a) = 3b - ab = 3b - 21 ----> 3b=30 ----> b = 10. Then the remaining computations are easy.