Can you solve this Logic Puzzle? | Fill the Boxes with the right values

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Can you solve this Logic Puzzle? | Fill the Boxes with the right values | Math Olympiad Training

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Жыл бұрын

Learn how to fill the Boxes with the right values. Add, subtract, and multiply the numbers. Step-by-step tutorial by PreMath.com
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Пікірлер: 126

@bigm383 Жыл бұрын

Thanks, Professor, for some Monday night fun!🥂😀

@PreMath Жыл бұрын

Excellent! You are very welcome! Thank you! Cheers! 😀

@bigm383 Жыл бұрын

@@PreMath 😂❤️

@OmPrakash-cq1wt Жыл бұрын

Only 17×3=51 or 3×17=51 or 1.7×30=51 or 30×1.7=51 and no other way contains this result. In this way it can solved by working out first row only.

@songweimai6411 8 ай бұрын

Very easy. U don’t need a pen to solve it. First of all, u should think that which step is the most important step. There are 4 numbers there . 51, 18, 23, 99. And there two kinds of arithmetic, add and multiplication. Generally, Multiplication will be harder to figure it out or dominate. So we find numbers to fit the multiplication first. By the way, u should know that 51 is very special, (if u have feeling, when u see these 4 numbers, 51 is the most special one).51compared with 99: only one kind combination: 3*17. 99: can be 3*33,9*11.But u will find that , it is impossible to be just integer #. And can not be negative #.so only can be fraction. Then 3*17 divides by 10. That’s the answer. Then everything works out immediately.

@jean-yvesPrax Жыл бұрын

No real need to convert into an equation: 51 is only divisible by 3, and 99 too, so B=3!😄 we find a=17, b=3, c = 33... in 10 seconds, after that gives c=6 which is wrong... at that time, we guess that it cannot be whole numbers, so a becomes 1.7 - b becomes 30, d becomes 3.3 and there is only a small addition left to find c at 21.3! in 20 seconds.

@MichaelRothwell1 Жыл бұрын

I went down that route too (but tried b=1,3,-1,-3. As none of these worked, had to solve the system of four equations in four unknowns.

@RexxSchneider Жыл бұрын

KISS. Put x in the bottom right (BR) box. Then the bottom left (BL) must be (x+18) and the top right (TR) must be (99/x). Now the top left (TL) and (x+18) add up to 23, so the top left is (5-x). Finally the top left times the top right = (5-x) * (99/x) = 51. Therefore 5*99 - 99x = 51x. That gives 495 = 150x and so x = 3.3. The boxes are: TL = 5-x = 1.7; TR = 99/x = 30; BL = x+18 = 21.3; BR = x = 3.3.

@alster724 Жыл бұрын

Super easy!!!! Solved this on my own after seeing the thumbnail and got the same answer. ab= 51 a+c= 23 bd= 99 c-d= 18 ============ a+c= 23 (c-d= 18)(-1) a+d= 5 a= 51/b d= 99/b (51+99)/b= 5 150= 5b b= 30 a= 1.7 d= 3.3 c= 21.3 Final answer a= 1.7, b= 30, c= 21.3, d= 3.3 Cheers from The Philippines 🇵🇭

@avroraaspasia7214 Жыл бұрын

For me more interesting to find whole numbers in a task, rather then fractional ones.

@Leberteich Жыл бұрын

Multiply all answers by 10.

@RexxSchneider Жыл бұрын

But that makes the problem trivial. If we knew that all answers were integers, we would immediately know the top right box could only be 3 or 1 (common factors of 99 and 51). It turns the exercise into mental arithmetic.

@anthonyc7045 Жыл бұрын

Yes, if all of the numbers had to be whole numbers you could figure it out using your MIND, not by a silly cancelling of "b" and "c" and "negatives"

@Uwidd Жыл бұрын

Than

@mrosskne Жыл бұрын

whole numbers are fractions

@marcosian820 Жыл бұрын

there's some joy to find the right answer that i can explain. thanks to made my day better :D

@laszloosvalt8935 Жыл бұрын

It can be solved more generally. Let the results (51, 18, 23, 99) be parameters: p1, p2, p3, p4. Substitutions: b = p1 / a, d = c - p2. Using the last two (vertical) equations: c = p3 - a, we get the equation p1/a (p3 - p2 - a) = p4 for a. Solving it to a, we get a = p1 (p3 - p2) / (p1 + p4), and we can substitute back the other values. (Certainly, we should also analyze all possible "zero division" cases, e.g. p1+p4=0 for the complete solution.) Now a = 51 (23 - 18) / (51 + 99) = (51 x 5) / 150 = 1.7, and we can substitute back b, c, and d from the substitutions above.

@MichaelRothwell1 Жыл бұрын

Very nice! If p1+p4=0, we either get infinitely many solutions (if p1=0, in which case also p4=0, or p2=p3) or no solutions (otherwise).

@meerabakshi2676 Жыл бұрын

the solution gave me an idea to think out of the box. I was trying to use only numbers and not the numbers with decimal fraction! Thanks. I love arithmetic as well as algebra. Thanks.

@PreMath Жыл бұрын

Great job, Meera ji!

@CJBrunt Жыл бұрын

It can be done by Logic: a and b need to end in ~3 and ~7 to make ~1, b has to be the 3 to multiply with d to make ~9. So a=17, b=3, d=33, c=doesnt work, so change by 10: a=1.7, b=30, d=3.3, and c is 21.3 and it's done.

@RexxSchneider 8 ай бұрын

Why change by 10? What's logical about 10? Why not "change" by 8 or 5 or 20 or any other number?

@CJBrunt 8 ай бұрын

@@RexxSchneider because changing only the decimals keeps the same necessary numbers.

@RexxSchneider 8 ай бұрын

@@CJBrunt That's nonsense. You took no cognisance of the bottom-left total of 23. If the bottom left total were 24.5 then "changing the decimals" wouldn't work. Instead of scaling up and down by 10, you would have to scale up and down by 8 to get the solution: 2.125 x 24 = 51 + x 22.125 - 4.125 = 18 = = 24.25 99 I ask again: What's _logical_ about 10? Why not "change" by 8 or 5 or 20 or any other number? There's nothing *logical* about guessing an answer.

@allanfrank291 Жыл бұрын

No fun. Anyone can solve simultaneous equations. I expected the puzzle to be whole numbers.

@gabrieloliveirafreitassant8752 7 ай бұрын

Oh my god! My exact same thought! Garbage puzzle

@aba12385 7 ай бұрын

I thought the same.

@billywytka56 7 ай бұрын

100% agree, these are garbage if they involve decimals. Defeats the purpose.

@TheMail518 Жыл бұрын

I did A x B = 51, A + C = 23, C - D = 18, and B x D = 99. I put boxes as A, B, C, D (left to right): A = 17/10 B = 30 C = 213/10 D = 33/10

@L8rCloud Жыл бұрын

xy = 51 by = 99 a - b = 18 x + a = 23 Solve using simultaneous equations x = 23 - a b = a - 18 Find y y(23 - a) = 51 y(a - 18) = 99 y = 51/(23 - a) y = 99/(a - 18) find a 51/(23 - a) = 99/(a-18) 51(a-18) = 99(23 - a) 51a - 918 = 2277 - 99a a = 639/30 Thus a = 21.3 b = 3.3 y = 30 x = 1.7

@yalchingedikgedik8007 Жыл бұрын

Thanks Prof. Thanks PreMath Thats very nice.

@soumyasharma5318 8 ай бұрын

Sinpler and a faster 60 second solution. 51 has only 2 factors. 17 and 3 both prime, cant be broken down further. Putting these in boxes quickly rells us that value of C is 6, which doesnt staidfy C-D equation. Hence, A and B are either 1.7 and 30 or 170 and 0.3. It is obvious that A and B are 1.7 and 30. Now calculate D as 3.3 and C as 21.3. Simple.

@misterenter-iz7rz Жыл бұрын

ab=51, bc=99, d-c=18, a+d=23, so c/a=33/17, a+c=5, then 17(5-a)=33a, 85-17a=33a, 50a=85, a=17/10, c=5-17/10=33/10, b=51x10/17=30, d=23-17/10=213/10.😃

@PreMath Жыл бұрын

Excellent! Thank you! Cheers! 😀

@kaushikbasu9707 Жыл бұрын

very nice math. Thank you.

@SAHIRVLOGCLP Жыл бұрын

More interesting problem... thank you Proffesor

@PreMath Жыл бұрын

Glad to hear that Thanks for your feedback! Cheers! 😀

@amazonianm8876 Жыл бұрын

I was assuming that the solutions were all integers!

@aichaboumezrag4642 Жыл бұрын

شكرا جزيلا لك يا دكتور 🎉🎉

@MrFeronias Жыл бұрын

it means, Thank you very much, Doctor

@HappyFamilyOnline Жыл бұрын

Very interesting 👍 Thanks for sharing😊

@marinacheckina2588 10 ай бұрын

Первый квадрат х. Справа 51/х. Снизу слева 23-х. Снизу справа: 99*х/51 и 5-х. Приравниваем и решаем уравнение. х = 1.7

@rishudubey1533 Жыл бұрын

tq v.much dear ❤❤️

@PreMath Жыл бұрын

You are very welcome! So nice of you. Thank you! Cheers! 😀

@TxRealtorSA Жыл бұрын

LOve your work. Logical, step by step. You are a credit to the human race! Blessings to you!

@Firouzeh444 Жыл бұрын

Nice , I solved in 5 minutes , thank you 😊

@PreMath Жыл бұрын

Great 👍

@tahirmirza8521 7 ай бұрын

Wish the heading to the puzzle mentioned not necessarily whole numbers.

@KAvi_YA666 Жыл бұрын

Thanks for video.Good luck sir!!!!!!!!!!

@PreMath Жыл бұрын

Thank you too, dear

@davidfromstow Жыл бұрын

Got a, b, and d right but made a stupid arithmetic mistake and had c as 21.7 - very annoyed with myself! Thank you for a puzzle I could (almost) do.

@jesusantoniocarhuashuerta4662 Жыл бұрын

Great explanation.

@PreMath Жыл бұрын

Glad you think so!

@harisvictory2712 10 ай бұрын

a x b = 51 + x c - d = 18 " " 23 99 The answer are : a = 51/30 b = 30 c = 639/30 d = 99/30

@ranjanprakash2521 11 ай бұрын

Excellent!

@gideonlapidus8996 Жыл бұрын

Great question

@bonginhlanhlangema7125 8 ай бұрын

Just Use Simultaneous Equations. Name the boxes A, B, C, and D. A * B = 51 C - D = 18 A + C = 23 B * D = 99 Make A, B, C and Subjects of the formulae and substitute accordingly

@dave9456 Жыл бұрын

Seems like making the rules as go along, far less complicated studying maths 40 years ago and managed to travel this far

@roscius6204 Жыл бұрын

it's math... the rules are not changing....

@deepakdeshpande529 Жыл бұрын

Very nice and interesting

@platinum9825 8 ай бұрын

Could've figured it out in my head if there were no decimals, basically did, if you round the numbers then you come almost exactly to what I was thinking

@kaushikbasu9707 Жыл бұрын

1.7 and 30 and 21.3 and 3.3 ans

@gmjackson1456 Жыл бұрын

a=1.7, b=30, c=21.3, d=3.3.

@Copernicusfreud Жыл бұрын

Yay! I Solved it.

@vidyadharjoshi5714 Жыл бұрын

Let top row letters be A & B and bottom be C & D. After trial & tries. A = 1.7, B = 30, C = 21.3 & D = 3.3

@Funkybassuk Жыл бұрын

Ahhhh. I was getting stuck working on the assumption that these should be integers. Well done. 👍🏽

@vidyadharjoshi5714 Жыл бұрын

@@Funkybassuk First I tried with 3 & 17 at top and 20 at C but that will not give me 99 & 18. Then I went 1, 51 & 22 and that will not give 99 & 18. Then 2, 20.5, 21 thus it made way to results.

@Funkybassuk Жыл бұрын

@@vidyadharjoshi5714 I had 17 and 3 at the top and then 33 bottom right - but then I could not find anything to fit in the bottom left to make the bottom row work. Makes total sense the way you’ve done it.

@roscius6204 Жыл бұрын

@@FunkybassukI started there then realized the 17 + 33 =50...and needed to = 5. so /10 bingo.

@user-ch6rg1bt3v 8 ай бұрын

99の時点で33*3と思ったら3.3か、なるほど😊直感より計算しろと

@aminex3519 Жыл бұрын

Geometry pls sir

@rezaalahyary2050 Жыл бұрын

Very nice

@renscience Жыл бұрын

4 variables, 4 equations, 2x4 matrix. Or grind through it the slow way

@user-cx4gc5lh4b Жыл бұрын

It's not a linear algebra problem because you have variables multiplied by each other.

@renscience Жыл бұрын

@@user-cx4gc5lh4b i stand corrected. Just glanced at quickly🤪

@A123A56D 11 ай бұрын

Puzzle are supposed to be short and solved in a whole number.

@akrambraiwesh5220 Жыл бұрын

We can solve it using one variable : the c

@davidhiggen3029 11 ай бұрын

4 variables. 4 Values. This is just basic simultaneous equations, right?

@nagendrathakkar4601 Жыл бұрын

brain tonic

@user-mj4js7vt3w Жыл бұрын

ちょっとやっただけで、答えが自然数ではないことだけはわかった。その時点でめんどくさくなっちゃった・・・

@wackojacko3962 Жыл бұрын

I'm not completely sure but @ 2:08 a + c = 23 and c - d = 18 looks like a system of consistent linear equations . 🙂 I'd like too get the terminology correct.

@PreMath Жыл бұрын

Thank you! Cheers! 😀

@alster724 Жыл бұрын

Multiply the second equation by -1 then you will get a+d= 5 and the rest will be easy

@jeffreese5977 Жыл бұрын

I thought I had it solved mentally until I realized my solution for d wasn’t correct, which threw off things.

@Timelether 9 ай бұрын

A = -5.3125 B = -9.6 C = 28.3125 D = -10.3125 This is not false so another solution group

@roscius6204 Жыл бұрын

I did this all wrong. I got to a+d=5 but only after starting with thinking a=17 and d=33 etc and having c-d not work after seeing a+d added to 50 realized the obvious. /10 So I got there, just badly.🙄

@jnsougata Жыл бұрын

4 variables & 4 equations … what’s challenging about it?

@kam7056 Жыл бұрын

❤

@PreMath Жыл бұрын

Thanks You are awesome. Keep it up 👍

@geirmyrvagnes8718 Жыл бұрын

Much simpler solution: 51=17*3. 3 is a factor of 99. So assuming integer solutions, a = 17, b = 3, d = 33 and c is impossible. So the answer must be boring. 🤣

@peculiartv8962 11 ай бұрын

This is quadratic equation not logic puzzle😂😂😂

@flowerflower3437 11 ай бұрын

鮮やかな解法があるのかと思ったらゴリゴリ方程式解くだけかい。答えも別に綺麗じゃないし

@Poshypaws 7 ай бұрын

It equals 99. Nothing about "it equals to 99". No preposition is required.

@trueindian887 Жыл бұрын

I solved it in 3 minutes

@patbrennan6572 Жыл бұрын

Now I see why I hated algebra so much.

@kavishwarmokal124 11 ай бұрын

It should have been better to keep the puzzle limited to whole numbers.

@mohamadtaufik5770 Жыл бұрын

A=upper left box = 1.7 B=upper right box = 30 C=lower left box = 21.3 D=lower right box = 3.3

@emmanuel_zunguero Жыл бұрын

Así no se vale, tenían que ser números enteros

@user-bh1nw7dk9d 7 ай бұрын

数学としては納得するが、パズルの解答に小数点アリは美しくない。

@JSSTyger Жыл бұрын

Quite strangely, I got no solution.

@JSSTyger Жыл бұрын

....because i made a dumb error and said 5b=150 and b=3...sigh.

@rods6405 Жыл бұрын

@@JSSTyger so did I 41*51= 2019 wrong should have type 2091

@michaelvillanueva233 3 ай бұрын

This is all soooo confusing

@lookingforahookup 7 ай бұрын

Your way is too complicated

@stevehuffman7453 Жыл бұрын

I cannot solve a math "logic" puzzle. I do not see ANY "logic" in math. "Math is logical" is an oxymoron. Just like "Military Intellegence" and "Honest Politician". ("Honest Politician" is also a lie.)

@orangeannet5250 8 ай бұрын

整数以外有りとか糞前提にも程がある

@zdrastvutye Жыл бұрын

it took mee more than 1 hour sitting in a concrete tower, games people play! 10 print "ein rechenquadrat":dim vz$(3),n(3):read a1$,a2$,a3$,a4,a5 20 sw=.5:a=sw:goto 140 30 for e=1 to len(au$):if mid$(au$,e,1)="+"then else 50 40 vz$(f)="+":goto 130 50 if mid$(au$,e,1)="-" then else 70 60 vz$(f)="-":goto 130 70 if mid$(au$,e,1)="*" then else 90 80 vz$(f)="*":goto 130 90 if mid$(au$,e,1)="/" then else 110 100 vz$(f)="/":goto 130 110 next e:if vz$(f)=""then else 130 120 print "ungueltiger operator":end 130 return 140 f=0:au$=a1$:gosub 30:au$=a2$:f=1:gosub 30:vz$(2)=mid$(a3$,1,1):vz$(3)=mid$(a3$,2,1) 150 au$=a1$:gosub 160:goto 170 160 n=instr(au$,"="):ne=val(mid$(au$,n+1,len(au$)-n+1)):return 170 print a1$,"§",ne:n(0)=ne:au$=a2$:gosub 160:n(1)=ne:print ne:rem stop 180 print n(0),n(1):n(2)=a4:n(3)=a5 :goto 320 190 nu=a:e=0:f=0:goto 290 200 if vz$(e)="+" then else 220 210 ne=n(f)-nu:goto 280 220 if vz$(e)="-" then else 240 230 ne=n(f)+nu:goto 280 240 if vz$(e)="*" then else 260 250 ne=n(f)/nu:goto 280 260 if n(f)=0 then print n(f),f:stop 270 ne=nu/n(f) 280 return 290 gosub 200:b=ne:e=2:f=2:gosub 200:c=ne:nu=b:f=3:e=3:gosub 200:d=ne 300 e=1:f=1:nu=d:gosub 200:dg=(ne-c)/(n(1)+n(0)):rem print a,"%",b,"%",c,"%",d :print vz$(0),n(0):stop 310 return 320 gosub 190 330 dg1=dg:a1=a:a=a+sw:if a>100 then stop 340 a2=a:gosub 190:if dg1*dg>0 then 330 350 a=(a1+a2)/2:gosub 190:if dg1*dg>0 then a1=a else a2=a 360 if abs(dg)>1E-9 then 350 370 print a,"%",b,"%",c,"%",d:print "probe" 380 print a*b/n(0),"%",a/b/n(0),"%",(a+b)/n(0),"%",(a-b)/n(0) 390 print a*c/n(2),"%",a/c/n(2),"%",(a+c)/n(2),"%",(a-c)/n(2):rem stop 400 print b*d/n(3),"%",b/d/n(3),"%",(d+b)/n(3),"%",(b-d)/n(3) 410 if d=0 then 430 420 print c*d/n(1),"%",c/d/n(1),"%",(c+d)/n(1),"%",(c-d)/n(1) 430 print "Loesung":print a;vz$(0);b;"=";n(0):print " "; vz$(2);" ";vz$(3) 440 print c;vz$(1);d;"=";n(1):print " =";n(2);" =";n(3) 450 data a*b=51,c-d=18,+*,23,99 ein rechenquadrat 1.7% 29.9999999% 21.3% 3.30000001 probe 1% 0.00111111112 % 0.621568626 % -0.55490196 1.57434783% 0.00347009594 % 1% -0.8521739 13 1% 0.0918273642 % 0.336363636 % 0.269696969 3.90500001% 0.358585858 % 1.36666667% 1 Loesung 1.7*29.9999999=51 + * 21.3-3.30000001=18 =23 =99 > have fun with bbc basic sdl. copy and paste is less harassment than uploading. if the operators are changed and there are 1 digit numbers given, there will be a possibility of "division by zero" so used "instr"

@aichaboumezrag4642 Жыл бұрын

شكرا جزيلا لك يا دكتور 🎉🎉

@PreMath Жыл бұрын

جزاك الله

@aichaboumezrag4642 Жыл бұрын

@@PreMath where are you from doctor !

@PreMath Жыл бұрын

@@aichaboumezrag4642 I'm from the USA! Take care 🙏