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

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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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5 сен 2024

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Комментарии : 126

@bigm383 Год назад

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

@PreMath Год назад

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

@bigm383 Год назад

@@PreMath 😂❤️

@OmPrakash-cq1wt Год назад

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 9 месяцев назад

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 Год назад

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 Год назад

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.

@alster724 Год назад

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 🇵🇭

@RexxSchneider Год назад

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.

@marcosian820 Год назад

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

@allanfrank291 Год назад

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

@gabrieloliveirafreitassant8752 8 месяцев назад

Oh my god! My exact same thought! Garbage puzzle

@aba12385 8 месяцев назад

I thought the same.

@billywytka56 8 месяцев назад

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

@meerabakshi2676 Год назад

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 Год назад

Great job, Meera ji!

@avroraaspasia7214 Год назад

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

@Leberteich Год назад

Multiply all answers by 10.

@RexxSchneider Год назад

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 Год назад

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 Год назад

Than

@mrosskne Год назад

whole numbers are fractions

@L8rCloud Год назад

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

@laszloosvalt8935 Год назад

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 Год назад

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).

@TheMail518 Год назад

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

@CJBrunt Год назад

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 9 месяцев назад

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

@CJBrunt 9 месяцев назад

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

@RexxSchneider 9 месяцев назад

@@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.

@yalchingedikgedik8007 Год назад

Thanks Prof. Thanks PreMath Thats very nice.

@kaushikbasu9707 Год назад

very nice math. Thank you.

@misterenter-iz7rz Год назад

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 Год назад

Excellent! Thank you! Cheers! 😀

@soumyasharma5318 9 месяцев назад

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.

@aichaboumezrag4642 Год назад

شكرا جزيلا لك يا دكتور 🎉🎉

@MrFeronias Год назад

it means, Thank you very much, Doctor

@SAHIRVLOGCLP Год назад

More interesting problem... thank you Proffesor

@PreMath Год назад

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

@rishudubey1533 Год назад

tq v.much dear ❤❤️

@PreMath Год назад

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

@amazonianm8876 Год назад

I was assuming that the solutions were all integers!

@TxRealtorSA Год назад

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

@Firouzeh444 Год назад

Nice , I solved in 5 minutes , thank you 😊

@PreMath Год назад

Great 👍

@KAvi_YA666 Год назад

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

@PreMath Год назад

Thank you too, dear

@marinacheckina2588 11 месяцев назад

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

@HappyFamilyOnline Год назад

Very interesting 👍 Thanks for sharing😊

@jesusantoniocarhuashuerta4662 Год назад

Great explanation.

@PreMath Год назад

Glad you think so!

@gideonlapidus8996 Год назад

Great question

@bonginhlanhlangema7125 9 месяцев назад

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

@davidfromstow Год назад

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.

@ranjanprakash2521 Год назад

Excellent!

@harisvictory2712 11 месяцев назад

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

@deepakdeshpande529 Год назад

Very nice and interesting

@tahirmirza8521 8 месяцев назад

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

@gmjackson1456 Год назад

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

@Copernicusfreud Год назад

Yay! I Solved it.

@dave9456 Год назад

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

@roscius6204 Год назад

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

@kaushikbasu9707 Год назад

1.7 and 30 and 21.3 and 3.3 ans

@user-ch6rg1bt3v 9 месяцев назад

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

@vidyadharjoshi5714 Год назад

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 Год назад

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

@vidyadharjoshi5714 Год назад

@@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 Год назад

@@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 Год назад

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

@platinum9825 9 месяцев назад

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

@rezaalahyary2050 Год назад

Very nice

@aminex3519 Год назад

Geometry pls sir

@renscience Год назад

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

@user-cx4gc5lh4b Год назад

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

@renscience Год назад

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

@akrambraiwesh5220 Год назад

We can solve it using one variable : the c

@A123A56D Год назад

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

@davidhiggen3029 Год назад

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

@Timelether 9 месяцев назад

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

@nagendrathakkar4601 Год назад

brain tonic

@user-mj4js7vt3w Год назад

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・・・

@flowerflower3437 11 месяцев назад

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

@jnsougata Год назад

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

@roscius6204 Год назад

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.🙄

@trueindian887 Год назад

I solved it in 3 minutes

@jeffreese5977 Год назад

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

@geirmyrvagnes8718 Год назад

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. 🤣

@mohamadtaufik5770 Год назад

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

@kam7056 Год назад

❤

@PreMath Год назад

Thanks You are awesome. Keep it up 👍

@wackojacko3962 Год назад

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 Год назад

Thank you! Cheers! 😀

@alster724 Год назад

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

@Poshypaws 8 месяцев назад

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

@patbrennan6572 Год назад

Now I see why I hated algebra so much.

@peculiartv8962 Год назад

This is quadratic equation not logic puzzle😂😂😂

@kavishwarmokal124 11 месяцев назад

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

@emmanuel_zunguero Год назад

Así no se vale, tenían que ser números enteros

@user-bh1nw7dk9d 8 месяцев назад

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 Год назад

Quite strangely, I got no solution.

@JSSTyger Год назад

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

@rods6405 Год назад

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

@michaelvillanueva233 3 месяца назад

This is all soooo confusing

@lookingforahookup 8 месяцев назад

Your way is too complicated

@stevehuffman7453 Год назад

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 9 месяцев назад

整数以外有りとか糞前提にも程がある

@zdrastvutye Год назад

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 Год назад

شكرا جزيلا لك يا دكتور 🎉🎉

@PreMath Год назад

جزاك الله

@aichaboumezrag4642 Год назад

@@PreMath where are you from doctor !

@PreMath Год назад

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