ABCD x 4 = DCBA. Can you find the values ?? Challenge for you !!

[@danmimis4576]

A more elegant and faster solution: It immediately pops up that A=2 and D=8. So now we have ABCDx4 = DBCA 4000A + 400B + 40C + 4D = 1000D + 100C + 10B + A and since A=2 and D=8 we'll have: 2C - 13B = 1, so C = 7 and B = 1

[@pierpaoloabbruzzese3601]

I think so, too!!! With the power of the linear equations!!!

[@salehdarraz2297]

.

[@rsie.sweets3798]

Thanks 👍👍 for your help 😸

[@manudude02]

I took a different route to the same answer. A must be

[@LOGICALLYYOURS]

Perfect !!

[@tamirerez2547]

Just great! Well done.

[@050138]

I used the same approach 👌

[@Estrada999]

@@LOGICALLYYOURS Friend, how can I perform this type of mathematical operations with the help of a software or tool that you know to do the operations automatically and not manually: 321- 123 = 198+ 891 = 1089 that is the trick of 1089 greetingssss ....

[@MrVIP1993]

I actually came to the conclusion on my own, I’m so proud of myself!! Although, I spent an awful long time deliberating, but it just came down to eliminating invalid answers like you said.

[@ShivamMishra-td3jz]

Really Very logical...

[@moonchild7032]

Amazing! Love your riddles.

[@AyrtonTwigg]

You can also do ABCDE*4=EDCBA. Solution: 21978*4=87912

[@AbhinavXevents]

Yes,but how did you knew it?

[@DeepakKumar-gd1wg]

@@AbhinavXevents use same procedure as stated in the video or use your own logic to solve the problem in a different way

[@AyrtonTwigg]

I AM A TECHY GUY I used a computer program to check.

[@050138]

By corollary, 219999....999978 x 4 = 879999....999912

[@yurenchu]

@@050138 Also consider the problem: 4 times x = reverse of x where x is an integer, and the digits of x don't need to be all different. if x = Y is a solution, then x = YY is also a solution if x = Y and x = Z are solutions, then x = ZYZ is also a solution x = 0 is a solution x = 2178 is a solution, x = 21[99...99]78 = 22 * (10^(k+2) - 1) is a solution (for any non-negative integer k)

[@Crockist]

This was a fun one to do. I found myself grabbing a paper and working it out myself.

[@DeepakKumar-gd1wg]

I had done some similar questions related to addition. And I really enjoyed solving this one. Thanks. Keep uploading such HOTS questions.

[@critisizerr245]

I am 16 i got this question ABCD=DCBA/4 BA MUST BE DIVISIBLE BY 4 AND A MUST BE EVEN BUT IN RHS A IS IN THE THOUSANDTH PLACE A MUST BE LESS THAT 3 AND SHOULD BE EVEN (IF A IS GREATER THAT 3 THEN D WILL BE DOUBLE DIGIT ) WE GET A =2, 4×D =xA 4×D =x2 From here we get D can be 3or 8 For B this is very tricky 2BCD ×4=DCB2 THE DIGIT D IS DEPENDENT D CANNOT BE 3 BECAUSE IF YOU MULTIPLY 4 WITH 2 YOU WILL GET VALUE>=8 IN THE PROCESS OF MULTIPLICATION WE HAVE TO MULTIPLY B WITH 4 AND GIVE CARRY next step that is 4×A + carry =D D CAN BE 3 OR 8 BUT WE HAVE ALREADY ELIMINATED E SO D IS 8 WHICH MEANS 4×B DOESNT GIVE CARRY SO IS MUST BE LESS THAT 2 IT IS ODD AND 2 IS ALREADY USED SO B=1 21C8×4=8C12 AFTER THIS I USED COMBINATION

[@mawat5207]

@@critisizerr245 i thought the same

[@pedrosil77]

Im so proud to have solved this one. Here is my process: If D=0, then ABCD*4 ends in zero, so A=0. Impossible because all values are different. If D=1, A=4, impossible because then ABCD>DCBA. And so on. Found D=8 and A=2. Then found C and B with the same thematic of "this number will end in x, so...". I was dumbfound that i found with just very basic math and 99,9% logic. Thanks for the great challenge!

[@sundaresank1485]

Sir, Hats off to ur logics. Really awesome

[@vaibhavmevada9668]

Nice i am very lucky to see your logical vodeo

[@ridhamshah3294]

I have done it.. Really happy..

[@arpitkesarwani6123]

I solved this question. This question belongs to Crypt Arithmetic concept

[@critisizerr245]

I am in 10 i am trying my best to do it

[@critisizerr245]

I am in 10 class i am trying my best to to solve it

[@Estrada999]

Do You have more information about cripto atithmetic concept, any software any system to check please, greatings.... :)

[@nandapeela1080]

You are amazing Fantastic questions....

[@LOGICALLYYOURS]

Thanks Nanda :)

[@gandhamsanthosh388]

ABCD : ( 1000A + 100B + 10C +D )×4 = 1000D + 100C + 10B + A {3999A + 390B - 60C - 996D = 0} MAIN EQ. A,B ARE SMALLER NUMBERS WHEN COMPARED TO C,D THEN WE DO TRIAL AND ERROR METHOD BY TAKING EXTREME VALUES IN THE SET {1,2,3,4,5,6,7,8} THEN THE ANSWER COMES TO BE 2,1,7,8...

[@areebahmad1172]

A very innovative one ...really requires out of box thinking ..nice explanation too

[@devondevon4366]

The key is to know what numbers are divisible by 4 since DCBA has to be divisible by 4 and DCBA last two digits MUST be divisible by 4 for DCBA to be divisible by 4 given that ABCD X 4 = DCBA. ABCD CANNOT be greater than 2500 since 2500 x 4 gives a five digit number but DCBA is a four digit number. and A is 2 or less (recall 2500 x 4 = 1,000) but A Cannot be 1 since no number that ends with '1' is divisible by 4 (eg 11, 21, 31, 41, 51, 61, 71, 81, 91, none are divisible by 4, yet for DCBA to be divisible by 4, its last two digits MUST BE DIVISIBLE BY 4) yet we know that DCBA is divisible by 4 , but if A cannot be '1' since DCBA must be divisible by 4 THEREFORE 'A' has to be 2 given that ABCD is less than 2500 (NOTE 'A' is the '2' in the '2500' ) , and therefore 'B' from DCBA has to be a number such that B2 is divisible by 4 (number such as B= 3, or hence '32' or B= 1 , hence '12 as 32 and 12 are divisible by 4 . B has to be either 3 or 1 (since 32 and 12 {the reverse of 23 and 21} are divisible by 4, however though 52 is divisible by 4, B CANNOT be '5' since as aforementioned ABCD is less than 2500. And B cannot 0, 2, or 4 since 02, 22, and 42 (are not divisible by 4.. recall B has to range from 0 to 4, since it can't '5' but has to be less than ''5' since 2500 x 4, gives a five digit number 1,000) So far, for AB we have either 23-- or 21-- . 4 x 2300 means that D has to be 9 but 9 x 4 UNIT digit is not '2' (it is '6') therefore AB is not 23 but rather 21 and D is 8 since 8 x 4 has a '2' UNIT digit , therefore for AB and D we have 21- 8. The question now is what is 'C' . 4 x 8 =32 . The question now is what number multiplied by 4 when added to '3' will give ten- digit of '1' and that of course is '7' as '7' x 4 = 28 + 3 = 31 and, therefore C is 7. So the answer is 2178 . and when multiplied by 4 gives a 'reverse' of 8712. So ABCD is 2187 and DCBA (the 'reverse' of ABCD) is 8712

[@arupchandra3415]

it was a difficult one ... i saw your other video too all are really helpful to us those are seeking for knowledges..👌

[@LOGICALLYYOURS]

Thanks Arup... this is all I need.. a little appreciation... I'll keep making more and more videos.

[@becalmandquiet881]

Great video

[@CultOfJ]

Very very great Ammar Bhai👌👌💪💪😎😎👍👍👍👍Plz Keep on Uploading this type of videos, really liked your each and every video. Even I like logical work and quiz.😇😇Keep it up..!!👍👍👍

[@LOGICALLYYOURS]

Thanks bro for your appreciation... really nice to see your comment :)

[@CultOfJ]

@@LOGICALLYYOURS You deserve it..!!😇😇👍👍

[@HomoIndoeuropaeophilus9766]

Wonderful! 😋

[@N8570E]

Excellent!

[@Prabhaharan.]

Hi this is interesting Please find a better solution here ABCD x 4 = DCBA The number can't be greater than 2500 bcoz 2500 x 4 = 10000(5 digit) Clearly A will be < 3 and even (0 or 2) and B is less than 5 A can't be 0 because D has to be 5 and B's remainder will be only 1 thus A = 2 options for B (0,1,3 and 4) Now D will be 3 or 8 can't be 3 because if B has remainder 1 then D will be 9 So D = 8 thus B will not have any remainder options for B now (0, 1) now forming equation 4000A + 400B + 40C + 4D = 1000D + 100C + 10B + D with A = 2, D = 8 equation will be reduced to 1 + 13 B = 2C we know B = 0 or 1 if B = 0 then C = 0.5 if B = 1 then C = 7 so answer 2178 x 4 = 8712

[@yurenchu]

- "Clearly A will be < 3 and even (0 or 2) and B is less than 5" Why should B be less than 5 ? We know B must be less than 5 when A = 2 . But when A = 0, then B might be greater than or equal to 5.

[@laluparija3487]

This one's my favorite till now!

[@fayallen6799]

So important i love math

[@neelmarzan860]

I've got ur channel & my days are going just Alhamdulillah.. All of ur puzzles I try to solve myself.. Which country r u from? NB: Im new (for the case u told in some video)

[@nishiki2753]

There is a variation to this that is only slightly harder. There is also exactly a 5 digit number with this property. ABCDE x 4 = EDCBA

[@sragav4559]

Solved it myself..... 😀😀

[@jogeshjoshi2535]

Very Nice

[@gizemeryilmaz3492]

N times ABCD = DCBA N>1 and N is a natural number find the resault of A+B+C+D This was our exam question how can we find the N in that question can you help me ? I struggled a lot with that 🌼

[@lukaeric3359]

Im 15 years old and I can solve most of the puzzles.I love logical puzzles,chess,mathematics and physics.I want to ask you what you think which job is good for me?

[@sabyasachirimpa]

A bomb disposal technician.

[@critisizerr245]

Go to FBI special agent I was earlier poor but now i am like you in reasoning and puzzles im 16 Ihave great understanding of physics and mathematics

[@anujkothari2134]

I like to thanks u ammar a lot....after watching your videos my horizon on logics has increased....as I was able to solve this problem by elimination method in single go(although I took some 5 minutes or so to crack it).....keep going buddy

[@flowbeus]

Great explanation!

[@_rtdp_]

a) A has to be

[@ShivamMishra-td3jz]

Great videos ammar sir

[@LOGICALLYYOURS]

Thanks a lot Shivam :)

[@critisizerr245]

I am 16 i got this question ABCD=DCBA/4 BA MUST BE DIVISIBLE BY 4 AND A MUST BE EVEN BUT IN RHS A IS IN THE THOUSANDTH PLACE A MUST BE LESS THAT 3 AND SHOULD BE EVEN (IF A IS GREATER THAT 3 THEN D WILL BE DOUBLE DIGIT ) WE GET A =2, 4×D =xA 4×D =x2 From here we get D can be 3or 8 For B this is very tricky 2BCD ×4=DCB2 THE DIGIT D IS DEPENDENT D CANNOT BE 3 BECAUSE IF YOU MULTIPLY 4 WITH 2 YOU WILL GET VALUE>=8 IN THE PROCESS OF MULTIPLICATION WE HAVE TO MULTIPLY B WITH 4 AND GIVE CARRY next step that is 4×A + carry =D D CAN BE 3 OR 8 BUT WE HAVE ALREADY ELIMINATED E SO D IS 8 WHICH MEANS 4×B DOESNT GIVE CARRY SO IS MUST BE LESS THAT 2 IT IS ODD AND 2 IS ALREADY USED SO B=1 21C8×4=8C12 AFTER THIS I USED COMBINATION

[@ahmedsorkatti6468]

I’ve come across a similar puzzle but with E instead of 4, is that even solvable?

[@aungthansein]

Find the numeric value of (A, B, C, D, E) from the following three formulas. However, A, B, C, D, E represent numbers between 0 and 9. Default: AC x 1E = 4CD (1) ADE x 1EC = AB (2) AB + E9 = 11A (3) Example: If F = 1, G = 2, then 3FG would be 312.

[@abhinavbhutada9b484]

I got it. In just 2 mins

[@prafullgeed7181]

Sir at 2:03 when we are considering a case when A=0 then why we need to think only about maximum number(i.e. 0987).. Can 't we think about any other posibility?

[@manudude02]

The point is that any other possibility would lead to a lower product, he is finding the maximum value of D (the thousands digit in the result)

[@fayallen6799]

WoW

[@fayallen6799]

Good

[@chrisg3030]

I must admit I knew it already and not because I'm good at this kind of puzzle but because I've taken an interest in the number 9801 which is also an integer multiple of its reverse, 1089. The next pair are the numbers in the video, 8712 and 2178, and the next integer pairs are 5445 and 5445 (at a multiple of 1).

[@wangzhp]

a can only be 1 or 2. since 4 * d = xxxa, a can only be 2, while d is 8. (2000 + 100b + 10c + 8) * 4 = 8000 + 100c + 10b + 2 ==> 390b + 30 = 60c, b must be smaller than 2 or c is larger than 10. b = 1 and a = 7.

[@chetanv2007]

Awesome!!

[@LOGICALLYYOURS]

Thanks a lot Chetan for watching all the videos and your comments are encouraging.

[@debarthabhattacharjee427]

I started from the point that if a 4 digit number multiples by 4 gives a 4 digit number so the 4 digit number on the left must be less than 2500. So we get a

[@yurenchu]

The condition "b < 5" isn't necessarily true. It's only certainly true when a = 2 .

[@kumarct6941]

Awesome bro.

[@shibajyotichoudhury8186]

2178*4 = 8712

[@LOGICALLYYOURS]

Perfect !

[@shibajyotichoudhury8186]

@@LOGICALLYYOURS please make some videos on interesting geometry problems.😃

[@amp_001]

Is there any other way to solve this problem much easier than this???

[@minhastrons69]

Thank you 😊

[@natnat1975]

Didn't watch the video to see if he gave the answer, but here it is: A=2 B=1 C=7 D=8

[@lalitkumarmehta1721]

c++ implementation of this may be ... #include using namespace std; int reci(int num) { int i=0; while(num) { i=(i*10)+(num%10); num/=10; } return i; } int main() { for(int i=1000;i

[@LOGICALLYYOURS]

Lalit... that's perfect... I verified your code and it's giving true result... thanks for sharing... :) I also made a minor change in your program for 5 digits (ABCDE) and I got perfect result.

[@ujjwalram3278]

Great

[@learnfromnature3975]

A = 2 B= 1 C=7 D= 8

[@고유환-c2q]

i solved it. In 1.9 seconds

[@sricharancharancharu7424]

Another solution 2008×4=8002😊✌

[@고유환-c2q]

You used two 0's, and 2008*4 =8032. It is a mistake. 😂😂🏁

[@mateenahmed6123]

I am speechless

[@rogerscottcathey]

nice

[@kuls43]

Seems all those 5% of the world ended up in this comment section.

[@050138]

Also 219999....999978 x 4 = 879999....999912

[@devondevon4366]

He said 5 % got it correct? or 95% got it wrong. The most important is to know that the last 2 digit must be divisible by 4 , and that four digit number ABCD) cannot be greater than 2500 since 2500 x 4 = 10000 ( a five digit number). It also cannot start with '1' since no number which ends with '1' is divisible by 4. Those are important things to take into consideration as you know basically that A starts with 2. and the number after 2 has to be such that it reverse is divisible by 4 such as 23 which reverse is 32 which is divisible by 4 or 21 which reverse is 12 and 12 is divisible by 4 these are the important things to know

[@AyushMishra-do5hk]

it took time but I solved it and it was same as yours thanks aamar.

[@jagankhethavath5771]

I got the answer by my own🤠🤠

[@yourdad5471]

I reached upto 2BC8 then get distracted and struck.. Anyway I got the path right that makes me happy

[@UNIVERSE-z1o]

There are so many aliment & entertainment - But no one like DARU friend ..🍷🍷

[@fmjubaier9585]

sir why u delete some of your videos?

[@LOGICALLYYOURS]

Jubaier... no i haven't deleted any videos... so far I uploaded 54.

[@fmjubaier9585]

@@LOGICALLYYOURS u gave a puzzal of 13 card fliping which is similar to dark room coin fliping. whare is the that video?

[@LOGICALLYYOURS]

It's here bro : kzbin.info/www/bejne/qpO8nIeViN9kjrM

[@desidoodperfect]

Still wating😗😗 For face....

[@LOGICALLYYOURS]

DDP... I owe you a nice give away bro (by the way give away session will be soon in the form of metal puzzles)... . will take little more time for facecam as I am preparing for a Hindi Vlog channel of same logic category but with entertaining contents.

[@desidoodperfect]

@@LOGICALLYYOURS you always say this But as a fan i will be waitings😶😶

[@sakshamtayal6534]

Found the solution within 5min

[@zayoudayoub1408]

I've solved this when i'm 14 y.o

[@AcaciaAvenue]

AT 4:35 there wasn't even the need to analyze the cases: C times 4 gives an even number; also, 3 is an odd number; since an even number plus an odd number always gives an odd number, and since 0 is considered even, there is no valid value of C.. EDIT: Ah i see you used the same layer for the correct C value, nevermind.

[@anewkind7636]

I know this maths puzzle 2178×4=8712 But logically yours

[@alkabindal4793]

Bhohut hard

[@amodpatel1]

I get this only in 1 hour..😅😅😅

[@OW0974]

Just as a side note I know that 21978*4 = 87912

[@sricharancharancharu7424]

I took 25 mins to solve this ✌✌✌

[@kaziemarieong9953]

I got 2148.. almost!! I challenged myself to not pause the video to solve so I only got a few seconds, got 3 out of 4 digits though

[@junhao4514]

i have one question in my homework that the same to that question.... 😁

[@muthukrishnanvikram440]

2178

[@V7B817]

I have said it before many times that ur videos are awsm, i really satisfied with your chery's birthday riddle explanation, I got the answer of this puzzle although it took time for me. I have to infer a query regarding of this as like A3 as a must equation as if we little manipulate the equation then it becomes ABCD= DCBA/4 and for any number less than 4 the answer must be a 3 digit number, i had done this problem by just comparing these two sides of the equation first of multiplication then simultaneously by division.

[@jasonparon235]

Technically as long as one of the values = 0 then it does not matter what the other values are because when you have letters next to each other that means multiply unless their is another sign so we take ABCD*4=DCBA. Broken down its A*B*C*D*4=D*C*B*A. So let's plug in numbers A=0, B=1, C=2, and D=3. So you have 0*1*2*3*4=3*2*1*0. Now we know anything time's 0 will equall to 0. 0*1*2*3*4=0, 3*2*1*0=0, 0=0. Solved. Took me 5 seconds to come up with that solution.

[@vaibhavmevada9668]

Video

[@yurenchu]

If the digits A, B, C, D are not necessarily different, then the only solutions are (still) ABCD = 0000 and ABCD = 2178 . Also consider the problem: 4 times x = reverse of x where x is a non-negative integer, and the digits of x don't need to be all different. if x = Y is a solution, then x = YY is also a solution if x = Y and x = Z are solutions, then x = ZYZ is also a solution x = 0 is a solution x = 2178 is a solution, x = 21[99...99]78 = 22 * (10^(k+2) - 1) is a solution (for any non-negative integer k) If I'm not mistaken, those five rules comprise all possible solutions. That would mean that the only solutions where all digits of x are different, are x = 0, x = 2178 and x = 21978 .

[@TJTrickster]

Mathematically one must be 0 the rest don't matter

[@utkarshmishra3133]

Lajawaab

[@smartclass27]

1111

[@chethanm.p2268]

0000 would work fine rit?.

[@고유환-c2q]

I can do ABCD × 9 = DCBA. Solution: 1089 × 9 =9801

[@anastasissfyrides2919]

1) Stop asking numerous times for subscriptions, it deters us from actually doing it 2) Use specific terminology when referring to math questions. Numbers between 0 And 9 OR numbers 0-9. Also integer numbers. Otherwise there are infinitely many numbers between any two real numbers. 3)Lack of the aforementioned necessities reveals that these problems are copy pasted from other videos(i have seen many of them before) and that you are barely understanding the problems and arent offering something new, to ask for subscriptions or ask us to "think logically" since you didnt do it in the first place. And what does it even mean to think logically? It isnt like we think irrationally everyday but for your videos we choose to be wiser. Jeez. 4)Feel free to sometimes be honest and say that you actually didnt solve those by yourself. Mindyourdecisions does it and earned my subscription and millions more.

[@devendraparmar8390]

??

[@marvinkitfox3386]

a OR b OR c OR d = 0, the rest don't matter. ABCD == A * B * C * D If any of them is zero, total is zero. Wot? Are you using some other mathemagical system where AB is not A multiplied by B? tsk, tsk, tsk.

[@HarshitSingh-nq4oy]

You are worng at 2:17 because you have multiplied 0987*4 that means D can be 7 also or why only 1,2 or 3 it cam be 5 also, you theory about it is wrong please make it right

[@arrowrod]

zero

[@talkingtal5369]

Very Nice

[@vivekatuitioncenter1529]

2178