ABCD x 4 = DCBA. Can you find the values? Challenge for you!
video description
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.
Date: 2023-11-15
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Comments and reviews: 24
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
reply
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
reply
yuri
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.
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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.
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Devon
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
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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
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varshit
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.
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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.
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Jason
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.
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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.
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Pedro
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!
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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!
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Acacia
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.
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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.
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Chris
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.
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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.
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lalit
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
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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
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Aung
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.
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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.
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Anuj
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
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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
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Marvin
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.
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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.
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Dan
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
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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
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---
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.
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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.
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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.
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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.
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Harshit
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
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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
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Cult
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! ---
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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! ---
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luka
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?
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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?
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Prafull
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?
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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?
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Debartha
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
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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
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Nishiki
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
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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
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Deepak
I had done some similar questions related to addition. And I really enjoyed solving this one. Thanks. Keep uploading such HOTS questions.
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I had done some similar questions related to addition. And I really enjoyed solving this one. Thanks. Keep uploading such HOTS questions.
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Kazie
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
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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
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Arup
it was a difficult one. i saw your other video too all are really helpful to us those are seeking for knowledges. -
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it was a difficult one. i saw your other video too all are really helpful to us those are seeking for knowledges. -
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