Can you solve it? Logic test ABCD x 4 = DCBA

[@dale3478]

I didn't use algebra. After finding out that A=2 and D=8, it means whatever Bx4 is, it must result in 1 digit. So B is either 0 or 1, because 2 has been used, and 3 above ends with 2 digits. If B=0, Cx4 must end with 7 to cancel out the 3 from 8x4=32, which is impossible because 4 is even. So B=1, and Cx4 must end with 8, so either 2 or 7. And since 2 has been used, C=7.

[@in-everything2330]

Same

[@benabedmhamed2360]

Yes, you resolve that by 100% logical reasoning by observing digits instead 50% logical and 50 % algebra.

[@dawnscript1]

Same

[@57thorns]

Same here, have been doing thousands of math puzzles in emacs, they are of the form a four digit number multiplied by a two digit number with the intermediate products and the final sum all given with one letter corresponding to the same digit.

[@angeluomo]

My method, as well.

[@MCLastUsername]

Fun facts: - This is one of only two 4-digits numbers that can be reversed by multiplying it by another number, the other being 1089 * 9 = 9801 - Adding any amount of 9s to the middle of the number generates a longer number that is also reversible in this way e.g. 2199978 * 4 = 8799912 - You can also generate longer reversible numbers by repeating the digits e.g. 21782178 * 4 = 87128712

[@EaglePicking]

Thanks for the info. Interesting stuff.

[@Fire_Axus]

I studied the problem further and it had some pretty interesting results.

[@puneetporwal]

It's something to do with alternate digits adding up to 9

[@espadadearthur1174]

And also: 1089*2=2178

[@novamc7945]

​@@espadadearthur1174what does that do

[@dvorszkydavid9140]

As this was a logic test, I avoided using equations and got the result like this: A and B can't be 0 because numbers in math don't have a leading zero. A can be either 1 or 2 because if it was 3 or larger the product would be 5 digits. We multiply D by 4, so we must get an even number, which means A equals 2. As A is 2, D could only be 8 or 9, but because D times 4 ends with a 2, it could only be 8. Now, B should be 0, 1, or 2 (otherwise, B*4 > 10, and we would go over 9000) We know that C*4+3=B (because we have the extra 30 from 8*4), which means B must be an odd number, therefore B = 1. Finally, we know C*4 ends with an 8 (because C*4+3 ends with a 1), which means C could be 2 or 7, but 2 is already taken, so C = 7.

[@daniellabe6838]

I did it the same exact way

[@number42iscool]

i used same method lol

[@arkadarkartist]

my exact thought about it. "they said it was a logical problem, so there must be a logical method for it"

[@jimlocke9320]

I did it pretty much the same way. Once you determine that A=2 and D=8, you know that B must be less than 3, because 3 or greater would produce a non-zero carry to the thousands digit, and the carry must be zero for D to be 8. Furthermore, B must be odd because you got an odd carry, 3, when you did the units multiplication, 8 x 4 = 32, and 3 must be added to an even number (all multiples of 4 are even). So, B = 1 and we look for values of C for which 4 x C + 3 will produce a 1 in the units position (10s position in the 4 digit product), and the only values are 2 and 7. However, 2 has already been used, leaving 7. So, we try 2178 x 4 = 8712 and obtain a valid solution, and we have ruled out any other possible solution.

[@jakistam1000]

I did the same thing, except I didn't assume that A can't be 0. Leading zeros aren't generally written, but it's not mathematically forbidden to write them. Instead, I eliminated A=0 as in the video.

[@BlazingLarvesta]

It took me a bit, but after figuring out A and D, I brute-forced my way through the other equations until I got ABCD = 2178. BC cannot be >= 25, as that would mean D would have to be 9 or greater. After going through the remaining possibilities, I got that BC = 17.

[@handanyldzhan9232]

Congratulations, you have evolved into Volcarona.

[@adamjohnson9188]

That is EXACTLY how I did hahaha

[@someone_2110]

Thanks for this video. Your channel is the most productive and useful one in my opinion!

[@massimomoro5895]

I got this exact problem in a high school math competition. I was the only one of my class who was able to solve it. I think it's a really good and clever problem to figure out

[@123vaibhavgupta]

i got this in GRADE 2 AMO and solved it bruh

[@cewla3348]

i mean, the way it is worded in the video means that it could be strings (variables representing plaintext) numbers (a(b(c(d)))*4 = a(b(c(d)))) for numbers, any combination with a 0 in as b,c, or d works

[@michael1]

That's surprising because it's pretty easy to solve by trial and error once you factor in a couple of constraints giving sensible values for A, B and D. Presh does well with figuring out 2 and 8 for A and D quickly just using logic, but then goes down a rabbit hole of more complicated maths even though it's obvious that B must be 1 - it can't be 2 because A is 2 and 2300 * 4 is over 9000 but we already know D is 8. So we have 21x8 and there are only a few to test. Albeit we can see that in the multiplication there'll be a carry of 3 from 4 x 8 = 32, so it's 4 x something + 3 = x1 and it's obviously 7. All that 1000a + 100b was overcomplicating it and not really using 'logic' as the puzzle suggested.

[@TheShox79]

Python: for a in range(10): for b in range(10): for c in range(10): for d in range(10): if a!=b!=c!=d: if (a*1000+b*100+c*10+d)*4==(a+b*10+c*100+d*1000): print(a,b,c,d)

[@R_V_]

This is not mathematical reasoning, this is brute-forcing the problem. That would had been perfect for a channel dedicated to elementary algorithmic, though (even if brute force is awfully inefficient and wasteful).

[@R_V_]

@@wernergamper6200 Tssk tssk tssk. It's shorter in SQL, which means you shouldn't persist in your error : your way is the wrong way.

[@Furious_Farts]

It is neat to read the different ways people problem solved this. I followed the logic of A having to be less than 3 finding A=2 and D=8 worked. For the next part I understood 8*4=32 where 2 is A but then used the 3 carrying over into B 3 ABCD 2BC8 x4 --> x4 DCBA 8CB2 so I used this to set up the next part. I brute forced 4(BC)+3 and looked for CB and landed on BC=17 as 4(17)+3=71.

[@alexr2347]

2:34 There is one more trick to make it simpler: to get D=8 in the first place from A=2 you need to be sure that B

[@danmerget]

Not only that, but B also has to be odd, for two reasons: (1) We know DCBA is a multiple of 4. We already know that A=2, and all multiples of 4 ending with 2 have an odd number in the ten's position. (2) We know D=8. So when we multiply D*4, we carry a 3. We know C*4 will be even, so when we carry the 3, it'll add up to an odd number. Given that B is an odd number and B

[@Donald_the_Potholer]

@@danmerget I am honestly surprised that Presh didn't consider that process in his solution.

[@anildeshkar906]

Way the logic is explained and developed is super

[@mystery5906]

before watching the video, just from the thumbnail, heres my thought process: to start, the letters must represent digits in an integer, not integers being multiplied, since order doesnt matter in multiplication. (that, or one of the letters is 0, which brings everything else to 0 through multiplication) in order to be a four digit number both before and after being multiplied by 4, the first number has to be less than 9999/4, which is ~2499. next, for ABCD x 4 to equal DCBA, A has to be an even number, because 4 is even, and any number multiplied by an even number is an even number. since its under 2499, A is either 0 or 2, and since whole numbers dont start with zeros, A=2. with 2BCD x 4 = DCB2, we can easily figure out that B is less than 5, because our maximum starting value is 2499. we can also deduce that because our minimum starting value is 2000, our minimum result is 8000, meaning D is either an 8 or a 9. then, we can see that any number ending in 9 multiplied by 4 ends in a 6, and any number ending in 8 multiplied by 4 ends in a 2, which matches our value for A. so, we now have 2BC8 x 4 = 8CB2, and B is less than 5. we can narrow that down to less than 3, because 2300 times 4 is 9200, which is too high. our result cant be higher than 8992, so our starting value cant be higher than 2248. at this point, we only have 25 possible answers (00-24 for BC). it would be easy enough to solve this through brute force, but ill try to find a more elegant solution. the next thing i noticed is that whenever you multiply a number ending in 8 by 4, the resulting number always has an odd number in the tens place. that means that B must be an odd number, and we already know it cant be 3, so B must be 1. now, we have 21C8 x 4 = 8C12, and you dont get a trophy for thinking harder instead of smarter, so we can just manually try 0-9 for C until we get 7 as the answer. 2178 x 4 = 8712. now, its time to watch the video and see the easy way to reach the answer. EDIT: nah his way seems way more complicated 💀

[@deerh2o]

At the beginning, eliminate A or D being 0 since they are the lead digit of a 4-digit number. Also, guided trial and error works out much better than the algebraic expansion you show in the video. Since 23 x 4 = 92, B must be 0, 1, or 2 and 2 is already taken . . .

[@OneWeirdDude]

I guess "zero can't begin a multi-digit number" isn't in his rules?

[@dale3478]

I also thought this at first, so I replayed the problem statement and it never said anything about "4 digit number". It only says ABCD x 4 = DCBA, and that each letter corresponds to a different digit between 0-9. It may also be the reason why the video tries to validate A=0

[@ShimrraJamaane]

You can always add a leading 0. There is nothing that rules this out in mathematics. However, it's easy to rule out A=0 as that would require: 1) If A=0, then D=5 1.a) (4 * D) mod 10 = A 1.b) 1.a can only be satisfied by the integers in the set (0, 5), as the remaining permutations result in A being in the set (2, 4, 6, 8) 2) B2 because 4 * D = 20, which carries a 2 5) There are no integers in the set (1, 2) that are greater than 2 6) Step 1 must be false, making A≠0

[@crtwrght]

I assumed A and D were non-zero since the question asked for 4 digit numbers. Then A

[@bjorncedervall5291]

Hurrah - I am 71 and solved this relatively quickly without taking notes - just exclusion principles - no equations - just a few trial and errors. Getting old is obviously not the complete bottom. Must however admit that I solved several such problems when I was about ten (thanks to one of my parents who gave me a lot of logical problems to tackle when I was a child) and in addition I always liked numbers (and still do basic calculations as part of my daily life). I am not some type of math "genius" - but see and exercise relatively trivial math all the time, trends, orders of magnitude and so on.

[@feelingzhakkaas]

ABSOLUTELY WONDERFUL

[@Georgggg]

2178 First thing you cap A above: its 0, 1, 2. because then number would be 5 digits. A is not 1, because D*4 is not ending by 1. A is not 0, because D*4 ends by 0 only when D = 5. 3 digits number multiplied by 4 cannot be more than 5000. So, we have 2BCD*4 = DCB2 Then D is either 8 or 9, because number is greater than 8000 and less than 10000. 9*4 ends with 6, should be 2(A) 8*4 end with 2. now 2BC8*4 = 8CB2 Then we can say B is less than 3, because 2300*4 = 9200 > 8999 Now just check for 1 and 0, because 2 is used and C is easily guessed because its only number left. 2178 * 4 = 8712

[@anonymoususer188]

I used basically the same logic for A and D (although I didn't bother accounting for the possibility of A being 0 because people don't usually include a leading zero). However, my method for solving B and C was slightly different. I realized that since D was 8, after multiplying 8 with 4 to get 32, a 3 would be carried. Since 4 x C + 3 = a number with B in the ones place and 4 x C has to be even, adding 3 makes it odd. Therefore, 3 has to be odd. Furthermore, because A is 2 and D is 8, 4 x B can't carry anything because that would increase D. Therefore, B has to be less than 3. The only thing it can be is 1, and figuring out that C is 7 is trivial once you know the other 3 numbers.

[@Whiplash_MUSIC-hj7vp]

I solved this by using a very similar process to find a=2 and d=8, and then used the same principal to say that b

[@MoTab78]

Before going to algebra after finding A and D, we could also find B with logical elimination, such as if B is 3 or more then the result DCBA should be 5 digits, so B can be 2, 1 or 0. Because A is 2, B can't be 2, also if B is 0 then there is no integer solution of C [since (C * 4) + 3 should be equal to C * 10 (where 3 is carried from D *4 = 8 * 4 = 32) ] , so B should be 1. Then it is much simpler to solve since (4 * (10 + C )) + 3 = (C * 10) + 1 where C should be 7

[@jasonw0823]

I remember this question in my grade 6 summer holiday homework! I can still remeber the excitement when I solved it! This is amazing!❤

[@Ultimate-Records]

That was amazing. Thank you Mr Talwaker.

[@quigonkenny]

All multiples of 4 are even, so the A in the product must be 2, 4, 6, 8, or 0. The A can't be 0, because the D in the multiplier would have to be 0 (which is already A in this hypothetical) or 5, and any 3-digit number multiplied by 4 is less than 5000. The A in the multiplier must also only be 1 or 2 because any multiplier larger than 2499 will give us a 5-digit product. So A = 2. The only numbers that give a final digit of 2 when multiplied by 4 are 8 and 3, so the D in the multiplier must be one of those. As we know A is 2 and there are no multipliers between 2000 and 2999 that result in a number between 3000 and 3999 when multiplied by 4, D must be 8. As we now know the product is between 8012 and 8972 inclusive, the B must be 1 or 0, as 2 is taken and any multiplier larger than 2249 gives a product of 9000 or more. 4×8 gives a carryover digit of 3, so the B in the product must be odd and therefore 1. The C in the multiplier must give a result of 8 for B to be 1, meaning C must be 2 or 7. 2 is taken, so C is 7. 2178 × 4 --------- 8712

[@kimsmoke17]

Nothing to see here people… move along. move along… ABCD x 4 = 2AF34

[@dvilardi]

for a in range(0,10): for b in range(0,10): for c in range(0,10): for d in range(0,10): abcd = d + 10*c + 100*b + 1000*a dcba = a + 10*b + 100*c + 1000*d if 4*abcd == dcba: print(str(abcd) + " x 4 = " + str(dcba))

[@MrLegendGaming]

This isn't realy a logic test, it is more of how good you are at analysing stuff.

[@Francis-Abdul]

Congratulation, you played yourself.

[@peppermintgal4302]

That... that requires logic lol.

[@peppermintgal4302]

Also fyi math is logic, and vise versa.

[@mjmeans7983]

I thought of this test differently, because I only looked at the initial problem without first listening to the requirement for the problem to have unique digits. More like a brain teaser that is designed to trick you into thinking it's something that it's not where X wouldn't mean multiplication, even through the entire structure of the presented problem looks like a classic math test. So, I thought if X is a transposition operator instead of multiplication, then it's a very simplified towers of Hanoi problem with no placement limits other than to find a solution that works with 4 transpositions instead of the optimal 2 (AxD, BxC). In that case, the solution with four transpositions is ((BxD, AxC), (DxC, AxB)) where the first two and last two transpositions happen to be commutative within the pair and the pairs are communitive to each other. So, there are 8 solutions that solve the problem if X means transposition.

[@deandelvin9924]

I solved it like this: 1. D cannot be 0, or else A would be 0 too, and all letters represent different numbers. That means DCBA is a 4-digit number (i.e. doesn't start with 0). 2. A is the last digit of a multiple of 4, so it's an even number or 0. So A is 0, 2, 4, 6 or 8. 3. The product is a 4-digit number, so A cannot be more than 2. So A is either 0 or 2. 4. If A=0, then D is 5, and ABCD is a 3-digit number. But there is no 3-digit number which you can multiply by 4 and get a 4-digit number starting with 5. Therefore, *A=2*. Other letters cannot be 2. 5. If A=2, then D is either 2*4=8 or more (9). But 9*4=36 which doesn't end in 2. 8*4=32 does end in 2, so *D=8*. Other letters cannot be 8. 6. B*4 must be less than 10, otherwise it would add a number to D at the beginning of the result and mess up 2*4=8. Therefore, B is 0, 1 or 2. A=2 so B is either 0 or 1. 7. If B=0: 8*4=32, C*4+3=a number ending in 0. C*4 is even, 3 is odd, so their sum is odd and cannot end in 0. Therefore, *B=1* (C*4+3 is odd so it can end in 1). C cannot be 1, 2 or 8. 8. 1*4=4, so C is between 4 and 9 (excluding 8). 9. (C)8*4 is a number ending in 12. 8*4=32, (C)12-32=(?)80. This means C*4 ends in 8. Between 4 and 9, only 7 fits this condition (7*4=28). *C=7*. Result: 2178*4=8712.

[@arkadarkartist]

I did the same method for A and D, but for B and C I kind of brute forced it by realizing Bx4=C couldn't become more than 9 because it would affect the result of 2x4=8, so it couldn't be 3 or higher again like it wasn't possible for A Knowing B=Cx4 was actually B=(Cx4) +3 (because of 8x4=32 on the final slot), I knew I needed a number that multiplied by 4 would end up on a 7 or 8 to become 0 or 1. 7 is not even, so it was impossible to get it from a multiplication of integers by 4, so the result had to end on an 8, which, coincidentally, was 7x4=28. +3 it becomes 31. 1x4=4, add the new +3, becomes the final 7 and everything match.

[@JDC2890]

Here's a way similar to how I did it. Start with the restriction on A. Both numbers are 4-digit numbers, so A cannot be 0. Furthermore, with A being multiplied by 4 and having a single-digit result with a possible carry, A < 3, otherwise you would get a 5-digit number. You need to figure out if A is 1 or 2. Next, you need to figure out the possible last digits of 4 X D. Going through the 4 times table will get you even integers every time, meaning the last digit will be even, and the last digit is A. This means A can only be 2. Consider 4 X A. This value, plus a possible carry, must be D. 4 X 2 = 8, and the carry can be at most 1, otherwise you would have a 5-digit number. This means D is 8 or 9. D = 9 would lead to a problem. 4 X 9 = 36, but that would mean A is 6, which it is not. So D = 8 must be the correct value of D. You have a 4-digit number between 2000 and 3000 in value, multiplied by 4 to give one between 8000 and 9000 in value. Focusing on the 2nd digit, B, yields the following restrictions. B = 3, then you'd have something like 23XX multiplied by 4 = 9XXX or something nonsensical. B is 0 or 1, and we can now simply try each one out. Trying 20C8 X 4 = 8C02 leads to instant trouble as the last digit of 4C + 3(carry from 4x8 = 32) has to be 0, which would mean the last digit of 4C would have to be 7, which is not possible with an even number. So B can't be 0. This leaves us with one possibility, B = 1. Trying 21C8 * 4 = 8C12 leads to 2108 * 4 + 10C * 4 = 8012 + 100C 8432 + 40C = 8012 + 100C 420 = 60C 7 = C Now you have a solution of 2178. Evaluating 2178 X 4 gives you 8712, which satisfies the equation.

[@nicemathproblems]

Very nice thumbnail. Love the use of colour👏 👍🥰

[@davidgillies620]

ABCD is < 2500. By inspection, A is even and less than 4, hence 2. D is therefore 8. Multiplication by 4 carries a 3 into the tens column. Now we are left with 4(10 x + y) + 3 = 10 y + x for the middle two digits. This yields 6 y - 39 x = 3, for y = 7 and x = 1. 4 2178 = 8712.

[@christopherg2347]

You can do it slightly quicker at two parts: 1:12 0x4=0. So it can not be 0. 2 is the only option left. 3:53 Right side must be uneven. Because any multiple of 2 must be even. And any even number -1 must be uneven. So only 13 is left.

[@amoghverma3108]

A common question for Business Schools now sorted 😊

[@gavindeane3670]

It seems to be mostly about thinking in terms of place value. A is

[@handanyldzhan9232]

Assuming there's no leading zero, A can only be 2 for it to be even and D to be a digit. Then 2BC8 x 4 = 8CB2, making B odd (carried 3 above the digits of 10s) and less than 3 (no carried numbers above the most significant digits), thus 1. The sums of digits are the same, so 21C8 must be divisible by 3, and C = 3k+1, and C can only be 2 or 7. Since it can't be 2, ABCD = 2178.

[@iminediamonds]

parity actually makes this faster A < 3 as Ax3 > 9999, which breaks the equation if Dx4 = A and A < 3, that means A must be an even number (x4) therefore A must be 0 or 2 but since A is a leading digit of ABCD, A cannot be 0 and therefore must be 2 2x4 = 8, but because of carryovers D might also be 9 however, Dx4 = 2 + 10x (x is an integral but unknown amount of 10s) and 9x4 = 36 =/= 2 + 10x, which means D = 8 2BC8 4 8CB2 there is a carryover of 3 from 8x4 = 32, and Cx4 + 3 = B + 10x, but this still makes B odd (Cx4 even + 3 odd = B odd) the maximum number DCBA can be is 8992, and 8992/4 = 2248, therefore ABCD can at most be 2248. therefore B

[@Maikelz93]

yea, I proudly say I can and did solve it. Thank you for the curious math riddle.

[@darbyl3872]

Solved it by thinking for A minutes. BA must be divisible by 4. A must be about one-fourth of D. The 3 carried over from 32 (4xD) must be subtracted from 1 (12) or 3 (32) or 5 (52) or 7 (72) or 9 (92). Tried the 12 (CD), giving 28 + 3, and 4x7 = 28. EZPZ.

[@2ndless57]

I use different logic for solving C and D. Since A is 2 and D is 8 that means 4B 4 which mean C=7

[@bakyra4683]

@MindYourDecisions this reminds me of my Math olimpiad excercise around 25 years ago... :) Find 3 digit number ABC, that when CBA is subtracted, equals to the digits A B C in any order.

[@olerask2457]

Less algebra: After realizing A=2 and D=8, you have 3 + 4*BC = CB. As before B must be either 0, 1 or 2, and as B must also odd, we have B=1. Thus 42=6*C, and C=7.

[@DiThi]

I tried to solve it from the thumbnail and gave up the moment I had two possible solutions (and one was correct) thinking I did something wrong. If I tested my two solutions I would have realized I forgot one step in my logic to check it.

[@nowymail]

I loved this kind of puzzles when I was a kid! You can also try some random letters and try to determine if there's a solution.

[@saisharan1132]

Nice question

[@Allangulon]

Would you like a beer is a nicer question.

[@DonaldKronos]

The first thing I noticed is that the answer A = B = C =D = 0 is a valid solution, and the second thing I noticed is that your not using the algebraic convention where ABCD means A times B times C times D because if you were, then it wouldn't matter what the values of those for variables were, since single-digit numbers are all real numbers and real numbers are commutative in multiplication. Actually I'm not sure what I might have recognized both of those things at the exact same time. The other solution was pretty obvious, a few seconds later, but was quite a bit more of a challenge. The easiest part of that solution of course was recognizing that the solution divided by 2 had to end in 4.

[@DonaldKronos]

I mean leaving this up here as is, including my now blatantly obvious error. LOL! I just noticed one reading someone else's comment that I failed to realize each digit was supposed to be different. I either missed the word different in the description entirely or must have taken it in a way that made it redundant, such as for example to say that in 0000, the first zero is a different digit then the second zero. That's pretty funny. I could delete or correct my comment, but I'll leave it there in the hopes that someone will get a good laugh out of it.

[@Neodynium.the_permanent_magnet]

After A,D I kept going with the logic: since 4B has no carry, B is 0 or 1 (2 is taken). But it cannot be even (0) since 4C+3 is odd. So B = 1. Then C can only be 7.

[@DreadKyller]

After finding that A=2 and D=8 I didn't go algebraic for the rest. 1) I multiplied D (8) by 4 to get 32, so a carry of 3. Any number multiplied by 4 would be even and since the carry was odd then B in the result must be an odd number. 2) Knowing that B must be odd now, we try solving for B in the top number. Multiplying anything above 2 by 4 would result in a two digit number, which would carry over in the B * 4 column, which would mess with the the first column A -> D equality, so B must be 2 or less, in addition 2 is already used so that leaves 0 and 1, and there's only one number that's odd in that pair, therefore B must be 1. 3) This meant that C * 4 + 3 (carry from D * 4) must end in 1, meaning D * 4 must end in an 8. Only two numbers 0-9 meet this criteria, 2 and 7, as 2 is already used and each letter must be unique this meant that C had to be 7. Interesting that there are so many ways to arrive at the answer.

[@teawithmilk747]

Exactly the same solution as I did. Elimination process and logic!

[@jagmarz]

I think the logic that set A&D also plays for B&C. Once we have A=2, we know that B = 4C*3, so it has to be odd. However, again we have no carry into the 1,000's place, so that means that B must be 0,1, or 2, so it has to be 1. Now, we know C must be 4,5,6 or 7, (4 * 1 + carry of 0..3), and we see that the only one of these which satisfies 4C+3 = X1 is 7.

[@sparshsharma5270]

For a 4 digit number, A>0 Since Dx4 ends with A, A is even (2,4,6,8) Since DCBA=4 digit no, so A=4, so D=8 Now 4*C+3 ends with B {B=1,3} Since B < 5 & since 4*B+[carry] < 10, so B=1 Since 4*B+carry=C, so C=7 Therefore, ABCD = 2178 & DCBA = 2178*4 = 8712

[@hetwitblad6544]

I started with A which indeed had to be 0 or 2 Then I went to B. If a number is divisible by 4 the last two digits are divisible by 4. This means BA is either 12, 20, 32, 40, 52, 60, 72, 80 or 92. However, if A is 2, B cannot be higher than 2 otherwise D>9. So B is 1, or an even number higher than 0. As D would have to be 5 for A to be 0, which is impossible with an A equal to 0, B would have to be 1 and A is 2. For A to be 2, D has to be 3 or 8, and if A is 2, D is 8 or 9. So 8 is the only option here And for C I used algebra, a linear equation with one variable is pretty easy to solve

[@u9vata]

After concluding how A and D should be (because there cannot be "carry" over from A multiplication) you can also just as well work backwards and realize that you cannot have "carry" over from the multiplication of B. Why? Because otherwise the only possible calculation for A would be wrong as it does NOT have any carry as you can see! That is an other way to solve this - at least that is how I go for it for example.

[@mehill00]

Fun! I began immediately with an equation using multiplication times powers of ten for the digits to get: 3999A+390B=60C+996D with 0

[@marcinkusmierzak991]

When we know A and D we already know that B can be only 0 or 1. 4B is smaller than 10 and we already used 2. 4C is always even. 4C +3 is odd which means B is also odd - it needs to be 1. 4C+3 needs to end with 1 - C can be only 2 or 7 and we already used 2. This leaves 7 without any calculation

[@MUJAHID96414]

About 5 years ago I found this, and this is written on my diary

[@SiqueScarface]

I was starting the same finding A=2 and D=8. But then I argued that B has to be 1, as Bx4 has to be  łess than 10, which means that B, as A, can only be 0,1,2. With the 2 already taken, B has to be 0 or 1. We can rule out B=0, because B has to be an odd number, because C*4 will be even, but D = 8 and D*4 = 32 causes B to be odd. Thus B=1. With B = 1, we know that C can only be 2 or 7, because those are the only digits for which Cx4+3 ends in 1. As B=1, and C larger than Bx4, we know that C  has to be larger than 4, which gives C = 7. Hence, ABCD = 2178 and 2178x4 = 8712 = DCBA.

[@Ima-hoot]

Pretty easy. I was able to solve without pen and paper ! A has to be 2 (even and can only be 1 or 2 ). D has to be 8 ( only possibilities are 8 or 9 and only 8x4 results in 2. B can only be 0 or 1. 0 does not work because with carry for 3 from 8x4 we cannot get 0 in tens digit for B. Therefore B =1. Cx4 must end in 8 (because of 3 carryover and B=1). Only possibility is C=7. The multiplication works and answer is 2178 for ABCD

[@MiahooJunk]

in excel or google sheets, A= 1000-9999 (u can up to 2499, since 2500*4 is 10000) B= =A2*4 C= =MID(A2,4,1) & MID(A2,3,1) & MID(A2,2,1) & MID(A2,1,1) D= =TO_TEXT(B2)=TO_TEXT(C2) look for true in D to get 2178

[@rudrodeepchatterjee]

My method: I realised ABCD

[@bradybirdy]

I found it by saying that because 2000*4 = 8000 and 8 is guarenteed to be the leading number, BC8*4 < 1000. Which means B is 0, 1, or 2. I then said that because 8*4 would carry the 3 then I need a number that when you multiply by 4 and add 3, it has to end in either 0, 1, or 2 so it can be equal to B. So that means C has to be 2 or 7 to satisfy this. So then because nothing can multiply by 4 and get 3 added and end in a 0 or 2, so B has to be 1. And if B is 1, then C has to be 7, because if C is 2, then that ends up making D = 5, which it isn't. So I used that way to find B = 1 and C = 7. I'm glad there were multiple ways to see it.

[@copernic7511]

¤ (A,D) can only be (2,8), because: - A*4 < 10 => A A even => A=2 - if D=9 then A=6, so D=8 ¤ (B,C): - B*4 -> C, no remainder => B B odd => B=1 - testing C=2 or 7, only the latter works => ABCD = 2178

[@Семён-ч7б]

To solve c, b i use equations B=c*4-30 C=b*4 B=0》c=0》0=-30(wrong) B=1》c=4》1=-14(wrong) B=2》c=8》2=2 I know that 2 is already =a, but this way dont give solution where all letters are different digits

[@walterengler5709]

A and D were easy. Then I went the route that B had to be 0,1, or 2 as anything greater would be carry over onto the ax4=D so not possible. Then 4xC+3 (4x8 = carry 3) is always going to be an odd number. Hence B had to be 1. Finally with B as 1 only4x2+3 and 4x7+3 produce a 1. 4x1 + that carry = C only works if C is 7.

[@TonVenhuizen]

Solved it but by approximation. Looking at the formula (if you simplify the equation) you can see D/A is approximately 4 and then C/B is approximately 6.5. There's no good solution for A=1 since if A would be 1, B would be 2 and thus C would be 12 or 14, so A=2. D would then become 8. B would then be 1 so C = 6 or 7...

[@albajasadur2694]

We can reach the step 2 B C 8 x 4 = 8 C B 2 follow the video. Without solving equations, For the 2 B C 8 number, we know that B x 4 must be a single digit, so only choice is 0 and 1 (2 has been used) the 8 in 2 B C 8 x 4 =32, so the B in 8 C B 2 cant be 0, set B = 1 For the 8 C B 2 number, now becomes 8 C 1 2. There shall be a "3" bring forward to tens digit of 8 C 1 2. We need a number end with 8, so it add 3 to generate 1 at ones digit. As "2" has been used, so the only choice is "7", 4 x 7 + 3 = 28 +3 = 31, end with a 1. C = 7 We don't need to solve equations

[@cherkovision]

Easier way to figure out B: If B is greater than 2, then 2B00 * 4 is greater than 8999. We know that the product is not greater than 8999, therefore B is 0, 1, or 2. If B is 0 or 2, then the product ends in 02 or 22. No multiple of 4 ends in 02 or 22, therefore B is not 0 or 2. Therefore B is 1.

[@robinlindgren6429]

ABCD * 4 = DCBA A, B, C, D are different single-digit non-negative integers. if A was 3 or greater then the resulting number would be at least 5 digits, therefore, A is at most 2. any number multiplied by an even number can only become an even number, therefore A must be even, combined with not being 3 or greater A must be 2. (A being 0 implies that D is also 0 which is not allowed) 2BCD * 4 = DCB2 D*4 must be a number that ends with 2, that means D must be either 3 or 8. the D digit in the product must be at least 8 because A*4 is 8, combined with D being either 3 or 8 means D must be 8. 2BC8 * 4 = 8CB2 the B digit in the factor does not contribute to the D digit in the product, so B must be at most 2, but since A is already 2, B must therefore be at most 1. the B digit in the product is equal to the smallest digit from C*4, which is an even number, plus 3 carryover from D*4, so B is odd. this combined with B being at most 1 gives us that B is 1. 21C8 * 4 = 8C12 the B digit in the product became 1 after recieving 3 carryover, so before the carryover it was 8. thta means C*4 has a smallest digit of 8; so C is either 2 or 7. it can't be 2 (A is already 2), so it must be 7. 2178 * 4 = 8712

[@pierreabbat6157]

Since the carry of 8×4 is 3, B must be odd, and since 2×4=8 with no carry from B×4, B

[@hippophile]

Yes, got this. It is a nice one. I got the 2 and 8 the same way, and noticed that B < 5 was no good, which yielded the answer quickly without algebra.

[@CardinalRed925]

I figured it out and didn't even notice that two letters can't be the same number. I just found as many constraints as I could without just assuming it to be 0000. In the end I actually didn't even use algebra, I just experimented with whether or not C=2 & C=7 would work with D=8 & A=2, since with a list of constraints too complicated for me to explain, those two were the only ones that could possibly work with D=8 & A=2.

[@mike1024.]

A = 2 and D = 4 is easy enough, but I feel it it easier to then say BC × 4 + 3 = CB (the + 3 coming from D × 4 = 32 requires a carry over) and use similar reasoning as A and D. B can't be 3 or higher, and it can't be 2 because A = 2. B can't be 0 because 4C would have to end in 7 to get the + 3 to end up ending in 0, and 4C is even. Thus B = 1, so C is at least 4. Now, BC × 4 + 3 must end in 1, so C × 4 ends in 8. Only 7 satisfies these two restrictions. Trial and error or the much simpler equation 4(10 + C) + 3 = 10C + 1 could also be used to quickly find C.

[@ccllau]

2178. A could only be 1 or 2 or result becomes 5 digits. Then get D as 8. Then try C and tie with B. Will result as 8712

[@sylvainmongeau]

This is like the old tale of a son off to college repeatingly asking for money. His parents got fed up and said if you can solve this, we'll send you more (same rules as this one, no repeats. You posted this one a few years back): SEND + MORE = MONEY

[@bobjordan5231]

Well done!

[@idminister]

If you like problems like this: there is a book called Sideways Arithmetic from Wayside School (or similar name) that is filled with word based arithmetic problems of letters = single digit numbers problem 1: elf+elf=fool and it goes into division

[@Jmerzio]

Wow i havent heard of that book in a long time, i read it as a kid. glad its still around

[@gregoryconran66]

0000, 1111 and 2222 work also, if there aren’t more restrictions in the order of digits (that is, that ONLY DCBA has to be the answer and others like ABDC or DABC, etc. are allowed

[@adrikmaciel641]

The numbers can't be the same because the problem isn't AAAA * 4 = BBBB

[@michaelspence2508]

I solved it similarly to others but slightly differently. D can't be 0 since it's the first digit of a 4 digit number, so A isn't either. That means A = 1 or 2 and D = 4 or 8. 4x4=16 and 6 is neither 1 or 2. 4x8=32 so D=8 (Yes I missed that D has to be even) 4B+the carry of 4C+3 has to be less than 10 so B also = 0,1or 2. 4C+3 for 0 through 9 are: 3,7,11,15,19,23,27,31,35,39 Two of those numbers have a ones digit of 0,1,or 2 (namely 11 and 31) so B =1 and C=2 or 7 (with carry 1 or 3 respectively) (11 and 31 are the C=2 and C=7 terms) 4(1)+1=5 and 4(1)+3=7 -> C can't be 5 so C is 7

[@michaelobrien1106]

I skipped the algebra and went with logical reasoning. 4A cannot carry another place value so A=1,2 (0 is out for being a leading digit). Since A is in the unit solution, A must be even so A =2 Since 4D produces 2, D = 3, 8; but since 4A=D with possible carry, 3 is eliminated and D = 8. Since 4A=D has no carry, 4B

[@crueI]

A must be between 0 and 2 (otherwise result exceeds 10000), A is even since it is the ones digit of an even number. A can’t be 0 since D would have to be 5 and the multiplication would result in a number lower than 4000. A is 2 then. D must be 8 or 9. D can’t be 9 since 9 * 4 would mean A is 6, but A is 2. D is 8 then. BC * 4 + 3 = CB with no carry (D is 8). B is between 0 and 2 otherwise the above calculation would have a carry. B is odd since it is the ones digit of a sum of an odd and an even number. B is 1 then. (10 + C) * 4 + 3 = 10 * C + 1 6 * C = 42 C is 7 then. This was how I solved it.

[@Xzyum00]

A = 0 B = 0 C = 0 D = 0 0000*4 = 0000 Sometimes there's more than one answer. 0 is such a powerful number. 0000 is just 0.

[@unoreyno129]

After I calculated A and D I didn't use algebra to solve the rest. From the D the number 3 moved to the left. BC cannot be larger than 24, otherwise 1 will move to the left side, added to the A and destroying it. So B must be 0, 1 or 2. C after multiplying should end up with 7, 8 or 9 so B will be as I said 0, 1 or 2. From 7, 8 and 9 only 8 is possible because that's the only even number, and by multiplying by even number we'll get an even number. So if C ends up with 8 after multiplying, it'll mean that B is 1, because 8 + 3 = 11, it'll end up with 1. Only C is left. Only 2 numbers end up with 8 after multiplying, 2 and 7. 2 doesn't fit as an answer but 7 did

[@flameknightkiana]

3,999a + 390b - 60c - 996d = 0 (a, b, c, d | 0 ≤ a, b, c, d ≤ 9 and a, b, c, and d € integers)

[@RoderickEtheria]

A must be 2, because it must be even, below 3, and cannot be 0. D must be 8 as it is 8 or greater and multiplies to a number that ends in 2 when multiplied by 4. B must be odd, and less than 3. C must be 7. 2178×4=8712.

[@maynameismay]

I suppose the solution could be way simpler. A and D could only be 2 and 8. (In that order) Because A < 3 and A = 1 wouldn't work cause then D should be 4 but 4 × 4 doesn't give us A = 1. Also we can't use 0. (repetition isn't allowed) B has to be 1; we've already used 2 and +3 would add something to A = 2. And the only number we can put in 4C+3 to have B stay 1, is 7.

[@poldelepel]

There is a trivial answer: 0000 x4 = 0000

[@RainbowDashShadesOfApproval]

Well A can only be 1 or 2 cause the number of digits eemain the same. Since A needs to be even for the equation to work, A is 2. That makes D 8 or 9. Seeing as 8 X 4 is 32 and 9 X 4 is 36, D is 8 cause it must end in a 2. B has to be either 1 or 2 cause it cant result in a two digit number, increasing D. 2 is already used so B is 1. The next part is probably the hardest. We need to make the ones digit of (C X 4) + 3 to a 1. To do that we need to subtract the 3 and get the ones digit of the result an 8. And the only two single digit numbers that can do this is 2 (8) or 7 (28). Since 2 is already used, C is 7. ABCD=2,178 using a calculator quickly confirms that this is correct.

[@TS-cs2pe]

2178 is also the key to extend this. 21782178, 217802178, 2178002178, ... (repeat 2178 and add n times of 0s in between them). Or, 21978, 219978, 2199978, ... (21 then n times of 9s then 78)

[@TophTheMelonLord]

It's such a weird shift to use logic on A and D and then algebra on B and C. Filling in the 2 and 8 tells us that a 3 is carried from the ones place to tens place, while nothing is carried from the hundreds place to thousands place. Then by *exactly* the same logic as before, we know that B is 0, 1,or 2 and B is odd.

[@twelvexii9474]

I did A and D the same way, but after that my solution went a different path. I knew with D=8 A*B had to be less than 10, so 0 1 are my options as 2 is occupied. From that C*4+3(due to the 3 from D*4) has to be negative so B must be 1. Thus C must be 2 or 7 as they have to have an 8 in the 1's digit of C*4 as that would produce 1 for B in the C*4+3. Again 2 is occupied so it's 7.

[@rcb3921]

a fun little puzzle from precious show Walker.

[@litcrit1624]

Haven't read any comments, but can't you LOGIC the values of B and C, as opposed to creating an equation? If A is 2 and D is 8, then B must be 0 or 1 (to avoid carrying any number to the 1000s place). But B cannot be 0 because (4xC)+3 cannot be even (i.e., cannot give a 0 in the 10s place) So, B is 1, with 7 being the only available value left for C

[@jakistam1000]

After solving it normally, I solved it in Python. But I accidently put 1 too much zero in the loop, and found the solution to ABCDE*4=EDCBA. (It's 21978)

[@devondevon4366]

Answer 2178 4 ---------------------------------- 8712 A is less than 3 but greater than 0 since 3*4 = 12 and 0*4 = 0 would put D out of place A is 2 since 4*D (an even number) = A Hence D= 3 or 8 (4*3 = 12; 4* 8=32) Using D as 8 will put D in place at the Product and the Multiplicand. 8 * 4 = 32 The 3 from (32) must be added to a number to form B, but B is less than 5 (since 5*4 = 20 and 2 + A would = a product with more than 5 digits). 7 is the only number less than 9 when multiplied by 4, which gives a number large enough '8' that, when added to 3 will give a smaller (as 8 + 3 =10 + 1). Let try 7 2 . 78 by 4 ------------- 8 . 12 2178 4 ----------- 8712 Hence, 7 must be placed in the product, and 1 must placed in the Multiplicand.

[@Djellowman]

Nice. Quite tricky. I got halfway there on my own.

[@monroeclewis1973]

ABCD must be < 2500; otherwise, x4 produces 5 digit #. A must be 1 or 2. But, A must be 2 because 4 x D is even. Turning to D, only 4x 3 and 8 yield # ending in 2. Can’t be 3 because too small. 4 x 2 = 8. The rest, B and C, is algebra as explained.

[@jimv9210]

If you consider that 4xABCD = DCBA, then DCBA/4 = ABCD. No? If so, in order then to avoid a remainder generating a new low order digit, A must be 0, 4 or 8. But since A must be 0,1 or 2 when it's at the high end (to avoid generating a new high order digit), it has to be the common value of 0. The same argument applies successively to the other digits and thus the only solution, which violates the rule of distinct values, is 0000. I.E. it has no solution.

[@kolelokaram8541]

As soon as you went ahead to use algebra to solve for B and C, I just mentally tuned out. Yeah it absolutely works, but it wasn't fun anymore and would take much longer than the more hacky method you already used for A and D.

[@teeteringonthebrink.305]

It took me less time than this video to work it out (about 2 - 2.5 minutes) despite maths not being a strong subject for me. Knowing that any number multiplied by an even number results in an even number, the last, and therefore first numbers had to be even. Finding them was easy, 4 x 2 = 8 and 4 x 8 = 32. I then thought it obvious that the two middle digits had each to be an odd number.

[@GretgorPooper]

Pretty easy, but still quite fun! I like this kind of puzzle that is easy enough to be accessible, yet multi-layered enough to get you into that flow where you discover one fact at a time, until you have all the facts needed for the solution. So while the puzzle itself is not hard, it is "designed" in such a way to be really fun to solve. I wonder if there is a theory that determines the "fun factor" of a given math puzzle, this kind of puzzle would score high.

[@dwiadidharma3834]

ABCDE x 4 = EDCBA A = 2 B = 1 C = 9 D = 7 E = 8 ABCDEF x 4 = FEDCBA A = 2 B = 1 C = 9 D = 9 E = 7 F = 8 Demikian seterusnya bisa dikembangkan lebih lanjut, cukup dengan menyisipkan angka 9 di antara angka 21 dan 78

[@jacobrast126]

How about the trivial solution ABCD = 0000?