Can you solve it? Logic test ABCD x 4 = DCBA
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- Опубліковано 28 вер 2024
- Thanks Manav, Devang, Martynas for the suggestion! The problem is ABCD x 4 = DCBA. Each letter is a different digit from 0 to 9. What is the value of each letter?
@LOGICALLYYOURS video comment by @danmimis4576
• ABCD x 4 = DCBA. Can y...
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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.
Same
Yeah. I did something similar. I knew A had to be 1 or 2. Tried A is 1 and then D didn't work because Dx4 can't end in a 1. So then A=2, D=8 is the start. Then yeah, as you said, B must be 0 or 1 because 2 is taken. Can't be 0 because of that 3 from the 8x4. So then you just look for numbers that when multiplied by 4 end in an 8 for C, which, like you said are 2 and 7. 2 taken. Done.
I think algebra actually takes a heck of a lot longer than this.
Yes, you resolve that by 100% logical reasoning by observing digits instead 50% logical and 50 % algebra.
Same
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.
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
Thanks for the info. Interesting stuff.
I studied the problem further and it had some pretty interesting results.
It's something to do with alternate digits adding up to 9
And also: 1089*2=2178
@@espadadearthur1174what does that do
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.
I did it the same exact way
i used same method lol
my exact thought about it. "they said it was a logical problem, so there must be a logical method for it"
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.
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.
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
i got this in GRADE 2 AMO and solved it bruh
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
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.
Thanks for this video. Your channel is the most productive and useful one in my opinion!
I assumed A and D were non-zero since the question asked for 4 digit numbers. Then A
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.
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)
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).
@@wernergamper6200 Tssk tssk tssk. It's shorter in SQL, which means you shouldn't persist in your error : your way is the wrong way.
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
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
@@danmerget I am honestly surprised that Presh didn't consider that process in his solution.
Way the logic is explained and developed is super
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 . . .
I guess "zero can't begin a multi-digit number" isn't in his rules?
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
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
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.
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.
Congratulations, you have evolved into Volcarona.
That is EXACTLY how I did hahaha
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
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 💀
A common question for Business Schools now sorted 😊
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.
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.
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))
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.
yea, I proudly say I can and did solve it. Thank you for the curious math riddle.
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.
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.
When i first saw the puzzle as a kid the question was a forgotten phone password and the only clue was the multiplication of 4 to abcd get dcba.
I was happy when i did this then but my friends then were all literature nerds and didn't enjoy like i did.
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.
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
@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.
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
Nice question
Would you like a beer is a nicer question.
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
The numbers can't be the same because the problem isn't AAAA * 4 = BBBB
I looked at the thumbnail, solved it, then clicked on the video to write this comment. I don't even have to watch the video.
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.
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.
About 5 years ago I found this, and this is written on my diary
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
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.
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.
Since the carry of 8×4 is 3, B must be odd, and since 2×4=8 with no carry from B×4, B
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
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.
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.
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.
OK... got to the A and D but did not think to try iwth an equation like this. And this was so quick and easy... Need to watch this in the morning afrer my coffee not before bed time ;)
GAMES MAGAZINE when Will Shortz was the Editor sometime in the Summer of 1984 had a similar problem. ABCDE * 4 = EDCBA. Similar rules except 1-9 only, zero is not a possibility.
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...
I did a spreadsheet and brute force solved for the 4 digits, as integer solution of the equation mentioned in the video. The coefficient for each unknown is in very different scale (996, 60, - 390, - 3999) hence playing only for about a minute with the values I found an integer solution. I didn't know that it was the only though.
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.
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.
Why can't the result not just be 2AF34 - would still be a challenge for some people I assume
Very nice thumbnail. Love the use of colour👏 👍🥰
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.
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.
No wonder MENSA won't take my calls or even answer my knock at the backdoor!
All the first 9 multiples of 1089 look fun in decimal.
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.
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.
Well done!
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
Wow i havent heard of that book in a long time, i read it as a kid. glad its still around
At the nearly end, i did the opposite of you, and put: C=(13B+1)/2 and conclue that B has to be odd for C can exist. And so B=1, and C =7 as the only possibilitie.
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
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.
i nearly went insane trying to solve this
My method:
I realised ABCD
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
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.
A and D is fairly easy to get. 2 is the only number that can be A. This is because you need a single digit number as the solution to A*4. That being 8.
0 or 1 won’t work because the solution to D*4 has to be a double digit number.
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.
I logiced my way to A and D, but from there I just brute forced it because I assumed there could only be 30 possible values for reasons I've forgotten and how long could it take to do 30 simple equations? Probably faster than thinking of the solution, that's for sure!
Fun!
I began immediately with an equation using multiplication times powers of ten for the digits to get:
3999A+390B=60C+996D
with 0
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
Solved it in my mind in 5 mins (Got lucky as I didn’t consider A to be 0 as that becomes a 3 digit number).
Watched the explanation and remained confused for 15 mins.
Thank you
This is my favorite one to solve
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.
Exactly the same solution as I did. Elimination process and logic!
I solved A and D similarly (A had to be
Paused the video this time, fiddled with the question for an hour or so and after a few tries, I got 2178. No algebra involved, just basic logic. 👍
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
It'd be nice if these were done as two videos about 20 minutes apart, that way we could work it out and offer a solution in the comment section before seeing the solution.
Seems like a nice idea.
He could put a question at the end of a video and provide the solution in the next video with another question at the end. And then keep repeating the cycle.
Makes no difference. You can still do that when there is only one video with the problem and the solution.
Having to watch two video's for one puzzle is something a click-baiter would do.
you can pause the video
This video ^is^ the solution.
The thumbnail states the question.
@@neuralwarp
The thumbnail is attached to the video and appears as the same time, also the comment section cannot be accessed without starting the video!
this would be immposible unless 0 could work every time, as a*b*c*d*4 could never be a*b*c*d in that case, as if we divide by both sides, we would get 4=1
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.
It took me an hour to find equations and try to solve thru that but after trying a lot and finally giving up on that method since it was too lengthy and my brain couldn't comprehend it enough to find the first 3 terms, so I substituted the values logically like you did and got the values as 2(A) and 8(D), then B has to either be 0 or 1 (cuz 2 is taken) and 0 wouldn't work cuz {(40C + 30)/10} = (4C + 3) has to be an odd number since 4C has to be even, so I got 1 as the answer there and thus got 7 after trial and error.
Feels good to get it right by myself, even though it wasn't a hard question to begin with lol. Although I think it's still possible to get it with the equation method, just really lenghty tho. (If someone does get it that way please reply to me with the equations, I really wanna know how you would do these kinda questions with that method)
You have to guess and check at some point but you can get pretty far with equations. With a b c d x y z all positive integers < 10:
1. 4d = a + 10x
2. 4c + x = b + 10y
3. 4b + y = c + 10z
4. 4a + z = d
Then
5. 16a + 4z = 4d = a + 10x
6. 15a + 4z = 10x
7. 16b + 4y - 40z = 4c = b + 10y - x
8. 160b + 40y - 400z = 10b + 100y - 15a - 4z
9. 150b - 396z = 60y - 15a. Solving mod 15 gives z = 0 mod 15, which means z = 0 because it is less than 10.
Therefore
10. c = 4b + y
11. d = 4a
12. 15a = 10x
13. 6y - 15b = x
Trying b = 0 gives y = 0, x = 0, a = 0, c = 0, d = 0 which sure enough is a (trivial) solution. Or y = 1, x = 6, a = 4, d = 16 which isn't. Trying b = 2 gives y >= 5, c >= 13 which also fails. However b = 1, y = 3, x = 3, a = 2, c = 7, d = 8 works: 2178 * 4 = 8712.
thanks@@NAM-ih3lb
Nice. Quite tricky. I got halfway there on my own.
I just made this as a python coding test few hours ago, really weird that I get this now....
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
I've confused a little while you are finding the value of D can u explain it in a brief
D times 4 must equal 10 times some integer plus A to fit the equation. Given the limits on A, that means D must be 8 (8 times 4 equals 32)
Actually there's two possible solutions the other is A=0, B=0, C=0, D=0 so ABCD×4=DCBA=0000×4=0000.
Nothing to see here people…
move along. move along…
ABCD x 4 = 2AF34
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.
Looking at the video without playing it and getting the other instructions, just did binary multiplication 0011x100, but video said they are different
A can‘t be 1 because Dx4 is always even; A also can‘t be 3 because D would then be 12 > 9
*A = 2*
and D can be 3 or 8 while 3 is too small (< 2x4 = 8), so
*D = 8*
B can‘t be higher than 2 because it would make D 9 or higher, so
*B = 1* (the only remaining of 1 and 2)
Dx4 = 8x4 = 32, so Cx4 = N8 => C is either 2 or 7; Bx4 = 4 which can‘t be increased to N2 with the help of Leading Digits, so
*C = 7*
2178 x 4 = 8712
8000
+400
+280
+ 32
(L100)
8712
I wouldn't be able to solved it without using table to simplify these very tiring elimination. Here is my method:
1. A must be an even number, but ABCD can't be more than 2500, therefore A must be 0 or 2.
2. DCBA must be a multiple of 4, therefore BA must be a multiple of 4, therefore BA must be 20, 40, 60, 80, 12, or 32.
3. Any number must end with 0 or 5, so that if said number is multiplied by 4 the result end with 0. If A=0 then D=0 or D=5, but because ABCD must have different digit therefore D can not be 0. Therefore if DCBA=DCB0 then ABCD=0BC5. 5CB0 can not be 4 times 0BC5, thus eliminate BA=20, BA=40, BA=60, and BA=80, therefore A must be 2 and B must be 1 or 3.
4. Any number must end with 3 or 8, so that if said number is multiplied by 4 the result end with 2. Because A must be 2, therefore D must be 3 or 8. So the possibilities are down to: 21C3, 21C8, and 23C8.
5. If ABCD=21C3, then DCBA=3C12. 3C12 can not be 4 times 21C3, thus eliminates 21C3.
6. If ABCD=23C8, then DCBA=8C32. 8C32 can not be 4 times 23C8, thus eliminates 23C8.
7. ABCD must be 21C8, then by trial we have 2178.
Thank you very much for this logic test. More of these please!
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.
Huh, I didn't even think to use algebra for this. Once you know A = 2, you know B < 5, since 2500 x 4 = 10,000.
Once you also know D is 8, since A is 2 and 4 x 2 is 8, B x 4 must be a one digit number, so B < 3, and since A = 2, B = 1 or 0.
From there, since D is 8, and 8 x 4 = 32, we know that B is the digit in the ones place for the result of C x 4 +3. Therefore, C x 4 must have 8 or 7 as its ones-place digit. Since the result must be even, 7 is out, so the ones-place digit is 8. This means that C must be 2 or 7, and since A = 2, C = 7.
This means C x 4 + 3 = 28 + 3 = 31, so B = 1.
A = 2, B = 1, C = 7, D = 8.
ABCD = 2178
ABCD = 2178
and 2178 times 4 equals 8712 .
A = 0 leads to the trivial solution ABCD = DCBA = 0000 . (A=0 ==> D=0 or 5. But 5CB0 cannot be equal to 4 times a 3-digit number, so D=0. Then B = 0, 1, or 2. B=2 would imply C=8 or 9, but neither works. B=1 would imply C10 is a multiple of 4 , which isn't possible. So B=0, in which case also C=0.) Otherwise:
A = 1 or 2, or else ABCD multiplied by 4 would be a 5-digit number. But since DCBA is a multiple of 4 , A must be even. So A = 2.
Then either D = 8 or D = 9 . But 2BC9 times 4 would end in a 6 . So D = 8 , 2BC8 times 4 equals 8CB2 .
Since 2BC8 times 4 is less than 9000, B < 3 ; so B = 0, 1, or 2 . However, in the tens' column we have 4*C + 3 must end in a B; since 4C is even, 4C+3 is odd, so B must be odd. Therefore B = 1. So 4C must end in an 8; this means C = 2 or C = 7 . C = 2 doesn't work (because 4 times 2100 = 8400 < 8C12 , so C must be greater than 4), hence C = 7 .
2178 times 4 equals 8712 .
A is the first digit I figured out. I knew it had to be even (due to the last digit of the product) and less than 3 (due to the product being only 4 digits). So it had to be 2. This means that D must be 3 or 8 to make the last digit of the product a 2, but 3 is clearly too small to be the first digit of the product. So right off the bat we know that A=2 and D=8.
Now we'll figure out B. It must be a small digit-- less than 3-- to keep the first digit of the product an 8. We know it's not a 2, so it must be a 0 or a 1. We know already know that D times 4 equals 32, so a 3 has to be carried over and added to C times 4 to give us the third digit of the product, which is B. Since no number can be multiplied by 4 and have 3 added to it to get something ending in 0, B must be 1.
For C, we know that multiplying by 4 and adding 3 produces a number ending in 1. So C times 4 must end in 8. Possible values are 1 and 7, but 1 is already taken by B, leaving 7 as the only possibility for C.
And 2178 times 4 indeed equals 8712.
got it!
For more fun, here is another one. abcde/4 =edcba There are no zeroes. This was on a Mensa tesy
I followed similar logic to a point but then decided to just focus on BC, working ut that 4 x BC = CB + 3 and then it was a matter of working out the combination of 2 digits for which that worked, excluding 2 and 8 as these were already assigned. Also since there was no carry to the 4t column this mean BC < 25 and with the 3 carried from the 1st column, BC < 22. Also, 02, 12, 20, 21, 08, 18, 00 and 11 were out so that only left 14 possible combinations to try.
These questions always make me feel worthless.
A has to be 1 or 2 because other wise the product will have 5 digits. So let's assume it's 1, you soon see that you need to find D so that D times 4 has a last digit of 1, which no such number exist. So you know A is 2, and that means that you need to find D so that D times 4 has a last digit of 2, which means D is 8. Now it gets interesting. B also has to be either 1 or 2. So let's go with 2 this time. This means that you need to find a C so that the last digit of C times 4 plus 3(since D times 4 gives us 32) is 2. This mean that C times 4 need to have a last digit of 9, which no such number exist. As a result, you know B is 1, and now similarly you need to find C so that C times 4 plus 3 has a last digit of 1, which means that C times 4 will have a last digit of 8, and the only number that satisfy such is 7. As a result, ABCD is 2178.
Also B is 1 and can't be 2 because it's been used haha
The answer is four numbers long, so that means that A has to be 1 or 2 (because if A was 3 or more it would result in an answer that was five numbers long).
So let's go back to the first thing that we multiply .. 4 x D .. the answer to which ends in an A .
A must be an even number because D has been multiplied by an even number (4) .. so A must be 2 .. because the answer isn't going to start with a 0.
So .. D x 4 gives you an answer that ends with a 2 .. so D is 3 or 8.
3 (giving the answer 12) would be too small, so D must be 8 .. giving the answer 38.
We now have a question number that starts with a 2 and ends in an 8 .. and an answer number that starts with an 8 and ends in a 2 .. and we are carrying a 3 over to make B.
Now .. if 2 x 4 = 8 and 8 is the first number of the answer that means that nothing was carried over from B x 4.
What we need to remember now is that we are only dealing with four letters/numbers .. and in the question number the BC is reversed to make CB .. in fact the whole number is reversed !!
I'm just going to try B = 1 now .. and YES that works if C = 7
A = 2 B = 1 C = 7 and D = 8
ABCD x 4 = DCBA
2178 x 4 = 8712