Well, that was a very brute-force approach to this problem. I feel like we should let a computer do this kind of repetitive work if we're going to go through this much trouble. I wonder what elegant methods may get us here? Some were mentioned early knocking out possibilities.

So, my mental reasoning for this one went: - Final result must be under 100 - We don't have a zero - Number after multiplication must leave us digits that can add to 98 or below You can eliminate the vast majority of options from this, so what next... - Multiplier cannot be 1; it would cause repeat digits. - Multiplier cannot be five; it would cause repeat digits. - Multiplier cannot be above five, as this doesn't leave any way of leaving you a second two-digit number that will keep you under 100. - ((Multiplier must be 2, 3 or 4)) - Second digit of initial value cannot be 5; it would cause repeats, or a zero. - Second digit of initial value cannot be 1; any result would require another 1 and be invalid. - First digit in product cannot be 1; this is impossible mathematically (we are multiplying at least a two digit number by at least 2). - First digit in the final answer also cannot be 1, for the same reason. - Second digit in product could only by 1 if the multiplier is 3 AND second digit in initial value is 7; no other combination allows this. (at this point I'd mentally eliminated a lot, but changed tack slightly without an immediate path of eliminating to a definite position for the 1) - If the multiplier must be 2, 3 or 4, we can begin checking some combinations, because we'll quickly be able to eliminate whole branches of inquiry without needing to check. - If we start with a multiplier of 4, we can only possibly have to check initial values in the 1(?) and 2(?) range - anything higher prevents us staying 98 or under. - Of the x4 multiplier options, we can eliminate any combinations where the second digit is 4. - Of the x4 multiplier options with an initial value of 2(?), we eliminate 21, 22, 24 and anything over 24 (product over 98) automatically. 23 can also be eliminated (4 x 3 = 12, 2 is used) - Of the x4 multiplier options with an initial value of 1(?), 12, 14, 15 and 16, are eliminated automatically for repeat digits, leaving just 13, 17, 18 and 19. - 13, 18 and 19 all leave us with remaining digits for which even the smallest possible addition will take us over 98. (52 with [6, 7, 8, 9] left, 72 with [3, 5, 6, 9] left and 76 with [2, 3, 5, 8] left) - I couldn't think of an immediate way to eliminate 17, so I checked it manually. - 17 x 4 = 68 [2, 3, 5, 9] Lead digit could only be 2, as anything else goes over 98. Second digit could only be 5, as anything else leaves an invalid unit digit. - 17 x 4 = 68 + 25 = 93 - Success! 1 Valid Solution, then, seems to be: 17 x 4 = 68 + 25 = 93 == I've been told there's only one solution, and this solution is valid, but I've not *proven* that there is only one solution; my tester might be mistaken, so I continued. We've eliminated all other x4 multipliers, so the only place where another solution could exist would be in the x2 and x3 sets. Still working down, I started thinking about what else I could eliminate from the x3 set. - Of the x3 multiplier options, we can eliminate all initial values over 2(?) as well, just like for x4 multipliers; we can't use 3(?) for repeat digits. - If we start with a multiplier of x3, we can only possibly have to check initial values in the 1(?) and 2(?) range; we can't use 3(?), and 4(?) puts us over. - Of the x3 multiplier options, we eliminate any combination where the second digit is 3. - Of the x3 multiplier options with an initial value of 2(?), we've already eliminated 21, 22, 23 and 25, and can quickly eliminate 24 and 28 (3 x 4 = 12, repeat, 28 x 3 = 84, repeat), as well as 27 and 29 (no valid value we could add to the product and stay under), leaving 26, which I tested. - 26 x 3 = 78 [1, 4, 5, 9]; 14 gives 92, failure, no other valid combinations for 2(?) values. - Of the x3 multiplier options with an initial value of 1(?), we've eliminated 13 and 15, and can quickly eliminate 12, 14 and 17 (products contain repeat digits), leaving 16, 18 and 19. - 19 can be eliminated because its product is odd, but it leaves only even digits to add, so the answer cannot be valid. - I didn't think of a quick way to eliminate these last logically, so tested them. - 16 x 3 = 48 [2, 5, 7, 9] Lead digit could only be 2. Second digit could only be 7 (answer uses 5) or 9 (answer uses 7), but sum will be in the 70s, so both are eliminated. - 18 x 3 = 54 [2, 6, 7, 9] Lead digit could only be 2. Second digit could also only be 2 (answer uses 6), so this set is eliminated. - Multiplier cannot be 3. Just x2 set to eliminate now. - If we start with a multiplier of x2, we would theoretically need to check the 1(?), 3(?) and 4(?) possibilities; 2(?) is out, and anything 5(?) and up is as well. - Any initial value ending in 6 is automatically eliminated, as it would leave us with a second 2. - Of the x2 multiplier options with an initial value of 4(?), we've already eliminated 41, 42, 44, 45 and 46. Any product over 85 will not leave digits that can keep us under 98, since our minimum addition would be 13 - which eliminates the rest as well. No need to test here. - Of the x2 multiplier options with an initial value of 3(?), we've already eliminated 31, 32, 33, 35 and 36, and can knock out 37 (product is 74, repeat) quickly as well. - We can also logically knock out 39 without testing, as the highest digit it leaves is 6, while its product is 78; we can't form a valid answer. - We can quickly eliminate 34 and 38 as well; their remaining digits means that the lead digit must be 1, to stay under 98, but this means the answer would be in the 80s, regardless of other combinations; we have no 8 for either, so they are eliminated. No need to test here. - Of the x2 multiplier options with an initial value of 1(?), we've already eliminated 11, 12, 15 and 16, and can eliminate 13 and 14 quickly without testing (products in the 20s, repeat). - I didn't think of any quick way to knock the remaining 3 out logically, so I tested them. - 17 x 2 = 34 [5, 6, 8, 9] Lead digit could only be 5, anything else goes over 98. Second digit could also only be 5. Eliminated. - 18 x 2 = 36 [4, 5, 7, 9] Second digit could only be 9 (answer uses 5); 49 gives 85, failure; 79 gives 105, failure. Eliminated. - 19 x 2 = 38 [4, 5, 6, 7] Lead digit could only be 4 or 5, Second digit could only be 6 (answer uses 4) or 7 (answer uses 5); 56 gives 94, failure, 47 gives 85, failure. Eliminated. - Multiplier cannot be 2. All other possible configurations have now been eliminated, thus 17 x 4 = 68 + 25 = 93 is the only possible solution. Q.E.D. ((For time-sense purposes, reasoning this out, with the small handful of quick scribbles that I tested, this process took about three minutes for me, which seems.... reasonable(?) without a 'clever' solution?))

Just from the video length, the solution process is clearly "try everything, with simple filtering of impossible cases". In that case, it's only a minor difference to convert that to "hey computer, try everything"

Really cool that for a puzzle with 362,800 possible arrangements, there's only one possible solution. Fun puzzle.

You can eliminate a ton of possibilities from the get go by noting that the multiplication has to be less than 76. Then, instead of cycling through constraints on the double digit number, look for constraints on the single digit. This gets you down to that number having to be 2,3 or 4. Then you can proceed checking those three iteratively

A small Python program : from itertools import permutations perm = permutations([1, 2, 3, 4, 5, 6, 7, 8, 9]) for p in list(perm): (a, b, c, d, e, f, g, h, i) = p if c*(10*a + b) == 10*d + e and 10*d + e + 10*f + g == 10*h + i : print(p)

I was thinking he would just do the logic elimination of cases, guess not today 😢

I haven't watched the video yet to see the solution, but my initial thought is "man, that's gonna be a ton of guessing and checking!"

Before adding the smallest combination of remaining numbers after the multiplication to check if it is greater than 99, a simple "is DE bigger than 86?" check would speed things up by a lot. You see, the highest number that HI can form is 98 (not 99), and on top of that, the smallest FG possible is 12, so it means that the largest DE possible is 86. And if you're checking arrangements with A = 1, the maximum DE is even lower because F can't be 1, so in this case, considering that the maximum HI is 98 and the minimum FG is 23, the maximum DE is 75. This also means that: if AB = 12, 2

A slight improvement to this brute force method is to realize that the addition of two 2-digit numbers will result in a 3rd 2-digit number. If the product resulting in the 1st 2-digit number is anything greater than 86, you don't have to check what the addition will be because the smallest 2nd 2-digit number you could add is 12 which results in 98 (the largest possible final result). This will quickly eliminate any 40's checks because 86 is the smallest possible product (43*2) that falls within the rules of the problem (any multiplier greater than 2 makes the product's 10's place at least 12).

If the only solution is brute force, then it's not a puzzle.

9:27 that actually jumpscared me so hard XD

Aah! and there I was expecting some elegant mathematical solution. However, It taught me some logical thinking and that sometimes, the only way to solve a problem is to yomp though it. 🤔--->🤯--->🤕--->🤓👍

An outstanding example that sometimes a solution requires work But That an organized approach reduces the work greatly!

This is brute force, but it’s also a very bad brute force. Although the calculations are minimal, a lot of time is wasted checking digits against 6/7/8/9 for c. As you said, 1 can’t be c, nor 5. But given the sum adds to a 2 digit number, 9 is out since it’s too big, 8 same reason, 7 has a chance with 12 but that’s 84 and you can’t add a 1z number. 6 loses 12 with 72, 13 is 78 but the smallest is 24, and 14 has the same problem as 7x12. Now you’ve eliminated 1,5,6,7,8,9 for c. 4 is fairly simple as the upper limit on it is 4x19, and you can find 4x17 fairly easily. If you’re told this has 1 solution, this is the fastest way to deductively solve. If you need to prove it, you can eliminate the 4s this way, the 3s have an upper limit at 29, and the 2s at 44. 46-49 are needless checks, and any multiplying of 2 or 3 after their limits is unnecessary.

I would give you a big and special thanks to you Presh because you helped me think more about how to solve problems like these, something I was unable to do some years ago in high school. I watch your videos since 2017 and I always find myself learning and improving. Thank you!

Given letters to the squares from a to i from top to bottom: c1, because times 1 repeats the digits => contradicts different digits (cdd) c5, because times 5 terminates by 5 => cdd or 0 => not possible For the same reason, b 5 c9, because then 10a+b cdd c8, because then 10a+b too much c7, because then 10a+b too much for addition 13*7=91=>cdd 12*7=84, leaving 35 as the smallest to add => too much c6, because then 10a+b cdd 13=6=78, smallest to add = 24 => too much That leaves c = 4 or 3 or 2 c=4 => 10a+bcdd 13*4=52, leaves 67 to add => too much 14*4=>cdd 16*4=64=>cdd 17*4=68 68+23=91=>cdd 68+25=93 => !! SOLUTION !! 68+29=97=>cdd 68+32=100=> too much 18*4=72, leaves 35 to add => too much 19*4=76, leaves 35 to add => too much 21*4=84=>cdd 23=4=92=>cdd The rest of the reasoning is more of the same somewhat tedious case by case examination showing there's no other solution than the one we found.

I started with the multiplicand and the elimination process is fastet.

WOW! 362,880 possible ways, I solved it in about 10 minutes on my 6th try! 😁

That's so much brute-force. First, just obvious check that -ab- < 44, because -de- can't be 9x because otherwise you can't add 2 digits number, brute force for the other 4 numbers to realized that the entire 4x row is invalid, narrow it down to -ab- < 40. Then, b cant be 1 or 5, as you said, and c cant be 1, 5 and 9. Brute force c by putting 8,7,6 on the 1x numbers and those weren't fit too, make the c possibility goes down to 2, 3 or 4. If you made a sheet outside before, there will be 2 -ab- "boxes", the one box that ended in 2, 3, 4, and the other box that ended in 6, 7, 8, 9. You can brute force the 234 box first, it's small enough and the c number will make the elimination faster, and then check the other box, which is just 12 numbers to do manually

Did it pretty much the same way, except I took c as my base for it, since it only has 3 options. 1 is out for repetition, 5 for getting a 5/0, 6+ is out because the smallest option (13*6=78+24=102) is too big. Leaving 2, 3 and 4 as options. If c=4, A can't be 3+ obviously. So we need to check the remainder (12, 13, 16, 17, 18, 19, 23. Anything bigger overflows). 12 doubles, 16 doubles, 23 doubles, so 13/17/18/19 remain to check the way you did. If c=3, A can't be 4+, nor can it be 3. So we have to check 12,14,16,17,18,19,24,26,27,28,29. Without doubles, that's 16, 18, 19, 26, 27, 29. If c=2, then A can't be 2 or 5+, so it must be 1, 3 or 4. So we have to check 13, 14, 16, 17, 18, 19, 34, 36, 37, 38, 39, 43 (46+ makes 92+ so it'd overflow even with +10). 13, 14, 16, 36, 37 double, leaving 17, 18, 19, 34, 38, 39 and 43. So while it looks daunting, simple elimination only leaves us 4+6+7=17 options to bruteforce check. Which, once you get the hang of it, is pretty fast: options for f and h tend to be limited, if any exist; and g and i can sometimes be eliminated by an odds/evens problem outright. Also note that many of the options to check use the 7. 8 or 9, which eliminates key options for h. So it takes mainly diligence to check and find 17*4=68; +25=93

I found a few more limitations...like "c can only be 2,3 or 4" and "a can only be 1 or 2". de cannot be prime, end in 5 or 0. This narrows it down to 24 options for de, with 30 corresponding possibilities for ab*c. It's easy to whittle those down, if you remove ab has to be 2 digits and a,b,c,d,e cannot be repeated. I got 11 such combinations and then brute checked those.

For people who want the answer, here it is: 17 x 4 = 68 + 25 = 93

1,7,4,6,8,2,5,9,3 f(from a to i). This has nothing to do with brute force. It's a successive process of elimination. After investigating combinations with a high likely hood of digits being generate multiple times, and zeros being produced, they were dropped. The main contestants for the initial multiplication (contained in the first 2 fields) are therefore either 17, 18, or 19. The multiplier in the 3rd field can only be 2,3, or 4. 4 bombed out immediately, one case produced a double digit and one went past the boundary of 98. That left merely 7 combinations with the first 5 digits being in order, plus 4 defined digits to play around with. Easy as pie. You combine 2 of those leftover digits and add it to what's already known and see if the result matches any combination of the remaining two digits. You only have to do that 7 times! 6 times I failed and on the very last option left being (17x4=68), voila! 68+25=93.

Another cool trick: Any product that is 88 or above can be eliminated because you have to add a 2 digit number to it that would make the result 3 digits and the smallest possible 2 digit number without repeating is 12

I will say a slight shortcut to elimiting answer is that any possible that has d = 9 will not work as d+anything > 9 so there is hard lose cap of de < 90 (you can expand the logic a little more to say it has to be less then 85 because the small number you can therotically ever add is 12 and you can only go up to 97 because of repeat digits) I am sure there are other cool limitations to the varibles out there that i still have not thought of

My intuition was that "c", being a multiplier on a two-digit number that must result in a two-digit product, would be the most restricted. So I tried each digit 1-9 as "c", and found that most of the digits led to immediate dead ends. c=1 obviously forced duplicate digits. c=5 forced either a duplicate e=5 or e=0. c≥6 forced three-digit sums at minimum. So "c" must be 2, 3, or 4. Then, for each c=2, c=3, c=4, I checked all possible "b", and computed "e" modulo 10. This led to only 16 b-c-e triplets with no 0s and no repeated digits: 3x2=6, 4x2=8, 7x2=4, 8x2=6, 9x2=8, 2x3=6, 4x3=2, 6x3=8, 7x3=1, 8x3=4, 9x3=7, 2x4=8, 3x4=2, 7x4=8, 8x4=2, 9x4=6 (all modulo 10). 16 candidates seemed a reasonable amount to brute force, so brute force I did. Each triplet has few, if any, valid values for "a", and "d" can be computed. So it's not much effort to check each triplet. Only the 7x4=8 triplet leads to a solution.

I appreciate this channel, this channel is very good. I am from India and learned many ways to solve problems. I usually see video of this channel. This is a good ❤channel❤ for student.

In the multiplication section you can eliminate any number over 90, since the addition number must be 2 digits. This also lets us stop checking at 45x2.

A way to reduce the verifications if the sum is bigger than 100: Suppose the fifth number is 98 and the fourth number is 12, then the maximum value for the third is 98-12 =86. Then, we search only three values (a,b,c) such (10a+b)*c

d can't be 1( cause a must be 1 too), d can't be 2 ( c and a must be 2 and 1) and d can't be 9 ( because f >=1 , 90+10 >= 100). And we know, that de = ab*c, so de is not prime. Also we know that de and e5 I think easier to check all variants for de.

Easy! Just try all the 362880 possibilites. No big deal

You could have stopped sooner by keeping in mind that the final step requires a 2-digit non-repeating number which still adds to 98 or less since 99 is repeating. So the final step can't be 1-9/10/11 so the max middle result is 86, assuming 12 for the last step. Although 12-16 would be eliminated too due to repeating 8 for 98, they might work at lower totals. So as soon as 43x2, using all the lowest factors in the first 3 blocks, came out to 86 and the 2 was a repeat, anything past that would have forced an end result above 98. Work it from both ends.

Awesome puzzle ,thank you so much for smart solution from Pearland,Texas USA.

Commenting before seeing the solution: I've brute forced in my head and got 17x4=68+25=93, I wondered if there's any more solutions. Update after watching: Let's FREAKING go. I'm not joking, I literally worked out the one and only solution, without any papers and pens. That's incredible!!! 😊 Bonus: There's a few more equalities of the same kind that I know on top of my head 54 x 3 = 27 x 6 = 18 x 9 29 x 6 = 58 x 3 = 174

I think you can also eliminate numbers that give you higher than 84 after multiplication. You need to add a two digits number, and the lowest one possible to add is 12. The number after addition can not go above 98, so 98-12=86. But in this case there is a repeated 8, so the number in question goes 1 lower, and we can't have a number with a 5 on the end, so our final number is 84. This could eliminate quite a few possibilities without a need to check them. For example you don't need to multiply by 4 any number in 20-s

Wow that was chaos

I looked for how I might solve this more analytically. The following steps were the best I came up with: 1. Prove that if de > 75 then either f = 1 or g = 1 2. Prove 1 < c < 5 3. List the possible combinations of a and c (three where a = 1, three where it isn't) 4. Prove d 5 7. Prove c = 4 8. Complete the solution A couple of steps (#4 and #7) require individual cases to be completely worked through but only a handful.

Beautiful reasoning!

Dude! That was awesome.

Well that was disappointing. I was just trying to solve it from the thumbnail and used logic to eliminate several possibilities, but was still left with a lot of options. I figured there had to be some other logical deduction I was missing, but it turns out this was just a 'run through the remaining options until you find something that works' problem. More of a tedious waste of time than a fun puzzle. It's cool that it only had one solution, but extremely uninteresting in how one must find that solution.

Checking most combinations like in the video is a valid way to do this There are clever eliminations that can be used 1. Noticing we must add another 2digit number, so the product 20 afterward) check: 4 21,23 don't work. So 12~19*4=48~76, eliminate wrong combinations, left 52 68 72 76 (72+3X>99, 76+2X>85) brute force through 52,68 to find answer 17*4+25=93 check:3 if we use 20s to multiply 3, the equation should be 2A * 3 + 1B = CD (obviously cannot add 4X) looking at multiplication, only 26*3=78 and 29*3=87 works, checking the addition, no answers here 1A * 3 + BC = DE multiplication: only 16*3=48, 18*3=54, 19*3=57 works, brute force to find no answers (Brute force, start from the smallest) check: 2 30s, 3A * 2 + 1B = CD 34*2=68, 38*2=76, 39*2=78 (68+1X=8Y, fail; 76+14or15 fail; 78+1X

Notating top left as A5 and bottom right being B1. a few obvious logic points, A5 has to be 1-4, (likely 1 or 2). B4 has to be 2-4, A1 has to be 6-9 (likely 8 or 9). 5 can't be in the multiplication digits (gives 0 or 5 as a product) etc. Other points- B5 is a nice place to put a big number, as is B3. I don't think I tried more than 10 permutations before I got the answer.

I knew it would be A LOT FASTER to solve this problem by the kind of method used by Presh but I couldn't resist trying to solve it with an Excel spreadsheet. It took "a few" hours to create, but it worked.

the smallest number you can make for the addition is 12, so no value larger than 87 (because 88 is double) from the multiplication has to be checked. saves some minor time of this bruteforcing

The part where you say any number over 98, I would say cut any number in the multiplication over 85 as the smallest 2 digit number is 12 that also leaves you with a 2 digit number at the end that does not repeat any with 8 is 97. Would help with the multiplication part.

I used a similar start to eliminate 1 and 5 from being B or C. However, I then showed that C could never be greater than 5 because if C is greater than 5, A would have to be 1 (otherwise the multiplication would have a 3-digit result). 12x6=72 and 72+34>100. Similarly, 13x6=78 and 78+24>100, and anything else just results in an even larger 3-digit number after the addition so C could only be 2, 3, or 4. After that, it was easy to prove through brute force that the only solution was 17x4=68 and 68+25=93 similar to how you did.

instead of just eliminating results over 99, you can immediately eliminate any result where the multiplication is greater than 89. There is no 2 digit number possible to add to 90 to get back a 2 digit number.

The approach could have been simplified significantly by saying that the multiplication result cannot be higher than 85, because 85 + the smallest allowable two digit number would be above 97 which is the highest allowable result when dealing with the 80s.

Man, I'm not sure if 20 minutes of brute force calculations is why I watch this channel....😊

You can ignore all cases resulting over 86. The smallest number you can add is 12. 87 + 12 is 99 which is already too big. So based on that there are lot more cases you didn't have to check

I narrowed down the multiplier to be one of 2, 3 or 4. Then I checked the numbers individually with 2, 3 and 4. Starting with 4 to get 17*4=68+25=93

Neat puzzle. After some reasoning and doodling on a scratch pad for 2-3 minutes, I had concluded that c must equal 2, 3, or 4 and that a must equal 1, 2, 3, or 4. From there I did go to a more "brute force / eliminate possible answer" approach and came up with the solution. I had worked in the opposite order as Presh, tackling a = 4 first, then a = 3, et cetera since those have much more restrictions on what c could be. In my opinion, this 22 minute video could have been more like 9-10 minutes, if a bit more time had been spent on the "front end" of eliminating possible digit combinations. As others have said, too much of a "try all combinations" approach ultimately isn't a very satisfying strategy.

Great patience! 😂

Nice puzzle. But you could have stopped with the maths way earlier. The biggest number in the end could only be 98. And you have do add a 2 digit number to it, which is at least 12... therefore the the difference is at ma 86... This 86 is the maximum for the product (and not 99). - But I understand your solution for this video to keep the numbers simple.

"In a remote and mysterious land, there exists a temple hidden deep within a dense jungle. The temple's entrance is guarded by a series of ancient statues placed at strategic locations. One particular statue, known as the Guardian, casts a shadow that holds the key to unlocking the temple's secrets. The Guardian's shadow is cast on a sacred stone platform, creating a triangle of shadow. The temple guardians believe that the shadow triangle is similar to a smaller triangle inscribed on the stone platform. The smaller triangle represents a map leading to the temple's inner chambers. The lengths of the sides of the shadow triangle are given by the quadratic equation s^2 - 6s + 9 = 0, where s represents the length of one side of the shadow triangle in meters. Similarly, the lengths of the sides of the smaller triangle are given by the quadratic equation t^2 - 4t + 4 = 0, where t represents the length of one side of the smaller triangle in meters. To decipher the temple's secrets, the guardians must determine the ratio of the perimeter of the shadow triangle to the perimeter of the smaller triangle. Find the ratio A) 3:2 B) 4:6 C)1:8 D) 2:5 Find the correct answer

Well done

You keep eliminating anything over 99 but you could use anything over 88 - the smallest 2 digit number is 12. In practice, for the cases where you are using 1, the smallest 2 digit possible is 23 and you could eliminate anything over 77.

Normally, I just proceed with the video because I can't solve it. This time I did and I'm glad I did because the video just checked everything instead of skimming the possibilities through logical eliminations

The logic way is to set the smallest digit at the top and largest at the bottom So we have 1 _ x _ _ _

Your references to "99" should be references to 98, because 99 already contains a repeated "9". The initial multiplication cannot add up to more than 86, because the minimum later addition is 12. That means no initial number above 43 works. That also rules out 43 as 2 would be repeated, so much of the tail end analysis could have been omitted.

Through setting up a main logic grid, and then a smaller one once your C possibilities are dramatically reduced, you can narrow it down to about 12 possible answers for the "DE" product. Then just check those with the remaining digits. You checked so many unnecessary cases.

I found a sort of analytical approach but with lots of assumptions: Lets rewrite the question to ABxC=DE + FG=HI First, looking at the one's digit, there are 6 possible cases: 1. even×even=even +even=even 2. even×even=even +odd=odd 3. odd×even=even +even=even 4. odd×even=even +odd=odd 5. evenxodd=even +even=even 6. evenxodd=even +odd=odd Second, we can note that there are more odds(1,3,5,7,9) than even(2,4,6,8). Therefore, we can compare each cases and assume we want the one with the most odds/least even. (Cases 4 and 6) Another assumption is that we'll start with a small number and end with a large number. Thus, we will assume the A=1 and H=9. After all these assumptions you can reach the answer by yourself. The lines after this is for those who are stuck or want to continue the solution: We can note all the locations of the odd numbers (A,B or C,G,H,I) and determine the location for even (B or C,D,E,F). Next, since D and F are both even. The only way they can be 9 is if D+F=8 and then E+G>10 So, D and F are either 2 or 6 E is even, so it is either 4 or 8 Cases for E+G>10: If E is 4: G is 7 but 4+7=11 and 1 is already used This means E is 8: 8+3=11 (can't be because 1 is already A) 8+5=13 which is possible 8+7=15 which is possible In both possibles cases 5 is being used. Looking at the one's digit: BxC=E(8) Options remaining are 3,4,7 Only possible cases are 4×7 or 7×4 Therefore, I=3 and G=5 Since B and C are either 4 or 7: D cannot be 2 and must be 6, F=2. Thus ABxC=DE is 1(4 or7)×(7 or 4)=68 Which makes B=7 and C=4 Again, lots of assumptions but a more funner way than brute force.

You can eliminate number > 87 in the multiplication because the final result is a 2-digit number and zero (0) is not allowed.

Instead of checking for 99, you could've eliminated the multiplied values that are greater than 86 itself as a two-digit number of at least 12 is to be added to it to arrive at the final value of less than 99.

Here is how I found the right answer (I think much more quickly) I considered the single digit multiplier first. That combined with the two-digit result imposes a major limitation. Second thing to check is the smallest remaining digit to place in the tens digit of the added number. 12 is the smallest multiplier, then 34 is the smallest amount you can add. 9x12 too large 8x12 too large 7x12 too large 6x12 = repeat 2 6x13 = 78 +24 = too large 5 is impossible because of 5/0 ending 4x12 = repeat 4 4x13 = all the small digits used (52, the 10s digit of the added number needs to repeat) 4x16 = repeat 4 4x17 = solution

Sometimes brute forcing can still be a joy to watch, but I'd expected that there would be some more clever tricks involved in solving the problem... This feels more like doing a computer's job.

17x4=68 => 68+25=93 :) I went by testing with the lowest possible values for a and b. I also eliminated impossible solutions like c=5 and b=5 for example.

This was a horrible problem requiring brute force, I appreciate your patience with "big" numbers because a "xx+12" surprise solution may have existed. I was hoping for a math equation system, but my LIKE goes to the editing quality.

This is much shorter to brute force:: first thing: *c cant be 1* ,so c is 2 or higher, thus *a cant be 9,8,7,6,5* which will result in 3digit number. Also *b cant be 5* as if it was-then e will be 0 or another 5 which both cant happen, *c cant be 5 as well* if it was-then e will again be 0 or another 5 which both cant happen. *e cant be 5* if it was,then either b or c should be 5. Now with this thing: go through every possibility of e from 1 to 9(5 cant come) the above conditions such as *a cant be 9,8,7,6,5* eliminates many possibility. We end up with e=8, i.e. 17*4=68, now with remaining 2,5,3,9 we know f is 2 or 3 else it will give 3 digit sum,and we discover that it is 17*4=68,68+25=93

Didn't think it would be s brute force solution tbh

Really amazing

Brute forcing the combinations would be trivial on a computer, but the algorithm might look cooler if it used a few tricks that humans might apply on paper. Upper bounds on hi (98) and lower bounds on fg (12) to limit the range of de to a maximum of 84, plus some divisibility rules and using sets in combination to manage the remaining options so the whole process looks concise. It was fun on paper.

I love solving for a solution based on the thumbnail before watching these videos for alternate perspectives. Surprised to see only one for this problem, and error checking isn’t that great

I did the same general process but I eliminated more options for C before going through the teens. 9-6 got eliminated, because they were out of bounds. The best case scenario would be 13*c+24

no need to check after 43.. because plus number is more than 12 if you multiply 46x2 = 92.. then you need to only 7 to get 99... so it is not possible as you need to add 07 and zero is not an option..

This problem might have 2 interpretations. 1. This could be a 2 step multiplication algorithm with the final answer on the bottom. 2. It could be interpreted like in this video

22:14 I solved another problem There are 10 letters and they are 0 to 9. It is addition. TERESA+TEASES+CHASE+ME=CHARLIE

use a little logic to try to figure out the top, and it was a bit bruteforce at the end but got there. I mostly was like "top has to be small, bottom probably has a 9 as a target, and b&c cant be 1&5"

Me checking the length of the video: "Surely he won't check every possible combination" 22 minutes later: Ah this mf did it :D

I'm a little bit disappointed that I was left with a feeling of "well I could have done this too" because there wasn't any kind of "deeper insight" or a "quick" solution. It was basically just going case after case and discarding what doesn't fit. 🤷 For all it's worth, im still glad it was someone else doing the counting instead of me. 🤣

I was surprised you kept going so long. The first thing I did was work backwards and determined the largest possible number at the end is 98. So my equation looks like this: ab*c+fg=98 Next, the smallest number that fg could be is 12. ab*c+12=98 -> ab*c=86 The smallest number for c is 2 (yes it is against the rules, but ok for making a bound) ab*2=86 -> ab = 43 So 43 is the absolute maximum that I could quickly prove that could work. No need to go all the way to 50.

The 18×2=36 +54=90 and 38×2=76 +14=90 are right, because the bottom right digit CAN be 0.

I only checked for C term and started with the largest possibility and slowly went down I reached from 9,8,7,6,5 and then when I checked for 4, it worked out. I started checking with 12, 13, 17 and found the answer

I accidentally solved it first try loll 😭

Now let's do it for all base number system

I assumed the first digit would be a 1 or a 2 and the third would also be low. Guessed it very quickly making similar assumptions, like at least one of the 8 or 9 would be in a singles column in the addition section and the solution would likely start with 8 or 9. Wanting to save low digits for later, guessed 17 in the first spot. After stupidly trying 17 x 3 and seeing I already had a repeat, tried 17 x 4. With a last digit of 8, saw the 5 in the digit column of the next number would give the 3 and the rest fell into place. Bottom line is that I got lucky and got it with just that first 17x3 error.^.

17*4=68, +25=93

It is one of the downsides of Math: too mcuh work for too little results!

You could have restricted the values of the addition part significantly. The "DE"

Can we get Wolfram to solve it? tried to tell it: set {1,2,3,4,5,6,7,8,9} = {a,b,c,d,e,f,g,h,i}; 10ac + bc = 10d+e; 10d + e + 10f + g = 10h + i but it just asigned the a-i in order :-(

@phantomlogic6940 G’day mate. You seem cluey. Can it be solved using algebraic equations? Cheers, SG🇦🇺

Im really proud i solved this problem on my own lol

We can start from the digit at C position. Only 2, 3, 4 can go to the C position. Around 10 minutes is enough to get the solution. Thanks

You shouldn’t have looked at products over 87. 90 and an over when summing with a 2 digit number automatically will end up a three digit number. 89 would start with 12 added and becomes a three digit number. 88 repeated 8 so 87 it is

accidentally solved it while trying to eliminate 4 from the single digit multiplier, and 17 was the first possibility where i had something going

This is an example of method of exhaustion

Python solves it in a fraction of a second :) from itertools import permutations for digits in permutations(range(1,10)): if (digits[0]*10 + digits[1]) * digits[2] == digits[3]*10 + digits[4] \ and digits[3]*10 + digits[4] + digits[5]*10 + digits[6] == digits[7]*10 + digits[8]: result = digits break print(result)

Hello, Mind Your Decisions! I have a tricky one for ya. Today, we were studying circles, and I came across a unordinary problem. [ If a Circle has a radius of 14 centimeters, which of the following is correct? ] A: d (diameter) is less than or equal to 28 centimeters B: C (circumference) ≈ 87.96 centimeters C: A (area) ≈ 104.78 centimeters D: A (area) ≈ 615.75 centimeters Now obviously, B and D are correct. By what about A? My teacher said it was correct, although I doubted it. Your opinion?

Also with the addition part you take to long on the cases instead of comparing the smallest 2 digit addition vs the biggest 2 digit you can make it 19 x 2 =38 with 4 5 6 7, 76 being the largest 2 digit with 45 being the smallest 38 + 45 is 83 > 76 no need to continue checking. This dauntless dribble of having to check everything when clear defined rules can cut so much out in simple one step checks is mind numbingly slow.