for the case (a55) = a! + 240, you could instantly say that theres no solutions, because for the a > 1 a! is even, but a55 is odd, and a = 1 is not a solution too
Although I appreciate a rigorous approach to such problems, I think this one could’ve been done with less rigor. I think the following deductive steps lead to an answer fairly quickly. Note: DFS = digit factorial sum of abc. 1. Clearly, all of the digits must be less than seven, due to the three digit restriction. 2. No digit is 6, since this forces the DFS to include at least a 7. 3. There must be at least one 5 to achieve a three digit DFS. 4. Since 3(5!) = 360, the first digit cannot be 5, so there are less than three 5’s. 5. If there were exactly two 5’s, the DFS is bounded between 241 and 264. So if there are two 5’s, the solution must be 255, which fails since its DFS is 244. Hence, there is exactly one 5. 6. Since there is one 5, the DFS is bounded between 122 and 168, forcing the solution to be 15X or 1X5. 7. Since there must be digits of 1 and 5, this bounds the DFS between 122 and 145, forcing the solution to be 125, 135, or 145. Only 145 works.
@@DaveyJonesLocka i understand...just trying to see if ur method could actually be explained in 6:30 min or less for the portion of the prrof after realizing 6 is not a valid value. After all its math ...and rigorum is part of the deal even if one skips a few lines. Without using too much complex math a proof using fewer lines is pretty after all. Esp when all the verbosity is elim. I agree the num cases gets tedious and im surprised the author didnt use his trusted methods that he so often does in other proofs(which woulda mount to 8-10 line symbolic proof for this q?) Its like his cos trick that he didnt use the reversal of in the last video or prev
As we restrict a,b,c not to be ≥6 thus a,b,c can't take each 5 and even either both of a,b,c to be 5 so this means a,b,c either is utmost 5 to be occured + rest of digits to be either recurring or distinct.👍 Thus u get finally (1,4,5).
This is more detective work than number theory! I used a different series of tests, first working through the values of "a" rather like the video which revealed that only a=1 was possible. Then we know if a=1 then 100
An easier way to solve it is to realise that if a,b,c
Surprised you stopped bounding...2 5s means 240
What doesabc with a line over it mean though. .the average or a 3 digit number with digits abc..he didn't say..
@@leif1075 the written q? states it
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@@bait6652 amazing ...ruclips.net/video/dIbCw8ElA3o/видео.html
for the case (a55) = a! + 240, you could instantly say that theres no solutions, because for the a > 1 a! is even, but a55 is odd, and a = 1 is not a solution too
amazing ...ruclips.net/video/dIbCw8ElA3o/видео.html
Another approach is to see that 5 divides the LHS, so 5 divides the RHS, so 5 divides a!, so 5 divides a.
Although I appreciate a rigorous approach to such problems, I think this one could’ve been done with less rigor. I think the following deductive steps lead to an answer fairly quickly. Note: DFS = digit factorial sum of abc.
1. Clearly, all of the digits must be less than seven, due to the three digit restriction.
2. No digit is 6, since this forces the DFS to include at least a 7.
3. There must be at least one 5 to achieve a three digit DFS.
4. Since 3(5!) = 360, the first digit cannot be 5, so there are less than three 5’s.
5. If there were exactly two 5’s, the DFS is bounded between 241 and 264. So if there are two 5’s, the solution must be 255, which fails since its DFS is 244. Hence, there is exactly one 5.
6. Since there is one 5, the DFS is bounded between 122 and 168, forcing the solution to be 15X or 1X5.
7. Since there must be digits of 1 and 5, this bounds the DFS between 122 and 145, forcing the solution to be 125, 135, or 145. Only 145 works.
But can you explain it in less then 9min( video length)
Also 4) implies upper bound in 5) is 246
@@bait6652 I’m not trying to one-up the presenter. I was just offering a different perspective. But the answer to your question is yes.
@@DaveyJonesLocka i understand...just trying to see if ur method could actually be explained in 6:30 min or less for the portion of the prrof after realizing 6 is not a valid value.
After all its math ...and rigorum is part of the deal even if one skips a few lines.
Without using too much complex math a proof using fewer lines is pretty after all. Esp when all the verbosity is elim.
I agree the num cases gets tedious and im surprised the author didnt use his trusted methods that he so often does in other proofs(which woulda mount to 8-10 line symbolic proof for this q?)
Its like his cos trick that he didnt use the reversal of in the last video or prev
Minor mistake at the beginning, 9! = 362880, not 3628800.
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Interesting author typically likes bounding/modulo but instead used alot of case. When u get to
As we restrict a,b,c not to be ≥6 thus a,b,c can't take each 5 and even either both of a,b,c to be 5 so this means a,b,c either is utmost 5 to be occured + rest of digits to be either recurring or distinct.👍
Thus u get finally (1,4,5).
This is more detective work than number theory! I used a different series of tests, first working through the values of "a" rather like the video which revealed that only a=1 was possible. Then we know if a=1 then 100
Nice equation
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In Soviet Russia, the factorials expand you
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Lol
why didn't you even analize the case of a=0? there's no solution but i think we should check over it. Btw very nice video thank you
Problem gives that has to be a three digit number so a must be 1 or greater
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@@alboris8203 amazing ...ruclips.net/video/dIbCw8ElA3o/видео.html
@@alboris8203 well that's true, still i think it would have been amazing if there were an answer with a=0
40585 is the only other non-trivial case
nice
145
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Nice and easy
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asnwer=1 isit
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