I would like to see a question where there is one or a few solutions but they are not “small”, for example if the only solution was a=37, b=78 and c=901
Pell's equations for example. I don't know of any problems that have a finite number of solutions, but also a solution with relatively big numbers, say all >1000
Further solutions in N: c=0 is impossible, as a! >=1 and 5^b >=1. b=0 implies a!+1=7^c. If a>=7 this is impossible because the LHS wouldn't be divisible by 7, but the RHS would. Thus a
@@thephysicistcuber175 Woah, chill down, dude, the way you're replying to me indicates that you got a bit ruffled by my reply. I explicitly told you that you're in fact right with the (3, 0, 1) answer, but that's in the context for whole numbers greater than or equal to zero, while the problem in the video is explicit for natural numbers. On the other hand, if we define zero to be a natural number (which in my view it would make a lot more sense), then of course, put all the zeroes you want. So, how am I wrong when in fact zero is defined not to be a natural number? It all depends in the context.
@@thephysicistcuber175 Well, you're absolutely right. As you told me before, Michael explicitly asked us to find any extra solutions if 0 is to be in N, at minute 2:30. I guess I didn't pay enough attention there; my bad. Credit where credit's due 👏🏻😲
Firstly c=0 has no solution as a! or 5^b would need to be zero. a=0 has no solution because 1+5^b=7^c means 5^b=7^c-1=(7-1)(7^(c-1)+...+1) but the left hand side is a power of 5 and the right hand side is a multiple of 6 (for c>0), which is a contradiction. For b=0, to start with the condition that a >=5 gives no solution does not hold because 5^b=1, however we know a0). We are left with a∈{2,3,4,5,6}, c>0 and b=0. Easy to check each case: If a=2 then a!+1
This is good. It encapsulates Michael's use of modulo arithmetic in the other problems he's posted this last year or two. And what's more, I understood it!
About allowing 0 in here: c cannot be 0 since 7^0=1 and a!+5^b>=1+1=2. b=0 makes the formula a!+1=7^c. For a>=7, we get 7|a!, which means that a!+1 cannot be divisible by 7 to equal a power of 6. Checking by cases of a, we arrive at the only additional solution of (a,b,c)=(3,0,1) from 6+1=7. For a=0, this is equivalent to ar1 in that a!=1, giving no new solutions.
I watched your video on the Caesar cipher the other day. Can you do a video on how public keys and private keys are generated in the RSA algorithm, with a trivial example? I think it’s fascinating that the security of the Internet is built on number theory.
@@ritam8767 Actually many mathematicians define the Natural numbers by way of set theory as the numbers which are the cardinalities of finite sets, meaning the smallest Natural number is the cardinality of the empty set which is 0. (In fact all my professors used this definition.)
@@ritam8767 "Some definitions, including the standard ISO 80000-2, begin the natural numbers with 0, corresponding to the non-negative integers" en.wikipedia.org/wiki/Natural_number In my personal experience, the natural numbers N always included zero, while N* was the set of natural numbers excluding zero. It would appear that there are different definitions for the natural numbers, and only the people who don't include zero as a natural number even define "whole" numbers.
@@ritam8767 And that's a perfectly fine definition! It just turns out there isn't broad agreement on what should be included in the set of Natural numbers.
Like several other modulus problems, this hinges on knowing that m^k mod n can only take on a few values for particular m,n as k ranges. Here e.g. he used that 7^k mod 25 only takes on a few values. I don’t get how he is finding these magic values though.
My idea was similar. 5^b has for b>0 last digit 5. 7^c has last digit 2 or 4 or 6 or 8. a! has always last digit 0 for a≥5. If a=3, then 7^b=1+5(1+5^(c-1)), but 7^b has not last digit1.If a=4, then 7^b=5^2*(1+5^(c-2))-1, only for c=2 Is there solution b=2
He has a whole number theory playlist. This particular video and following ones might be what you're looking for: th-cam.com/video/QjTMbspAjsw/w-d-xo.html
I was able to prove that a can be only 2 or 4 immediately. But I couldn't deny the existence of other solutions than (a,b,c)＝(2,1,1),(4,2,2) in spite of pondering for some days.
I'm a little bit confused. Please, is 0 not a natural number? If so, then the set of natural numbers is lN* (not lN). Isn't it? Because, the equation above works when (a,b,c)=(3,0,1) for example. Thanks for helping me
The short of it: It is more correct to define 0 as natural number, but some do not. The original tasks do avoid calling it natural numbers bec of the ambiguity. Michael ususally uses N as a shortcut for positive integers. There have been long discussions before.
@@TechToppers Well it's a junior competition after all (for students under 15.5 years). And also N6 in this SHL doesn't nearly match up to NT6 in a ISL, but that's understandable.
Technical point but whether or not to include 0 as a Natural Number is a long open debate. Basically if you build up your definitions from set theory then the most “natural” (pardon the pun) definition of the Natural Numbers is that they are the cardinalities of finite sets. So 0 is the cardinality of the empty set, 1 is the cardinality of the set containing only the empty set, and so on. This also has the advantage that it means the Natural Numbers includes its arithmetic identity element 0. (So personally I prefer to include 0 in the Naturals.) (P.S. FYI ISO 80000-2 also includes zero in the Naturals, so if you want to go by ISO standards this is it.) Of course like I said some mathematicians define the Natural numbers as the strictly positive integers like Michael does in this video. The main thing, though, is to remember to specify which definition you’re using for the Naturals when you write down a proof involving them since not everybody uses the same definition.
We need a universal definition to avoid confusions. So your personal definition doesn't matter. I have been taught in schools that naturals start from 1 and the whole numbers start from 0.
@@wafimarzouqmohammad8054 Including 0 in the Natural Numbers is not my “personal definition”, it’s a standard definition recommended by the ISO and many mathematicians. It’s somewhat analogous to how Pluto used to be defined as a planet but now it isn’t by the modern definitions of what it is to be a planet. Similarly when set theory was introduced the modern definitions of the Natural Numbers likewise evolved to include zero. But like there are still a lot of people and textbooks that call Pluto a planet because “that’s what was taught when they were a kid” there are a lot of textbooks and people who use the older definition of the Naturals because “that’s what they were taught as a kid.”
@@wafimarzouqmohammad8054 I agree, that's what I said above. ("The main thing, though, is to remember to specify which definition you’re using for the Naturals when you write down a proof involving them since not everybody uses the same definition. ")
I would like to see a question where there is one or a few solutions but they are not “small”, for example if the only solution was a=37, b=78 and c=901
Can someone make a problem with solutions (69, 420, 1337) ? 😂
@@goodplacetostop2973 uff
@@goodplacetostop2973 lmao
Pell's equations for example. I don't know of any problems that have a finite number of solutions, but also a solution with relatively big numbers, say all >1000
there's putnam 2018 A1, there are only 6 solutions, but they are big (a = 673, b =1354114)
Further solutions in N:
c=0 is impossible, as a! >=1 and 5^b >=1.
b=0 implies a!+1=7^c. If a>=7 this is impossible because the LHS wouldn't be divisible by 7, but the RHS would. Thus a
Yes, that is true, but remember that we are talking about the NATURAL numbers here, and 0, by definition, is not a natural number.
Nice!
@@titan1235813 1. You're wrong.
2. Michael explicitly asked us in the comments to come up with the missing solutions.
@@thephysicistcuber175 Woah, chill down, dude, the way you're replying to me indicates that you got a bit ruffled by my reply. I explicitly told you that you're in fact right with the (3, 0, 1) answer, but that's in the context for whole numbers greater than or equal to zero, while the problem in the video is explicit for natural numbers. On the other hand, if we define zero to be a natural number (which in my view it would make a lot more sense), then of course, put all the zeroes you want.
So, how am I wrong when in fact zero is defined not to be a natural number? It all depends in the context.
@@thephysicistcuber175 Well, you're absolutely right. As you told me before, Michael explicitly asked us to find any extra solutions if 0 is to be in N, at minute 2:30. I guess I didn't pay enough attention there; my bad. Credit where credit's due 👏🏻😲
I think the only extra solution if a,b,c could be 0 would be a=3, b=0, c=1
In regards to your username: nice
Firstly c=0 has no solution as a! or 5^b would need to be zero. a=0 has no solution because 1+5^b=7^c means 5^b=7^c-1=(7-1)(7^(c-1)+...+1) but the left hand side is a power of 5 and the right hand side is a multiple of 6 (for c>0), which is a contradiction.
For b=0, to start with the condition that a >=5 gives no solution does not hold because 5^b=1, however we know a0).
We are left with a∈{2,3,4,5,6}, c>0 and b=0. Easy to check each case: If a=2 then a!+1
@@matthewryan4844 by that logic, 5 doesn't divide 30 because 6 divides 30 :P
Edit: erp, nevermind, I figured out what you meant
This is good. It encapsulates Michael's use of modulo arithmetic in the other problems he's posted this last year or two. And what's more, I understood it!
Again the idea is that a! is divisible by all numbers less than a+1, thus taking mod 5 breaks for a>=5 and b,c>0
HOMEWORK : Compute the last digit of 2^(3^(4^(...2014))).
SOURCE : SMT 2014 - Team Round
SOLUTION
*2*
The exponent of 2 is equivalent to 1 (mod 4). Since 2^x (mod 10) has period 4, we
have that 2^1 ≡ 2 (mod 10).
Can u tell what is smt?
@@metablaze3523 Stanford Math Tournament
@@goodplacetostop2973 thank you!!
How you know so many things ??
Are you 👽👽👽 alien ???
About allowing 0 in here: c cannot be 0 since 7^0=1 and a!+5^b>=1+1=2. b=0 makes the formula a!+1=7^c. For a>=7, we get 7|a!, which means that a!+1 cannot be divisible by 7 to equal a power of 6. Checking by cases of a, we arrive at the only additional solution of (a,b,c)=(3,0,1) from 6+1=7. For a=0, this is equivalent to ar1 in that a!=1, giving no new solutions.
I watched your video on the Caesar cipher the other day. Can you do a video on how public keys and private keys are generated in the RSA algorithm, with a trivial example? I think it’s fascinating that the security of the Internet is built on number theory.
There is also a solution (3,0,1) if we consider 0 as a natural number.
0 is by definition, not a natural number. It is a whole number
@@ritam8767 Actually many mathematicians define the Natural numbers by way of set theory as the numbers which are the cardinalities of finite sets, meaning the smallest Natural number is the cardinality of the empty set which is 0. (In fact all my professors used this definition.)
@@ritam8767
"Some definitions, including the standard ISO 80000-2, begin the natural numbers with 0, corresponding to the non-negative integers"
en.wikipedia.org/wiki/Natural_number
In my personal experience, the natural numbers N always included zero, while N* was the set of natural numbers excluding zero. It would appear that there are different definitions for the natural numbers, and only the people who don't include zero as a natural number even define "whole" numbers.
@@joeo3377 oh, didn't know that. I've been using N={1,2,3,...} forever. OK then.
@@ritam8767 And that's a perfectly fine definition!
It just turns out there isn't broad agreement on what should be included in the set of Natural numbers.
12:07
you commented 2 days ago 😱 howwww????
@@mil9102 Who dares summon the time wizard? 🧙♂️
Amazing!!!!
Very nice and easy method to solve
Like several other modulus problems, this hinges on knowing that m^k mod n can only take on a few values for particular m,n as k ranges. Here e.g. he used that 7^k mod 25 only takes on a few values. I don’t get how he is finding these magic values though.
If we admit a zero value, there's another obvious solution: a=3, b=0, c=1 (6 + 1 = 7).
My idea was similar. 5^b has for b>0 last digit 5. 7^c has last digit 2 or 4 or 6 or 8. a! has always last digit 0 for a≥5. If a=3, then 7^b=1+5(1+5^(c-1)), but 7^b has not last digit1.If a=4, then 7^b=5^2*(1+5^(c-2))-1, only for c=2 Is there solution b=2
Suppose p and q are distinct primes. Let n be a positive integer. As k ranges over positive integers, how many distinct values does p^k mod q^n take?
Is there a video/site I can read about that "mod" operation you performed and that "mod 25" and "mod 18" charts?
en.wikipedia.org/wiki/Modular_arithmetic
He has a whole number theory playlist. This particular video and following ones might be what you're looking for:
th-cam.com/video/QjTMbspAjsw/w-d-xo.html
Hi guys it’s ma-th dimension and this is JBMO 2019 SL N6( not suggested by me) but I swear it looked so familiar so I had to give credits:)
I was able to prove that a can be only 2 or 4 immediately.
But I couldn't deny the existence of other solutions than (a,b,c)＝(2,1,1),(4,2,2) in spite of pondering for some days.
This is a really nice viewer suggestion
It was my 2020 National Olympiad problem number 5 :)
I never knew math can be so fun
Can we do a number theory problem that doesn't involve working in mod N?
Litererally my suggestion:a!+b⁵=c⁷
Nice Viewer Suggested Problem At 0:01
I'm a little bit confused. Please, is 0 not a natural number? If so, then the set of natural numbers is lN* (not lN). Isn't it?
Because, the equation above works when (a,b,c)=(3,0,1) for example.
Thanks for helping me
zero may or may not be a natural number depending on the definition. in my experience, zero is usually not included.
The short of it: It is more correct to define 0 as natural number, but some do not. The original tasks do avoid calling it natural numbers bec of the ambiguity. Michael ususally uses N as a shortcut for positive integers. There have been long discussions before.
Yes, more number theory !
*Viewer suggestion*
Find all polynomials p(x) with real coefficients such that p(x^2) = p(x) × p(-x)
Nice problem. Looks easy but the devil hides in the details.
This is problem N6 from the 2019 JBMO Shortlist (a junior competition)
Bruh... N6? I got the finishing idea in 1 minute straight.
@@TechToppers Well it's a junior competition after all (for students under 15.5 years). And also N6 in this SHL doesn't nearly match up to NT6 in a ISL, but that's understandable.
@@MarinHristov-n8m I'm 15 too. But still N6 was pretty simple. IMO is is still a far cry...
Heyy mst michael good problem !!!
assuming 0 is natural number
a = 3, b=0 , c=1
a=4, b = 2, c = 2
My first two guesses
Idk why but everytime I see an random absurd equation like this my first instinct is to prove there is no solution.
Or just guess the year of the competition as the only solution.
nice. I found the obvious solutions and
I got as far as showing that a was in { 2 , 3 , 4 }
but couldn't finish it. I kicked myself when I saw mod25.
wonderful
Pretty awesome
8:28 b could be equal to 0 which gives c=1
I do have a question. Find theta where the cos(theta) = +- i.
I just write what's proportional if Im close.
Technical point but whether or not to include 0 as a Natural Number is a long open debate. Basically if you build up your definitions from set theory then the most “natural” (pardon the pun) definition of the Natural Numbers is that they are the cardinalities of finite sets. So 0 is the cardinality of the empty set, 1 is the cardinality of the set containing only the empty set, and so on. This also has the advantage that it means the Natural Numbers includes its arithmetic identity element 0. (So personally I prefer to include 0 in the Naturals.) (P.S. FYI ISO 80000-2 also includes zero in the Naturals, so if you want to go by ISO standards this is it.)
Of course like I said some mathematicians define the Natural numbers as the strictly positive integers like Michael does in this video. The main thing, though, is to remember to specify which definition you’re using for the Naturals when you write down a proof involving them since not everybody uses the same definition.
We need a universal definition to avoid confusions. So your personal definition doesn't matter. I have been taught in schools that naturals start from 1 and the whole numbers start from 0.
@@wafimarzouqmohammad8054 Including 0 in the Natural Numbers is not my “personal definition”, it’s a standard definition recommended by the ISO and many mathematicians. It’s somewhat analogous to how Pluto used to be defined as a planet but now it isn’t by the modern definitions of what it is to be a planet. Similarly when set theory was introduced the modern definitions of the Natural Numbers likewise evolved to include zero. But like there are still a lot of people and textbooks that call Pluto a planet because “that’s what was taught when they were a kid” there are a lot of textbooks and people who use the older definition of the Naturals because “that’s what they were taught as a kid.”
@@Bodyknock Well then people should just specify to avoid confusions.
@@wafimarzouqmohammad8054 I agree, that's what I said above. ("The main thing, though, is to remember to specify which definition you’re using for the Naturals when you write down a proof involving them since not everybody uses the same definition.
")
@@Bodyknock An
d he did specify it in the vid. "by the natural numbers we really mean the positive integers"
I take advantage of diagrams
Jbmo shortlist
Stop the ad. Stop the cap.
Well you seem to skip the solution a=3, b=0 and c=1 unless you don't count 0 as a natural number...
Actually there is one missing solution: a = 3, b = 0, c = 1
*Solve[{a! + 5^b == 7^c, 0 < a < 10, 0 < b < 10, 0 < c < 10}, {a, b, c}, Integers]*
Funny how the only solution he didn't mention at the beginning was the first one I thought of:
3!+5^0=7^1
Because the solution set has to be Natural numbers
a=4 case doesn't need to be any longer than a=2
a = 0 because its timed by factorial. b = 2, c = 3.57
"even" and "odd" are not in all languages. I recommend using 0 and 1 (mod 2), respectively.