He didn't mention it in the video for some reason, but φ(n) has a pretty easy to use formula, it's just n ( 1- 1/p₁)(1- 1/p₂)...(1- 1/pₖ) for all the prime factors of n. So for instance, the prime factors of 100 are 2 and 5, so φ(n) = 100 (1/2) (4/5) = 40. Similarly φ(40) = 40 (1/2) (4/5) = 16 .
There's another way to do it-- if you take powers of 97 (or -3) mod 100, you'll see it repeats after a cycle of 20. If you take powers of 98 (or -2) mod 20, it has a cycle of 4. 99 is congruent to 3 mod 4, and the mod 3 entry in the cycle of powers of 98 mod 20 is 12. The mod 12 entry in the cycle of powers of 97 mod 100 is 41.
This is the solution I wrote before watching the video, and avoids the "handwaving" about gcd(1998, 40)≠1. This looks like a job for the Carmichael λ-function we saw in Michael Penn's video "this number is "super divisible"". 1997ⁿ≡(-3)ⁿ (mod 100) will have a certain period mod 100, and we can take the residue of the exponent mod this period without changing the residue mod 100. Let's work out this period (but for 3, instead of -3) Powers of 3 reduced mod 100: 1: 3 2: 9 3: 27 4: 81 5: 43 6: 29 7: 87 8: 61 9: 83 10: 49≡-51 (mod 100) 11: -53 (mod 100) Clearly the pattern repeats but with 50 less the original value, so at 20 we'll get to 1. We can save doing this work - and in fact we shall solve the problem as if we hadn't done this - as we know (since 1997 is comprime to 100) this period is a divisor of φ(100)=φ(2²5²)=2×(5×4)=40, and, potentially, better still, using the Carmichael λ-function, a divisor of λ(100)=λ(2²5²)=2×(5×4)=40. However, both are suboptimal in this case. So we can reduce 1998^1999 mod 40 (or 20) without changing the value of 1997^1998^1999 (mod 100) In particular, we get 1998≡-2 (mod 40), so 1998^1999≡(-2)^1999 (mod 40)≡-2^1999 (mod 40) In this case, as 2 is not coprime to 40, we can't use φ or λ to reduce the exponent 1999 directly. However, as 40=2³5, we note that 2⁴≡1 (mod 5), as guaranteed by φ & FLT. So we get, for m, n≥3, if m≡n (mod 4) then 2ᵐ≡2ⁿ[≡0] (mod 8) and 2ᵐ≡2ⁿ (mod 5) so 2ᵐ≡2ⁿ (mod 40). So we can reduce 2ⁿ mod 40 by reducing n mod 4, as long as the reduced power ≥3 So 2^1999≡2³≡8 (mod 40) So 1998^1999≡-2^1999≡-8≡32 (mod 40) So 1997^1998^1999≡1997^32 (mod 100) ≡(-3)^32 (mod 100) ≡3^32 (mod 100) [from the initial work with powers of 3 we see this is same as 3^12≡-59≡41 (mod 100)] ≡9^16 (mod 100) Powers of 9 reduced mod 100: 1: 9 2: 81 3: 29 4: 61 5: 49=50-1 6: 41=50-9 Clearly the pattern repeats after 5 steps but with 50 less the original value, so at 10 we'll get to 1, and the pattern will repeat. [As expected based on what we saw earlier for powers of 3] So 9^16≡9^6 (mod 100) ≡41 (mod 100) So the last two digits of 1997^1998^1999 are 41
pick a starting base (b), pick a modular power (m, greater than b if you please) if you calculate powers of b mod m, it eventually starts to cycle: powers of 2 mod 10: 2;4;8;6;2;4;8;6;2 etc ad infinitum that loop repeats itself every 4 units, so if you needed to calculate 2^400000000001, its gotta be 2^1 = 2. now. different bases will have different loop lengths (powers of 9 mod 10 are 9;1;9;1 etc), BUT for all bases a sequence of length phi(m) is guaranteed to be a loop. phi(m) is euler's totient function- in this video phi(100) = 40 and phi(40) = 16 were both facts that made an appearance. why this works and more importantly the properties that let you compute phi are unpleasant to type in the comments section, but probably if you google it you can get a reasonable handle on phi(n) within ten minutes. hard to read, easy to understand; the mark of pure math everywhere and always. the way this stuff works is closely allied with RSA encryption, which is a standard unit in discrete math (first university proofs course). RSA is genuinely important so this all provides a lot of motivation to include this variety of problem in math competitions edit! if you're lost at a more basic level, you want to look up "what is modular arithmetic"; this problem is summarized as "calculate the value of [preposterously huge number] mod 100"
My reasoning was simpler (to me): 97^20≡1 mod 100, so the exponent of 1997 can be reduced mod 20. Since 100 is a multiple of 20, and reducing mod 100 is easier, I went with that. No power of 98 is going to be congruent 1 mod 100 (except the 0th power), but after the 1st 2 powers, every 20 powers gets you back to the same result. Since 1999>2, this works. 1999≡19 mod 20. 98^19≡12 mod 100. 97^12≡41 mod 100.
@@akifbaysal9141 No, the standard for power towers is to evaluate the exponents from the top down. It does **not** simplify to multiplying the exponents.
@@jursamaj thx for clarification, I noticed I misunderstood the reply I received. I was looking to delete my previous note postef here but was not able locate it before you posted, I deleted now..
38 and 40 are not coprime, so it's not correct to apply Euler's theorem when reducing 38¹⁹⁹⁹ mod 40 (though it does seem to work in this case). If this reduction were correct in general, then we could equally argue that 38²⁰⁰⁰ = 38⁰ = 1 mod 40, because 2000 = 0 mod 16. But that's clearly wrong because 38²⁰⁰⁰ is even, as is its residue mod 40. Powers of -2 mod 40: 1, -2, 4, -8, 16, 8, -16, -8, 16, 8, -16, ... So (-2)¹⁹⁹⁹ = -8 = 32 because 1999 = 3(4) and 1999 > 2. I *think* the reduction will always work for a^b mod n if the exponent is reduced to no less than phi(n)/gcd(a,n) (because that guarantees we stay within the eventual cyclic portion of a^b mod n) which is why it works here but not for 38²⁰⁰⁰, but I'm not sure that's the right condition. Is there a general result for when we can "use Euler's theorem" even though its precondition is not satisfied?
The general result is that we can reduce down to the least power r such that a^r shares no prime factors with m that have a lesser exponent in the factorization of a^r than in the factorization of m. So, to get a statement true for all a, choose r at least as large as the largest exponent in the prima factorization of m. Which is 3 in this case.
Yeah for 38^2000 you can then reduce to 38^16, which gives (-2)^16 = 2^16 = 65536 = 16, which is the correct result. (In fact, in the group of multiples of 8 mod 40, 16 is the neutral element, since it reduces to 1 mod 5--and indeed 16*8=128 reduces to 8 mod 40, etc.)
I had a question at 3:40. I thought that since 1999 mod(16) = 15, you could also use -1. But then you're left with 38^-1 mod(40). In other words, you need some number x s.t. 38x = 1 mod(40), but this is impossible since 38 is even so 38x is even. Is the general rule of thumb that you don't want to mess around with negative integers in the exponent and only the base?
I have an interesting question: Prove whether or not there exists a constant ‘p’ s.t. the limit as n tends towards infinity of: (n!)^(1/n) - n/p converges
@@DrR0BERTyou can find some justification on the Wikipedia page for the Carmichael function (though without proof) that modular powers must loop starting at r = the largest exponent in the prime factorization of the modulus. In this case, m = 40, the highest power is 3, so you can reduce powers mod 16 down to 3 but not necessarily lower. This case reduces to 15 so it's ok.
I won't write a full proof, but the reason for this is that once you take the r power, any prime factors that the base and the modulus shared are now have at least as large an exponent as in the modulus. Thus, we can reduce it all mod p_i^r_i. In this case, we could go from looking mod 40 to mod 5, since once the exponent of -2 reaches 3, every subsequent power will be divisible by 8, and we can reduce the whole modular equation down to modulo 5
Yeah the way I did it is with CRT. Since we want to eventually reduce 1998^1999 mod 40, let's separate the prime factors of 40 = 2³ · 5 and look at 1998^1999 mod 8, which is clearly 0 since it contains a factor of 2^1999, and 1998^1999 mod 5, which reduces to 3^3 = 27 reduces again to 2 (1998 = 3 mod 5, phi(5) = 4 and the exponent 1999 = 3 mod 4) so we have 1998^1999 = 0 mod 8, and 2 mod 5, which must mean it reduces to 32 mod 40 (one can quickly look through the multiples of 8 to see which one reduces to 2 mod 5).
Basics only: we have 2⁵ = 32 ≡ 2 mod 20, thus if k = 1998¹⁹⁹⁹ then k is even and k ≡ (-1)¹⁹⁹⁹∙2¹⁹⁹⁵∙2⁴ = -1∙2³⁹⁹∙2⁴ = -1∙2⁴⁰⁰∙2³ ≡ -1∙2¹⁶∙2³ = -1∙2¹⁵∙2⁴ ≡ -1∙2³∙2⁴ = -1∙2⁵∙2² ≡ -1∙2∙2² = -8 ≡ 12 mod 20. Next, 1997ᵏ ≡ (-3)ᵏ = 3ᵏ mod 100 (k is even). But then 3ᵏ = (3²⁰)ᵐ∙3ʳ where k = 20k + r, so r is a remainder which we've just found, meaning 3ᵏ = (3²⁰)ᵐ∙3¹² ≡ 1ᵐ∙3¹² mod 100 = 3¹² mod 100 = 41. And this is because 3²⁰ = ((3⁵)²)² = (343²)² ≡ (43²)² = 1849² ≡ 49² = 2401 ≡ 1 mod 100 and similarly, 3¹² = ((3³)²)² = (27²)² = 729² ≡ 29² = 841 ≡ 41 mod 100.
So (10a+b)^big will always end in some combination of 10a b^(n-1) + b^n, since all higher terms become irrelevant. Therefore we're looking at 97^1998^1999. If we work mod 100, we can actually count 97 as -3; since 1998 is even, the power is even and the minus sign vanishes; we're looking at 3^1998^1999. The even powers of 3, mod 100, progress... 9, 81, 29, 67, 3, 27, 43, 87, ... that's a lot, i quit
I love these. So if this determination can only be made when phi(a,n)=1 then are there towers of powers that are impossible to find the last digits of?
Not impossible, you'll just need an even more general theorem. In fact, the tower in the video is such a harder tower (Michael didn't take this into account, though)
Should we consider the thumbnail expression is for last two decimal digits of umber A = (1997)^(1998^1999).. When different interpretation is made to power of power, the result may come out diferent.
@@theupson But it's quicker than explaining how to calculate φ(n). Optionally, he could have said that 2 and 5 are the only primes that divide 40, and every other number is even so we can remove 20 numbers from the 1-40 list, so we have 1/2 of 40 = 20 left. Then every fifth number in the 1-40 list is divisible by 5, so we have 4/5 of 20 = 16 left. So φ(40) = 16. Similarly, φ(100) = 100 x 1/2 x 4/5 = 40. I still think writing out the lists as he did makes it easier to understand for anyone who has little number theory.
We follow Mr. Michael Penn's videos. I wondered what it would be like if he spoke in Turkish. let's see how he reacts. by the way, he has a lot of followers in Turkey, I wonder if we could somehow publish his translated videos online. would he be interested? ua-cam.com/video/1UB14Mm5TMM/v-deo.html
He didn't mention it in the video for some reason, but φ(n) has a pretty easy to use formula, it's just n ( 1- 1/p₁)(1- 1/p₂)...(1- 1/pₖ) for all the prime factors of n. So for instance, the prime factors of 100 are 2 and 5, so φ(n) = 100 (1/2) (4/5) = 40. Similarly φ(40) = 40 (1/2) (4/5) = 16 .
There's another way to do it-- if you take powers of 97 (or -3) mod 100, you'll see it repeats after a cycle of 20. If you take powers of 98 (or -2) mod 20, it has a cycle of 4. 99 is congruent to 3 mod 4, and the mod 3 entry in the cycle of powers of 98 mod 20 is 12. The mod 12 entry in the cycle of powers of 97 mod 100 is 41.
This is the solution I wrote before watching the video, and avoids the "handwaving" about gcd(1998, 40)≠1.
This looks like a job for the Carmichael λ-function we saw in Michael Penn's video "this number is "super divisible"".
1997ⁿ≡(-3)ⁿ (mod 100) will have a certain period mod 100, and we can take the residue of the exponent mod this period without changing the residue mod 100.
Let's work out this period (but for 3, instead of -3)
Powers of 3 reduced mod 100:
1: 3
2: 9
3: 27
4: 81
5: 43
6: 29
7: 87
8: 61
9: 83
10: 49≡-51 (mod 100)
11: -53 (mod 100)
Clearly the pattern repeats but with 50 less the original value, so at 20 we'll get to 1.
We can save doing this work - and in fact we shall solve the problem as if we hadn't done this - as we know (since 1997 is comprime to 100) this period is a divisor of φ(100)=φ(2²5²)=2×(5×4)=40, and, potentially, better still, using the Carmichael λ-function, a divisor of λ(100)=λ(2²5²)=2×(5×4)=40.
However, both are suboptimal in this case.
So we can reduce 1998^1999 mod 40 (or 20) without changing the value of 1997^1998^1999 (mod 100)
In particular, we get 1998≡-2 (mod 40), so
1998^1999≡(-2)^1999 (mod 40)≡-2^1999 (mod 40)
In this case, as 2 is not coprime to 40, we can't use φ or λ to reduce the exponent 1999 directly. However, as 40=2³5, we note that 2⁴≡1 (mod 5), as guaranteed by φ & FLT. So we get, for m, n≥3, if m≡n (mod 4) then 2ᵐ≡2ⁿ[≡0] (mod 8) and 2ᵐ≡2ⁿ (mod 5) so 2ᵐ≡2ⁿ (mod 40).
So we can reduce 2ⁿ mod 40 by reducing n mod 4, as long as the reduced power ≥3
So
2^1999≡2³≡8 (mod 40)
So
1998^1999≡-2^1999≡-8≡32 (mod 40)
So 1997^1998^1999≡1997^32 (mod 100)
≡(-3)^32 (mod 100)
≡3^32 (mod 100)
[from the initial work with powers of 3 we see this is same as 3^12≡-59≡41 (mod 100)]
≡9^16 (mod 100)
Powers of 9 reduced mod 100:
1: 9
2: 81
3: 29
4: 61
5: 49=50-1
6: 41=50-9
Clearly the pattern repeats after 5 steps but with 50 less the original value, so at 10 we'll get to 1, and the pattern will repeat.
[As expected based on what we saw earlier for powers of 3]
So 9^16≡9^6 (mod 100)
≡41 (mod 100)
So the last two digits of 1997^1998^1999 are 41
I understood nothing
pick a starting base (b), pick a modular power (m, greater than b if you please)
if you calculate powers of b mod m, it eventually starts to cycle:
powers of 2 mod 10: 2;4;8;6;2;4;8;6;2 etc ad infinitum
that loop repeats itself every 4 units, so if you needed to calculate 2^400000000001, its gotta be 2^1 = 2.
now. different bases will have different loop lengths (powers of 9 mod 10 are 9;1;9;1 etc), BUT for all bases a sequence of length phi(m) is guaranteed to be a loop. phi(m) is euler's totient function- in this video phi(100) = 40 and phi(40) = 16 were both facts that made an appearance. why this works and more importantly the properties that let you compute phi are unpleasant to type in the comments section, but probably if you google it you can get a reasonable handle on phi(n) within ten minutes. hard to read, easy to understand; the mark of pure math everywhere and always.
the way this stuff works is closely allied with RSA encryption, which is a standard unit in discrete math (first university proofs course). RSA is genuinely important so this all provides a lot of motivation to include this variety of problem in math competitions
edit! if you're lost at a more basic level, you want to look up "what is modular arithmetic"; this problem is summarized as "calculate the value of [preposterously huge number] mod 100"
My reasoning was simpler (to me):
97^20≡1 mod 100, so the exponent of 1997 can be reduced mod 20. Since 100 is a multiple of 20, and reducing mod 100 is easier, I went with that. No power of 98 is going to be congruent 1 mod 100 (except the 0th power), but after the 1st 2 powers, every 20 powers gets you back to the same result. Since 1999>2, this works. 1999≡19 mod 20. 98^19≡12 mod 100. 97^12≡41 mod 100.
@@akifbaysal9141 No, the standard for power towers is to evaluate the exponents from the top down. It does **not** simplify to multiplying the exponents.
@@jursamaj thx for clarification, I noticed I misunderstood the reply I received. I was looking to delete my previous note postef here but was not able locate it before you posted, I deleted now..
38 and 40 are not coprime, so it's not correct to apply Euler's theorem when reducing 38¹⁹⁹⁹ mod 40 (though it does seem to work in this case). If this reduction were correct in general, then we could equally argue that 38²⁰⁰⁰ = 38⁰ = 1 mod 40, because 2000 = 0 mod 16. But that's clearly wrong because 38²⁰⁰⁰ is even, as is its residue mod 40.
Powers of -2 mod 40:
1, -2, 4, -8, 16, 8, -16, -8, 16, 8, -16, ...
So (-2)¹⁹⁹⁹ = -8 = 32 because 1999 = 3(4) and 1999 > 2.
I *think* the reduction will always work for a^b mod n if the exponent is reduced to no less than phi(n)/gcd(a,n) (because that guarantees we stay within the eventual cyclic portion of a^b mod n) which is why it works here but not for 38²⁰⁰⁰, but I'm not sure that's the right condition.
Is there a general result for when we can "use Euler's theorem" even though its precondition is not satisfied?
The general result is that we can reduce down to the least power r such that a^r shares no prime factors with m that have a lesser exponent in the factorization of a^r than in the factorization of m. So, to get a statement true for all a, choose r at least as large as the largest exponent in the prima factorization of m. Which is 3 in this case.
Yeah for 38^2000 you can then reduce to 38^16, which gives (-2)^16 = 2^16 = 65536 = 16, which is the correct result. (In fact, in the group of multiples of 8 mod 40, 16 is the neutral element, since it reduces to 1 mod 5--and indeed 16*8=128 reduces to 8 mod 40, etc.)
I suggest see my comment for a solution that doesn't make this assumption. This illustrates the general rule stated by @TobyBW
I had a question at 3:40. I thought that since 1999 mod(16) = 15, you could also use -1. But then you're left with 38^-1 mod(40). In other words, you need some number x s.t. 38x = 1 mod(40), but this is impossible since 38 is even so 38x is even. Is the general rule of thumb that you don't want to mess around with negative integers in the exponent and only the base?
The theorem may fail when gcd(a,n) is not 1 (here gcd(38,40)=2). I think he just choose a specific expression that keep it working.
I suggest see my comment for a solution that doesn't make this assumption.
I have an interesting question:
Prove whether or not there exists a constant ‘p’ s.t.
the limit as n tends towards infinity of:
(n!)^(1/n) - n/p converges
Why can we reduce mod 16 in the exponent when gcd(38, 40) = 2 != 1?
He addresses at 5:28 it with a lot of hand waving. I would like to have had a discussion on that.
@@DrR0BERTyou can find some justification on the Wikipedia page for the Carmichael function (though without proof) that modular powers must loop starting at r = the largest exponent in the prime factorization of the modulus. In this case, m = 40, the highest power is 3, so you can reduce powers mod 16 down to 3 but not necessarily lower. This case reduces to 15 so it's ok.
I won't write a full proof, but the reason for this is that once you take the r power, any prime factors that the base and the modulus shared are now have at least as large an exponent as in the modulus. Thus, we can reduce it all mod p_i^r_i. In this case, we could go from looking mod 40 to mod 5, since once the exponent of -2 reaches 3, every subsequent power will be divisible by 8, and we can reduce the whole modular equation down to modulo 5
Yeah the way I did it is with CRT. Since we want to eventually reduce 1998^1999 mod 40, let's separate the prime factors of 40 = 2³ · 5 and look at 1998^1999 mod 8, which is clearly 0 since it contains a factor of 2^1999, and 1998^1999 mod 5, which reduces to 3^3 = 27 reduces again to 2 (1998 = 3 mod 5, phi(5) = 4 and the exponent 1999 = 3 mod 4)
so we have 1998^1999 = 0 mod 8, and 2 mod 5, which must mean it reduces to 32 mod 40 (one can quickly look through the multiples of 8 to see which one reduces to 2 mod 5).
@@TobyBW I understand that. My comment was for those who aren't number theorists. As an educator, I saw this was a missed opportunity.
Happy new abundant odd year
Note that Abundant odd numbers are very rare
Also a square year, and probably the only square year in our lifetime (next one is 2116).
@RecreationallyCynical I know, but I am more interested in abundant numbers
I was hoping the answer would be 42.
And! Before it ends: A Happy Old Year to everyone 🙂
Carmichael function would have worked better here.
Yes. I suggest see my comment.
Basics only: we have 2⁵ = 32 ≡ 2 mod 20, thus if k = 1998¹⁹⁹⁹ then k is even and k ≡ (-1)¹⁹⁹⁹∙2¹⁹⁹⁵∙2⁴ = -1∙2³⁹⁹∙2⁴ = -1∙2⁴⁰⁰∙2³ ≡ -1∙2¹⁶∙2³ = -1∙2¹⁵∙2⁴ ≡ -1∙2³∙2⁴ = -1∙2⁵∙2² ≡ -1∙2∙2² = -8 ≡ 12 mod 20. Next, 1997ᵏ ≡ (-3)ᵏ = 3ᵏ mod 100 (k is even). But then 3ᵏ = (3²⁰)ᵐ∙3ʳ where k = 20k + r, so r is a remainder which we've just found, meaning 3ᵏ = (3²⁰)ᵐ∙3¹² ≡ 1ᵐ∙3¹² mod 100 = 3¹² mod 100 = 41. And this is because 3²⁰ = ((3⁵)²)² = (343²)² ≡ (43²)² = 1849² ≡ 49² = 2401 ≡ 1 mod 100 and similarly, 3¹² = ((3³)²)² = (27²)² = 729² ≡ 29² = 841 ≡ 41 mod 100.
So (10a+b)^big will always end in some combination of 10a b^(n-1) + b^n, since all higher terms become irrelevant. Therefore we're looking at 97^1998^1999. If we work mod 100, we can actually count 97 as -3; since 1998 is even, the power is even and the minus sign vanishes; we're looking at 3^1998^1999. The even powers of 3, mod 100, progress... 9, 81, 29, 67, 3, 27, 43, 87, ... that's a lot, i quit
Humm, i need a bit help here. Why could we reduce 1998^1999 mod 40? To do this, didn't we need gcd(1998,40)=1?
5:29
@@eu7059 Thanks!
I suggest see my comment for a solution that doesn't make this assumption.
I love these. So if this determination can only be made when phi(a,n)=1 then are there towers of powers that are impossible to find the last digits of?
Not impossible, you'll just need an even more general theorem. In fact, the tower in the video is such a harder tower (Michael didn't take this into account, though)
I suggest see my comment for a solution that doesn't make this assumption.
Should we consider the thumbnail expression is for last two decimal digits of umber A = (1997)^(1998^1999).. When different interpretation is made to power of power, the result may come out diferent.
The standard interpretation of the power tower is from the top down, so x^y^z means x^(y^z)
Thx.. Even there may be some accepted convention, I hit different ways of power tower interpretations.
dont understand - but good presentation
There is an error in phi(40), it contains 5.
i think it's fine because he forgot the 9 so it acts as a replacement
phi(40) = phi(8)*phi(5) = (8-4)*(5-1). attempting to list the relatively prime integers was an idiosyncratic approach
phi(n) is multiplicative so phi(n) can be represented as the product of the prime powers in the canonical representation of n.
@@theupson But it's quicker than explaining how to calculate φ(n). Optionally, he could have said that 2 and 5 are the only primes that divide 40, and every other number is even so we can remove 20 numbers from the 1-40 list, so we have 1/2 of 40 = 20 left. Then every fifth number in the 1-40 list is divisible by 5, so we have 4/5 of 20 = 16 left. So φ(40) = 16.
Similarly, φ(100) = 100 x 1/2 x 4/5 = 40. I still think writing out the lists as he did makes it easier to understand for anyone who has little number theory.
Happy New Year ! 2025=(20+25)(20+25)
2025 = (0³ +) 1³ + 2³ + 3³ + 4³ + 5³ + 6³ + 7³ + 8³ + 9³ = ( (0 +) 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 )²
@rainerzufall42 beat me to it. I guess that's the benefit of squaring a triangle number.
@@xinpingdonohoe3978 You know, sum n³ = (sum n)² is always right, but I find it fascinating, that in this case, it's all the digits!
Thank you for giving interesting applications of number theory
I would use patterns of powers, not to use number theory mambojambo.
They are the same thing.
We follow Mr. Michael Penn's videos. I wondered what it would be like if he spoke in Turkish. let's see how he reacts. by the way, he has a lot of followers in Turkey, I wonder if we could somehow publish his translated videos online. would he be interested?
ua-cam.com/video/1UB14Mm5TMM/v-deo.html