Which of Them are Primes? | Putnam 1989 A1

[@littlefermat]

Nice problem! 😉 I have a different approach with no need to discuss 2 cases: We can express the numbers as 1+10^2+10^4+...+10^{2n} Which is (10^{2n+2}-1)/99 That is (10^{n+1}-1)(10^{n+1}+1)/99 And that is prime only when n=1 (101) And so we are done!

[@alainrogez8485]

I think you make a mistake. Why did you say Un isn't prime in the last result?

[@andrea-mj9ce]

Can you justify the very last statement, please ?

[@littlefermat]

The multiplication of two numbers and clearly both of them are greater than 99 for n larger than 1

[@andrea-mj9ce]

@@littlefermat Why does this imply that this number is not prime for n > 1 ?

[@jofx4051]

@@andrea-mj9ce I think I got it since this shows that in case of n>1, the number patterns showing that they are not a factor of itself and 1 Elegant solution by little fermat

[@NeelTigers]

Looks like the only prime is 101 let a(n) denote the 1 and 0 member with n 1’s. It’s easy to see that 101 divides a(2n). Therefore we need to look at a(2n+1) which is 1+10^2+10^4...10^4n=(10^(2n+1)+1)/11 times (10^(2n+1)-1)/9. The first factor is 1-10+10^2-10^3...+10^2n and the second one can be written out as a sequence of 2n+1 1’s and both of these factors are integers

[@zanti4132]

This is just an observation that I think is interesting: the base 10 number 10101... is equal to 111... in base 100, i.e. it is what's known as a repunit number in base 100. So, the argument in this video essentially proves that the only repunit prime in base 100 is 11₁₀₀. But, noting that there is nothing special about the number 10, we can make an even stronger statement: In any base n, where n is a perfect square ≥ 4, the only repunit number that might be prime is 11ₙ.

[@lexyeevee]

favorite part is that you never used any property of 10 here, so this is true in any base! (although 101 itself may still be composite)

[@udaysrivastava1957]

your videos are wonderful, though I'd appreciate if turn up the volume a bit.

[@JM-us3fr]

If n is the number of 1’s, then we have two cases. If n is even, then 101 is a factor, with complementary factor 10001000…10001 (n/2 1’s). If n is odd, then 11…11 (n 1’s) is a factor, with complimentary factor 9090…9091 ((n-1)/2 9’s).

[@simongross3122]

Interesting. If you read this in binary, the first number is 5 and all the other numbers end in 5. Therefore only the first one is prime.

[@lexyeevee]

you can't use divisibility rules across bases like that. 1101[two] ends in 101[two], but it's 13, which is prime.

[@alainrogez8485]

U0 = 1 = 100^0 U1 = 101 = 1+100=100^0+100^1 ... Un=100^0+100^1+...+100^n Check the remainder modulo 101. 100 = 101 - 1 = -1 (mod 101) Un =(-1)^0+(-1)^1+...+(-1)^n-1 + (-1)^n (mod 101) If n is odd : Un = 1-1+1-1+...+1-1 = 0 (mod 101) Meaning Un is a multiple of 101, so it isn't a prime. U1 = 101 is prime (check it isn't divisible by 2, 3, 5 or 7. Meanwhile 11^2=121>101) If n is even : Un is the sum of terms of a geometric serie of which ratio is 100. Un = 1*(100^(n+1)-1)/(100-1) Un = (10^2^(n+1)-1)/99 Un = (10^(n+1)-1)(10^(n+1)+1)/11*9 Check 10^(n+1)-1 is a multiple of 9. 10 = 1 (mod 9) 10^(n+1) = 1 (mod 9) 10^(n+1)-1 = 0 (mod 9) So (10^(n+1)-1)/9 is an integer which is equal to 1 if n=0. Check 10^(n+1)+1 is a multiple of 11. 10 = -1 (mod 11) 10^(n+1) = (-1)^(n+1) = -1 (mod 11) because n is even, so n+1 is odd 10^(n+1)+1 = 0 (mod 11) So 10^(n+1)/11 is an integer which is equal to 1 if n=0. Un is the product of two integers greater than 1 when n is greater than 0. So Un isn't prime. The only Un which is prime is U1= 101.

[@ImaginaryMdA]

It's much easier to see that 10101 is divisible by three.

[@zanti4132]

Easier but irrelevant. You've eliminated one easy case, when the goal is to eliminate every case.

[@beautyofinfinity2024]

🤣🤣🤣

[@leif1075]

Easier than what? How can you prove for the infinity of integers?

[@nicbajito]

@@zanti4132 dont be rude, isnt just one case, it eliminates a third of the posibilities. But i wonder how we eliminate the other 2 posibilities with a simple process like with 3

[@HoSza1]

This is not irrelevant at all! Indeed when the number of 1s are even, that rules out half of the cases and when is a multiple of 3 tjaz eliminates another third. Then the cases remaining are 6k+1 and 6k+5, that is 6k+3+-2. Then the expression (100^(6k+3+-2)-1)/99 can be factored *more easily* than the one shown in the video.

[@kenichimori8533]

Mersenne prime number is binary structure. description 0x2f

[@chanlyelee]

Nice. Thanks for sharing.

[@242math]

very well done, thanks for sharing

[@uwomisakirins3141]

do use mouse yo write or you draw on tablet? and which app do u use?

[@darkomarkovic3373]

Wonderful solution 👍

[@andrea-mj9ce]

6:37 Are they both geometric series?

[@heywhyisthisreallybas6246]

10101 is not a prime number, we can easily determine this by taking the digital root of the number, 1+0+1+0+1 = 3, as we can see it can be divisible by three.

[@WerewolfLord]

For 10101 = 10⁴+10²+1 Consider the polynomial x⁴+x²+1 Maybe the Romans were on to something.

[@sheafsmash142]

Nice

[@nicbajito]

Isnt funny that every even term with odd numbers of 1s is always divisible by the same number without 0s? For example: 10101 is divisible by 111 101010101 is divisible by 11111 and so on

[@beautyofmath6821]

Great!

[@isaacchan8159]

nice

[@samosavaglio2141]

Nice