36 factors, again!

Special thanks to my number 1 supporter, *ChiaChia*, and her two star cats, Oreo and Luna.

Find the smallest whole number that has 36 factors,
fun number theory problem,
how to find the number of divisors,
how to find the number of factors,

blackpenredpen,
math for fun,

[moskthinks]

I was actually thinking of it long ago after your special vid. My approach here, without checking all cases, is just like checking only 2 of them.

Step 1: Factorize the number of factors needed without powers in descending order. In this case: 36 = 3×3×2×2.
Step 2: Multiply the last number with the first. 36=6×3×2.
Step 3: Subtract the first number in Step 1 from Step 2's. 6-3=3.
Step 4: Compare 2^(Result of Step 3) with the last prime that would be taken in Step 1 (2^2×3^2×5^1×7^1, so 7 is that last prime). 2^3>7.
Step 5: If the power of 2 is greater, then the number gotten from Step 1 is the smallest. If the power of 2 is less, then the number gotten from Step 2 is the smallest. In our case, the power was greater, so 2^2×3^2×5^1×7^1=1260 is the smallest.

This is because we take the last prime and replace with a power of the smallest prime, 2, that satisfies our factor case. As for why only check these 2 cases, multiplication and powers shift stuff REALLY high. Like 10×2=20, but 2^20 is MUCH larger than 2^10. Just a simple multiplication by 2 can make stuff very big! You can try this with 8 factors to get 24 not 30 as possibly expected. You can also check for powers of 2 AND 3, but that rarely happens, unless you have a number with a lot of consectutive primes with high powers. As for factors, I tested the algorithm and isn't a number with consectutive primes, so the answer would be A quiet large number indeed, so I hope this is the correct answer and algorithm and yeah. This is it.

[Sid-ixqr]

Way to go, Professor 36!


ISN'T IT?

[alwysrite]

nice one. Once again congratulations on 100K

[benjaminbajd]

Congratulations on hitting 100k subscribers! You really are the best teacher there is and your explanations are simply the best! Keep up the good work!

[rafaelcampana]

Congratulations, my friend! Greetings from Spain.

[ffggddss]

My approach:
To get 36 as a prodcut of (α+1)'s, first find all the ways to factor 36 in a product of integers (including the "null" factorization, 36 = 36).
So for this, we factor 36 into primes: 36 = 2²·3², from which we find ways to split it into 1, 2, 3, 4 factors:
(α's)
36 = 36 (35)
36 = 2·18 (1, 17)
36 = 3·12 (2, 11)
36 = 4·9 (3, 8)
36 = 6·6 (5, 5)
36 = 2·2·9 (1, 1, 8)
36 = 2·3·6 (1, 2, 5)
36 = 3·3·4 (2, 2, 3)
36 = 2·2·3·3 (1, 1, 2, 2)

To get the smallest product N = Pª·...·Pª, we assign to each largest remaining factor of 36, the smallest unused prime.
It seems likely that the smallest N will be achieved with the smallest factors of 36, but let's go through the list.

2³⁵ = 34, 359, 738, 368
2¹⁷·3 = 393, 216
2¹¹·3² = 18, 432
2⁸·3³ = 6912
2⁵·3⁵ = 6⁵ = 7776
2⁸·3·5 = 3840
2⁵·3²·5 = 1440
2³·3²·5² = 1800
2²·3²·5·7 = 1260 - - - EDIT: Nobody caught my mistake here; I had exponents (3, 3, 2, 2) instead of the now correct (2, 2, 1, 1)

WINNER: 1260 [Sure enough, the hunch was correct!]

Next: I'll work out the whimsical question I posted to your previous video about factoring 100, 000; namely, what's the smallest number with 100, 000 factors?
Stay tuned . . .

Fred

[VerSalieri]

Congrats on 100k my friend.

About the video, my first thought was take the several variable function f(x1, x2, .., xm)=product of (xk+1), k runs from 1 to m.
Then just find its critical points. Then it hit me, i can probably know the upper bound for m first, it’s 4 because in the worst case scenario 36=2x2x3x3, which means there at most 4 distinct primes of powers 1, 1, 2, and 2.
I am gonna give it a go.

[majkgmajkg]

Congrats for 100k!!!! When we talk about smallest possible numbers with some factors (n) we can talk about something that might be called "anti-prime numbers". Numberphile (it isn't ad!!!) has a great video about these numbers where they explain how you can find such numbers.

Before watching I thought the answer is 7776 (2^5*3^5 as you mentioned). But then I saw correct answer and I realised that my point of view was good but my calculation was wrong. I thought about some factorization of 36 such that the sum of factors is the smallest. And I thought that that will be 12 (6+6 what gives us 7776). First I forgot about 6+3+2, which gives us 11 and the number would be 1440. And then I forgot about 3, 3, 2, 2 which give us 10 as sum. But we can see that the smaller the sum of prime factors is, the smaller number we get. It was pleasure to discover it with your great video! Thank you! <3

[LS-Moto]

Congrats from eastern Belgium my friend :)

[Filip-pdzc]

You are the best bro keep it up to the

[himanshumallick]

Ramanujan wrote about such smallest possible numbers with n factors in his paper "Highly composite numbers".

[yoavshati]

Can you do the integral of ln(x) by integrating e^ln(ln(x))?

[samarthsai]

Going to do aJava a computer code for it.

[Zonnymaka]

Congrats! You're really a marvellous teacher, no wonder ppl subscribe your channel
P.S. Next time i'll donate you a real cake...that's not a "lie" (cit.)

[omrimg]

Hey! What do you think about doing some videos about Euler's totient function and some of its (interesting) properties?
Love the channel.

[Treegrower]

Congrats on 100k BPRP! I've been a long time sub, it's awesome to see your channel blow up like this.
Thanks again for helping me through college Calc and inspiring me to start running.

Next up, 200k subs :)

[arghyadeepchatterjee]

Can you please upload a tutorial for the rational zeros theorem . Or if you have already uploaded one can you please give me the link.

[roddy]

My function f(x) (see previous comment) goes up and down, but in the end it should converge to zero, right?

a|b ⇒ There is an n∈Z with a·n=b (a, b∈Z)
For example:
x|10 ⇒ There is an n∈Z with x·n=10
x|10 gives us the solution x∈{-10, -5, -2, -1, 1, 2, 5, 10}, so for x>0 there are 4 solutions, so 10 has 4 factors.
Now zero:
x|0 ⇒ There is an n∈N with x·n=0
For each x∈Z we can put n=0 and that is true: x·n=0.
Therefore x|0 gives us the solution x∈Z, so x has infinite solutions (even for x>0), so 0 has an infinite amount of factors.

Another approach:
a|b ⇒ b ≡ 0 (mod a)
x|0 ⇒ 0 ≡ 0 (mod x)
For each x∈N the remainder of 0:x is 0.

That should mean: f(∞)=0

Am I right?

[tamircohen]

I designed some Python code to work out the smallest number that has x factors. It is very slow but does the job. It will take ages to test large numbers though.

[roddy]

Let’s say f(x) gives you the smallest natural number to have x factors.
These should be the values of the function for x = 1 to 20:
But let me know if I’m wrong here.

f(1) = 1
f(2) = 2
f(3) = 2² = 4
f(4) = 2·3 = 6
f(5) = 2⁴ = 16
f(6) = 2²·3 = 4·3 = 12
f(7) = 2⁶ = 64
f(8) = 2·3·5 = 30
f(9) = (2·3)² = 6² = 36
f(10) = 2⁴·3 = 16·3 = 48
f(11) = 2¹⁰ = 1, 024
f(12) = 2²·3·5 = 4·3·5 = 60
f(13) = 2¹² = 4, 096
f(14) = 2⁶·3 = 64·3 = 192
f(15) = 2⁴·3² = 16·9 = 144
f(16) = 2·3·5·7 = 210
f(17) = 2¹⁶ = 65, 536
f(18) = (2·3)²·5 = 6²·5 = 36·5 = 180
f(19) = 2¹⁸ = 262, 144
f(20) = 2⁴·3·5 = 16·3·5 = 240

10ⁿ = 2ⁿ·5ⁿ, so it has (n+1)² factors.
Let’s say g(x) = (log(x)+1)²
f(g(10⁰)) = f(1²) = f(1) = 1
f(g(10¹)) = f(2²) = f(4) = 2·3 = 6
f(g(10²)) = f(3²) = f(9) = (2·3)² = 6² = 36
f(g(10³)) = f(4²) = f(16) = 2·3·5·7 = 210
f(g(10⁴)) = f(5²) = f(25) = (2·3)⁴ = 6⁴ = 1, 296
f(g(10⁵)) = f(6²) = f(36) = (2·3)²·5·7 = 6²·35 = 36·35 = 1, 260
f(g(10⁶)) = f(7²) = f(49) = (2·3)⁶ = 6⁶ = 46, 656
f(g(10⁷)) = f(8²) = f(64) = 2·3·5·7·11·13 = 30, 030
f(g(10⁸)) = f(9²) = f(81) = (2·3·5·7)² = 210² = 44, 100
f(g(10⁹)) = f(10²) = f(100) = (2·3)⁴·5·7 = 6⁴·35 = 1, 296·35 = 45, 360

And for 10ⁿ we get that pattern:
f(10⁰) = 1
f(10¹) = 2⁴ · 3 = 16·3 = 48
f(10²) = (2·3)⁴ · 5·7 = 6⁴·35 = 45, 360
f(10³) = (2·3·5)⁴ · 7·11·13 = 30⁴·1, 001 = 810, 810, 000
f(10⁴) = (2·3·5·7)⁴ · 11·13·17·19 = 210⁴·46, 189 = 89, 828, 829, 090, 000
f(10⁵) = (2·3·5·7·11)⁴ · 13·17·19·23·29 = 2, 310⁴·2, 800, 733 = 79, 747, 968, 403, 032, 930, 000
f(10⁶) = (2·3·5·7·11·13)⁴ · 17·19·23·29·31·37 = 30, 030⁴·247, 110, 827 = 200, 961, 610, 708, 938, 459, 249, 870, 000
f(10⁷) = (2·3·5·7·11·13·17)⁴ · 19·23·29·31·37·41·43 = 510, 510⁴·25, 626, 846, 353 ≈
f(10⁸) = (2·3·5·7·11·13·17·19)⁴ · 23·29·31·37·41·43·47·53 = 9, 699, 690⁴·3, 359, 814, 435, 017 ≈
f(10⁹) = (2·3·5·7·11·13·17·19·23)⁴ · 29·31·37·41·43·47·53·59·61 = 223, 092, 870⁴·525, 737, 919, 635, 921 ≈