The power of binomial coefficients

If you are going to use an ad-blocker, considering using brave and tipping me BAT!

Books I like:

Abstract Algebra:

Differential Forms:

Number Theory:

Analysis:

Calculus:

My Filming Equipment:

[wesleydeng]

I liked the thumbnail. 5 consecutive numbers must contain one number divisible by 5, two even numbers with one of which divisible by 4, and two numbers divisible by 3 if the middle number is NOT divisible by 3, otherwise the middle number squared is divisible by 9. So the whole thing is divisible by 5*8*9 = 360.

[alainbarnier]

The trick of the second one is really impressive !

[mrflibble]

Nice application of Binomial Coefficients, and presented Professionally as usual, thanks Michael.

[pdpgrgn]

Since (a choose b) is only defined for a>=b, the solution only proves for n>=4 since we have (n+2 choose 6). Of course, the statement is true for n=1, 2, 3 as well, but shouldn't we check it separately ?

[cbbuntz]

It really is wild what you can do with binomial coefficients.

[professorx]

I always watch your videos even though I started studying medicine and I don't study math anymore.
Great video as always.

[drsonaligupta]

We could've directly checked mod 9 to prove divisibility by 9
And writing it as n*120* (n+2 choose 5) we get it divisible by 120 which together proves divisibility by 360.

[crustyoldfart]

As an alternative method, I recall that there is a theorem to the effect that the product of k consecutive integers is an integer multiple of k!. A simple proof relies on showing the difference between consecutive products produced by starting numbers differing by 1.
In the case of k=3, we have (n+1)*(n+2)*(n+3) - (n)*(n+1)*(n+2) = (n+1)*(n+2)*(n+3-n) = 3*(n+1)*(n+2) and since either n+1 OR n+2 is even the product is divisible by 3*2 = 3!

[tomatrix]

You can actually cut your solution shorter to Q2. We know (n+2)(n+1)(n)(n-1)(n-2)/ 6! can be written as (n+2 choose 5) / 5(6) . The N in total is written as the above times 2n times 360. We want to show that this is an integer. 2n will have no effect as that is always an integer. Also, we must show that dividing by 30 will still give use as integer. Notice for all n greater or equal to 3 we can generate 30 by doing 5.6 within the descending product as we can always make a multiple of 6 and 5 and thus a multiple of 30. This will always be there to cancel the 30 below and thus is always an integer. As for n =1 and n=2, , plugging those values into the descending product will generate a 0 in both cases (i.e. n-2 is 0 when n =2 and n-1 is 0 when n = 1) so that gives use N is 0 for n =1, 2 and 0 is a multiple of 360.

[hotchaddi]

I cant thank you enough for your playlist on number theory, i had little to no source to study but you helped my ace my exam! 💯❤️

[juanignaciodiaz]

For the second problem you could add that n >=4, or show the special cases where n + 2 < 6 for which it holds

[-dimesional_Cube]

I'll introduce some my notation related to base in number theory. Whenever you talking base of a number, like base 2 (binary), you have enough digits in numeric notation as from 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. So that means from base 1 to base 10, we have enough digits from numeric digits so that we don't need to invent some symbols. But then, when we talk about some high base like base 100 or base 256, we've to accepted that there isn't enough symbols to categorized all the digits. So what do we do?

We introduced a notation called:
Base-Divide notation (or you can called it as Base-Power notation for yourself)

The notation looks like this:

(|1|2|3|4|)13 with 13 is subscript decode what the "number" in notation should be in base n

So what's useful?

When we derive a base, we need convert it to decimal for our human purpose reading. So the way we'll convert the Base-Divide Notation to Base 10 is very easy (just sort of general formula to convert base n to base 10)

(|n1|n2|n3|n4|...n(k)|)m
Where k is the "digit" of the notation and m is the base. Note that the value contains in n1, n2, n3, n4 should not be larger or equal to the base that they have set to (kind of 1023, there is no single digit that is larger than the base "10")

General formula to convert the Base-Divide notation to Base 10:

Examples:
(|3|6|7|9|)11 = 3*11^3 + 6*11^2 + 7*11^1 + 9*11^0 = 4805 in base 10

(|1|2|3|4|5|)15
= 1*15^4 + 2*15^3 + 3*15^2 + 4*15^1 + 5*15^0
= 58115

Note to know:
You cannot write the notation like this:
(|12|32|11|5|)19 where there is/are number that's larger than the base

You need to rewrite whenever you write "wrong" Base-Divide Notation (Technically, you can write the wrong notation for yourself if you don't publish it in paper or share this notation to another human than you know)

How to rewrite the notation when you write it wrong:

Example:
(|34|12|55|14|)23
Compute 14/23 = 0*23 + 14
Compute 55/23 = 2*23 + 9 (the 2 here is the result when divide so in next step we add the result to the remainder)
Compute 12/23 = 0*23 +12+2 (it's like carry sums from the long multiply that learn in elementary
school)
Compute 34/23 = 1*23+11+0 (actually, you can +0 if you want becuase +0 means "no carry before". But it's meaningless)

Then the result are:
(|34|12|55|14|)23 -> (|1|11|14|9|14|)23
So we done here... Not yet

The thing that you can do with this notation is plug any kind of formula that you can imagine, quadratic, cubic, sigma, pi, e, definite integral, taking a derivative value... Anything

Take this as example:

You can write like this:
(If a digit equal to its base then we gonna add 1 to the left of its digit and put the 0 at its original spot)
Like this:
(|1|2|3|5|7|)7 -> (|1|2|3|6|0|)7
(Note that we doesn't use the equal sign when rewriting)

(Real numbers plug in base? Use the base convert formula to solve)

[christopherswanson]

micheal, i just wanna say that your videos are edited very subtly, but in ways that make them very fun to watch

[jkid]

Are we not worried about expressions like 3 choose 6 in the last problem for small n? I suppose it works out

[Nys]

You got me at 2n = (n+3) + (n-3). The kind of obvious things that unlock you godmode out of nowhere

[kwichmath]

Wouldn't we have to check the first few, low values of n explicitly?

[SlidellRobotics]

The second one can be easily shown similar to the first, without binomial coefficients. After factoring to the point of 5:00, we have the product of five consecutive integers, with the middle one repeated.
As there are five consecutive integers, at least two are even, and one of those must be a multiple of four, so we know the product is a multiple of eight.
As there are five consecutive integers, exactly one is a multiple of five.
If n is a multiple of three, we have the product being a multiple of nine because n is squared. Otherwise, there is a multiple of three less than n, and a multiple of three greater than n. Either way, we have two threes, or nine.
8 * 5 * 9 = 360 and the same argument gets us there.

[CM_France]

Hi,

I stopped the video at 0.45 and wrote : it is obvious that in the product n(n+1)(n+2) you have almost an even number, in other words a multiple of 2, and almost a multiple of 3.

And then I started the video again and Micheal said exatly what I wrote. What is the note when you find the solution before the teacher?
My bad, I think I don't get 20 because for 3 I said "almost" wheras it is "exactly".

Obviously I understand that the goal of Michael is to illustrate a property of the binomial coefficients.

3:14 : it was only a warm up :)

[mcwulf]

I started splitting into 3!*3!*((n+2)C3)*(nC3). That gives us the 36 on top but it relies on a 5 and a 2 coming from the products. Which are clearly there.

[SoonRaccoon]

For the first problem, rewrite it as (n-1)(n)(n+1). This multiplies out to n^3-n. Now, we need to show that this equals zero mod 6. It can be shown that (n^3 = n) mod 6. Therefore (n^3 - n) mod 6 => (n - n) mod 6 => 0 mod 6.