I thought you were claiming that if if 3 | (a+b+c) then 3 divides the product abc, which is not true (since 3 | 2 + 5 + 8 = 15 but 3 doesn't divide 2*5*8).
Dude I almost didn't open the video at first because i wanted to prove it by myself but then i got lazy. Now by watching the video I see that a,b and c are digits LOL.
All of this can be done much simpler using modulo arithmetic. You can easily test for divisibility by 2, 3, 4, 5, 8, 9, 11, & yes, also 7 & 13 & larger divisors.
@@dibeos Every divisibility test can be boiled down to determining "weights" to multiply each digit by. Let's do example of testing for 3. The weight of the units digit is always 1. The weight of the next digit is the weight of the previous one times 10 mod (whatever the divisor being tested is). For the 3 case, 10's digit is 1 (units digit's weight) times 10 giving 10. Apply mod 3 to 10 gives 1. You repeat this for each digit getting 1 for each. Hence, you simply add the digits up. Now for divisor of 11, unit weight is 1. 10's weight is 10 mod 11 = -1. 100's digit weight is (-1 * 10) mod 11 = 1. So the weights alternate between + & - 1. Is 176 divisible by 11? 1 - 7 + 6 = 0, so yes it is. Not only that, if result is not zero, that is remainder of the division. So 123 mod 11 = 3 - 2 + 1 = 2, which is the remainder. 123/11 = 11 remainder 2. For dividing by 2, weights (LSD to MSD) are 1, 0, 0... For 3 it is 1, 1, 1.... For 4, it is 1, 2, 0, 0 .... For 5, it is 1, 0, 0.... For 7 it is 1, 3, 2, -1, -3, -2.... For 8 it is 1, 2, 4, 0, 0,.... For 9 it is 1, 1, 1,.... for 11 it is 1, -1, 1, ... So contrary to popular opinion, divisibility by 7 & 11 are not hard to calculate! Other rules: is any weight is 0, all those that follow are 0's forever. If some weight is the same as one calculated for lower place, the weights loop in a cycle. If a weight is the negative of previous weight, it loops the negative, then positive, like Mobius strip. 7 is example of this looping with period of 6.
Switching to modular arithmetic and modulo symbol you can infer trivially that since 10 = 1 (mod 3) than a * 10 = a (mod 3), and then a*10^{n} = a (mod 3) for any n >= 0. Suming up different equations with the last expression gives you a desired result.
This exploration really drives home the arbitrariness of our digit system. If you break each place into a multiple of their power of 10, it becomes more obvious why this is true. I still need more exploration to fully wrap my head. But what I’m getting at is something like “all numbers are actually the sum of powers of 10”. Do some algebra with these powers of 10 and you can find divisibility patterns as well.
@@dibeos I’ve been discussing with ChatGPT about how this stems from the idea that all digit place numbers divided by 9 are x mod 9 = 1. 1000 mod 9 = 1 100 mod 9 = 1 10 mod 9 = 1 This is useful… because if this is so, the count of each place is actually meaningful when dividing by 3 or 9. The sum of the digits = the total remainder 1s for each place when dividing by 3 or 9. If this sum is also divisible by 3 or 9, then it also can divided out and the entire number is divisible. But say, if we use base-8, then the number with this feature would be 7. 10 (decimal 8) mod 7 is 1. When dividing by 7 in base-8, you can count the digits in all the places and directly count the remainder of the total number by 1s. If the remainders all add up to 9 (dec 7), you can divide by that 9 one more time and the whole number becomes divisible. In base-10, this is the same idea. 9 and 3 are special because when you divide any power of 10 by 9, you always get a remainder of 1. 10 mod 9 = 1 Also 10 mod 3 = 1 Wandered a bit to reach my conclusion because I wasn’t quite clear in my head. But the bottom line is this works for 3 and 9 because they are special cases where the digit is also the remainder after division by 3 or 9.
@@dibeos Ah. It seems my phone ate my answer. Good, I can write it better this time. What I realized after a lot of thought and discussion with ChatGPT is that 3 and 9 are special because their remainder after dividing any of the places is 1. 1000 mod 9 = 1000 mod 3 = 1 100 mod 9 = 1 10 mod 9 = 1 1 mod 9 = 1 And it matches the place’s digit. 500 mod 9 = 5 When we add up the digits in the number, what we are essentially doing is collecting the remainders of all the places after dividing by 3 or 9. If this remainder is also divisible by 9? It divides and nothing is left. Thus, the whole number is divisible. But see, this depends on the digit system’s base. For base-8, the special number for this becomes 7. 1000 (dec 512) mod 7 = 1 200 (dec 128) mod 7 = 2 So in base-8, adding the digits and finding out if the digits add up to 10 (dec 8) also can indicate divisibility. It is a feature stemming from the remainder being the same number as the place digit.
@@dibeos UA-cam keeps on eating my comments, but here we go again. What I realized is the reason this works is because when dividing by 3 or 9, the remainder of any power of 10 is the place's digit. So consider the number 1511. 1000 mod 9 = 1 500 mod 9 = 5 10 mod 9 = 1 1 mod 9 = 1 (For anyone wondering, mod here means "divide and return the remainder".) This means that ONLY FOR 9 or 3, summing the place digits is the same as adding up all the remainders for all the separate powers of 10. If the summed remainders is ALSO divisible by 9, then the whole number is divisible, as there is no remainder left. So, for example in 1511 above, the remainders of all the powers of 10 is the same as the sum of the digits. 1 + 5 + 1 + 1 = 8 8 is not divisible by 9, so 1511 is not divisible by 9. Because it has 8 remaining. (1512 is divisible by 9, because this adds the needed last 1.) As I said though, this is particular to the base-10 system! If I did base-8, then the special divisor would become 7. 231 base-8 (153 decimal) is separated this way. 200 base-8 (128 decimal) mod 7 = 2 30 (24 decimal) mod 7 = 3 1 (1 decimal) mod 7 = 1 2 + 3 + 1 = 6 6 is not divisible by 7, so the 231 base-8 (153 decimal) is not divisible by 7. But! The pattern says that if we add 1... it should become divisible! 232 base-8 -> 2 + 3 + 2 = 7 Is this divisible by 7? This is 154 decimal. Yep. Interestingly... you can also just play around with the digits here and instantly find many other numbers divisible by 7 in base-8. 232, 133, 331, 430, 34, 25, 16. These are all base-8, mind you. Fascinatingly, you can trick people into thinking you're some math savant for a few minutes with this trick.
For a second I thought it was talking about integers a, b and c...then I realize it's that digit-rule...it wouldn't be true for whole numbers, 3 | (1+1+1) but doesn't divide 1*1*1, etc, lol...
I thought you were claiming that if if 3 | (a+b+c) then 3 divides the product abc, which is not true (since 3 | 2 + 5 + 8 = 15 but 3 doesn't divide 2*5*8).
@@TepsiMorphic yes, I meant a,b and c as digits 😎
@@dibeosYou didnt write the statement correctly then
@@uggupugguyou sure? tell how it should be written
@@DanishHafiz-gt4sq well he is right it shall be given that a,b and c represent place values
@@DanishHafiz-gt4sq the final statement should be
3|(100a+10b+c)
the thumbnail is wrong, it should be 100a+10b+c, not abc.
Or ate least a•b•c (concatenation)
Dude I almost didn't open the video at first because i wanted to prove it by myself but then i got lazy. Now by watching the video I see that a,b and c are digits LOL.
@@DiegoTuzzolo yeah, I meant a, b and c as digits…
You can also give a bar above ABC like
______
abc
Fun fact: This also works for divisibility by 1. Though a simper test in that case is to just check that the last digit is divisible by 1. ;-)
All of this can be done much simpler using modulo arithmetic. You can easily test for divisibility by 2, 3, 4, 5, 8, 9, 11, & yes, also 7 & 13 & larger divisors.
@@bpark10001 interesting! Can you describe an example of one of these cases?
@@dibeos Every divisibility test can be boiled down to determining "weights" to multiply each digit by. Let's do example of testing for 3. The weight of the units digit is always 1. The weight of the next digit is the weight of the previous one times 10 mod (whatever the divisor being tested is). For the 3 case, 10's digit is 1 (units digit's weight) times 10 giving 10. Apply mod 3 to 10 gives 1. You repeat this for each digit getting 1 for each. Hence, you simply add the digits up. Now for divisor of 11, unit weight is 1. 10's weight is 10 mod 11 = -1. 100's digit weight is (-1 * 10) mod 11 = 1. So the weights alternate between + & - 1. Is 176 divisible by 11? 1 - 7 + 6 = 0, so yes it is. Not only that, if result is not zero, that is remainder of the division. So 123 mod 11 = 3 - 2 + 1 = 2, which is the remainder. 123/11 = 11 remainder 2.
For dividing by 2, weights (LSD to MSD) are 1, 0, 0... For 3 it is 1, 1, 1.... For 4, it is 1, 2, 0, 0 .... For 5, it is 1, 0, 0.... For 7 it is 1, 3, 2, -1, -3, -2.... For 8 it is 1, 2, 4, 0, 0,.... For 9 it is 1, 1, 1,.... for 11 it is 1, -1, 1, ... So contrary to popular opinion, divisibility by 7 & 11 are not hard to calculate!
Other rules: is any weight is 0, all those that follow are 0's forever. If some weight is the same as one calculated for lower place, the weights loop in a cycle. If a weight is the negative of previous weight, it loops the negative, then positive, like Mobius strip. 7 is example of this looping with period of 6.
Very interesting to see in this perspective, I remember learning the rule in like 7 grade, but now I see why
@@felipefred1279 yeah, it is really see how these simple arithmetic rules relate to the symmetry of the problem 😎
Switching to modular arithmetic and modulo symbol you can infer trivially that since 10 = 1 (mod 3) than a * 10 = a (mod 3), and then a*10^{n} = a (mod 3) for any n >= 0. Suming up different equations with the last expression gives you a desired result.
@@PunkSage oh yeah, it makes sense! Thanks for sharing it!! 😎
This goes for any factor of 10-1, regardless of the value of 10. I learned this years ago.
This exploration really drives home the arbitrariness of our digit system.
If you break each place into a multiple of their power of 10, it becomes more obvious why this is true.
I still need more exploration to fully wrap my head.
But what I’m getting at is something like “all numbers are actually the sum of powers of 10”.
Do some algebra with these powers of 10 and you can find divisibility patterns as well.
I think that since our system is base-10, some rule based on 9 makes sense.
And thus 3, as the sole factor of 9.
@@intptointp interesting, I see what you’re saying. Can you give me some other examples? 🤔
@@dibeos I’ve been discussing with ChatGPT about how this stems from the idea that all digit place numbers divided by 9 are x mod 9 = 1.
1000 mod 9 = 1
100 mod 9 = 1
10 mod 9 = 1
This is useful… because if this is so, the count of each place is actually meaningful when dividing by 3 or 9.
The sum of the digits = the total remainder 1s for each place when dividing by 3 or 9.
If this sum is also divisible by 3 or 9, then it also can divided out and the entire number is divisible.
But say, if we use base-8, then the number with this feature would be 7.
10 (decimal 8) mod 7 is 1.
When dividing by 7 in base-8, you can count the digits in all the places and directly count the remainder of the total number by 1s.
If the remainders all add up to 9 (dec 7), you can divide by that 9 one more time and the whole number becomes divisible.
In base-10, this is the same idea.
9 and 3 are special because when you divide any power of 10 by 9, you always get a remainder of 1.
10 mod 9 = 1
Also
10 mod 3 = 1
Wandered a bit to reach my conclusion because I wasn’t quite clear in my head.
But the bottom line is this works for 3 and 9 because they are special cases where the digit is also the remainder after division by 3 or 9.
@@dibeos Ah. It seems my phone ate my answer.
Good, I can write it better this time.
What I realized after a lot of thought and discussion with ChatGPT is that 3 and 9 are special because their remainder after dividing any of the places is 1.
1000 mod 9 = 1000 mod 3 = 1
100 mod 9 = 1
10 mod 9 = 1
1 mod 9 = 1
And it matches the place’s digit.
500 mod 9 = 5
When we add up the digits in the number, what we are essentially doing is collecting the remainders of all the places after dividing by 3 or 9.
If this remainder is also divisible by 9? It divides and nothing is left. Thus, the whole number is divisible.
But see, this depends on the digit system’s base.
For base-8, the special number for this becomes 7.
1000 (dec 512) mod 7 = 1
200 (dec 128) mod 7 = 2
So in base-8, adding the digits and finding out if the digits add up to 10 (dec 8) also can indicate divisibility.
It is a feature stemming from the remainder being the same number as the place digit.
@@dibeos
UA-cam keeps on eating my comments, but here we go again.
What I realized is the reason this works is because when dividing by 3 or 9, the remainder of any power of 10 is the place's digit.
So consider the number 1511.
1000 mod 9 = 1
500 mod 9 = 5
10 mod 9 = 1
1 mod 9 = 1
(For anyone wondering, mod here means "divide and return the remainder".)
This means that ONLY FOR 9 or 3, summing the place digits is the same as adding up all the remainders for all the separate powers of 10.
If the summed remainders is ALSO divisible by 9, then the whole number is divisible, as there is no remainder left.
So, for example in 1511 above, the remainders of all the powers of 10 is the same as the sum of the digits.
1 + 5 + 1 + 1 = 8
8 is not divisible by 9, so 1511 is not divisible by 9. Because it has 8 remaining. (1512 is divisible by 9, because this adds the needed last 1.)
As I said though, this is particular to the base-10 system!
If I did base-8, then the special divisor would become 7.
231 base-8 (153 decimal) is separated this way.
200 base-8 (128 decimal) mod 7 = 2
30 (24 decimal) mod 7 = 3
1 (1 decimal) mod 7 = 1
2 + 3 + 1 = 6
6 is not divisible by 7, so the 231 base-8 (153 decimal) is not divisible by 7.
But! The pattern says that if we add 1... it should become divisible!
232 base-8 -> 2 + 3 + 2 = 7
Is this divisible by 7? This is 154 decimal.
Yep.
Interestingly... you can also just play around with the digits here and instantly find many other numbers divisible by 7 in base-8.
232, 133, 331, 430, 34, 25, 16.
These are all base-8, mind you.
Fascinatingly, you can trick people into thinking you're some math savant for a few minutes with this trick.
can you do a video on tropical geometry next!!
@@mohammedfarhaan9410 yeah, sure!! I never heard about it, but we will definitely search and create a very cool explanation of it!!!
awesome video!!!
For a second I thought it was talking about integers a, b and c...then I realize it's that digit-rule...it wouldn't be true for whole numbers, 3 | (1+1+1) but doesn't divide 1*1*1, etc, lol...
Interesting
Thank You 🤔
@@JosephAzar-no3ks Does it make sense? Is it clear enough?
I didn’t watch but I downvoted for that annoying thumbnail