First year college student here, majoring in CS and I slacked off in math throughout school. This channel is helping me catch up so much its a lifesaver
After multiple professors and RUclips videos, this is the first technique and explanation I have finally been able to understand and use. Thaaaank you!
I am happy to help out the next generation. This trick was casually mentioned in one of my college classes and I remember thinking how I wish I had this back in the day. I wish your the 10 year old the best with this new trick!
Thank you so much sir, this was immensely useful for me. Even as a 12th grader taking Maths, I never actually thought there was a way where you could find all factors of a number, but I had to learn. And no video on RUclips was as clear as yours, not only that, but also, very easy to do. Unlike these complex methods on RUclips I found that I honestly could not understand at all or memorize. But this, this is it! Thank you again. (This would've been so useful to learn when I was younger, too. But at least I know now. )
Hello Vin, Currently I'm studying in class 5th and tomorrow is my Maths test so I searched that easiest way to find all factors of a number and after watching this video, I think there will be no doubt in my mind that how to list factors of a number and learnt few new thing from here that exponents and all that thing. It was really helpful no mater this video is uploaded 7 years ago.
you used to be my old math teacher in CSI back in like 2018-2019 you were mad cool and would always tell us to watch the ads I don't think this video had nowhere near this many views at the time. good luck bro and keep it up
Great to hear from you and I appreciate your kind words, Anthony! I enjoyed that time at CSI, I can't believe how fast the years are going. What career path did you choose?!
@@vinteachesmath i graduated with a soc/ant degree but that never was my passion I'm a sales guy so I have always been interested in shoes and have been reselling full time and enjoy making connections with people
I found this technique in a college combinatorics textbook. It was mentioned somewhat casually but I thought it was very interesting. I remember thinking that this topic should be taught in every elementary school!
The easiest way to _find_ the factors of a larger number, is to complete the prime factorization, then we can simply list all the unique prime factors as factors., then multiply every possible combination of that number. Note that where you have more than one of the same factor, for example, if you have 3 x 3 in your prime factorization, those should be multiplied together, however, if you have 2 x 3 x 3, you don't need to multiply 2 x 3, and the 2 x 3 again. Only unique combinations count. For 90, as an example, the prime factorization as we know is 2 x 3 x 3 x 5 This therefore means that 2, 3 and 5 must be factors on their own. We can also list 1, obviously. Now if we take every possible combination of numbers from there, and multiply them we get: 2 x 3 = 6 2 x 5 = 10 3 x 3 = 9 3 x 5 = 15 2 x 3 x 3 = 18 2 x 3 x 5 = 30 3 x 3 x 5 = 45 2 x 3 x 3 x 5 = 90 Then, if we simply sort these out in order, we get: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90 But what if you have a really large number? Well, it's simple. The less prime factors, the less multiplications you have to do, and vice versa For example, take 30030. I have chosen this number specifically, because its prime factorization is 2x3x5x7x11x13 Now, since there are no repeated factors, this makes it _even_ easier. Now firstly, we can obviously list 1, 2, 3, 5, 7, 11, 13, and then 30030 as factors Now let's multiply every combination: 2 x 3 = 6 2 x 5 = 10 2 x 7 = 14 2 x 11 = 22 2 x 13 = 26 3 x 5 = 15 3 x 7 = 21 3 x 11 = 33 3 x 13 = 39 5 x 7 = 35 5 x 11 = 55 5 x 13 = 65 7 x 11 = 77 7 x 13 = 91 11 x 13 = 143 2 x 3 x 5 = 30 2 x 3 x 7 = 42 2 x 3 x 11 = 66 2 x 3 x 13 = 78 2 x 5 x 7 = 70 2 x 5 x 11 = 110 2 x 5 x 13 = 130 2 x 7 x 11 = 154 2 x 7 x 13 = 182 2 x 11 x 13 = 286 3 x 5 x 7 = 105 3 x 5 x 11 = 165 3 x 5 x 13 = 195 3 x 7 x 11 = 231 3 x 7 x 13 = 273 3 x 11 x 13 = 429 5 x 7 x 11 = 385 5 x 7 x 13 = 455 5 x 11 x 13 = 715 7 x 11 x 13 = 1001 2 x 3 x 5 x 7 = 210 2 x 3 x 5 x 11 = 330 2 x 3 x 5 x 13 = 390 2 x 3 x 7 x 11 = 462 2 x 3 x 7 x 13 = 546 2 x 3 x 11 x 13 = 858 2 x 5 x 7 x 11 = 770 2 x 5 x 7 x 13 = 910 2 x 5 x 11 x 13 = 1430 2 x 7 x 11 x 13 = 2002 3 x 5 x 7 x 11 = 1155 3 x 5 x 7 x 13 = 1365 3 x 5 x 11 x 13 = 2145 3 x 7 x 11 x 13 = 3003 5 x 7 x 11 x 13 = 5005 2 x 3 x 5 x 7 x 11 = 2310 2 x 3 x 5 x 7 x 13 = 2730 2 x 3 x 5 x 11 x 13 = 4290 2 x 3 x 7 x 11 x 13 = 6006 2 x 5 x 7 x 11 x 13 = 10010 3 x 5 x 7 x 11 x 13 = 15015 2 x 3 x 5 x 7 x 11 x 13 = 30030 And so, if we list those in order, we get: 1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 21, 22, 26, 30, 33, 35, 39, 42, 55, 65, 66, 70, 77, 78, 91, 105, 110, 113, 143, 154, 165, 182, 195, 210, 231,273, 286, 330, 385, 390, 429, 455, 462, 546, 715, 770, 858, 910, 1001, 1155, 1365, 1430, 2002, 2145, 2310, 2720, 3003, 4290, 5005, 6006, 10010, 15015, 30030 Of course you can list them as pairs of factors: 1 | 30030 2 | 15015 3 | 10010 5 | 6006 6 | 5005 7 | 4290 10 | 3003 11 | 2720 13 | 2310 14 | 2145 15 | 2002 21 | 1430 22 | 1365 26 | 1155 30 | 1001 33 | 910 35 | 858 39 | 770 42 | 715 55 | 546 65 | 462 66 | 455 70 | 429 77 | 390 78 | 385 91 | 330 105 | 286 110 | 273 113 | 231 143 | 210 154 | 195 162 | 185
I don't know about you but honestly, I don't find it easy at all. Multiplying all possible combinations etc etc like how can I possibly do that, and sometimes there's even a certain order for it, it's just very complicated and it's easy to make mistakes doing it. I like the method shown in this video much more better. But you should do whatever you're comfortable with. What matters is that in the end the result is one, no matter how different a person's methodology might be.
@@_Dreamer_9306 Well, it kind of depends on how many unique prime factors there are. For example, with a number like 50, you already know 1 and 50 are factors. Then when you work out the prime factorization, you get 2 x 5 x 5. So, there's two more factors, 2 and 5. Now, when you multiply all unique combinations of these prime factors, you get 2 x 5 = 10, and 5 x 5 = 25. Those are the only ones you need to do. This gives you the factors of 50: 1, 2, 5, 10, 25 and 50. However, if you have to do something like 510,510 then even though you'll know that 1 and 510,510 are factors, when you work out the prime factorization, you'll end up with 2 x 3 x 5 x 7 x 11 x 13 x 17. While this does give you 7 more factors, i.e. 2, 3, 5, 7, 11, 13 and 17, you'll then need to multiply every unique combination of these 7 factors. In total there are 21 pairs, 35 sets of 3, 35 sets of 4, 21 sets of 5 and 7 sets of 6. This makes a total of 119 more multiplications. Quite a lot more than just 2.
Introductory Combinatorics Book by Richard A. Brualdi, 5th edition. I believe there is a pdf copy of the book somewhere online. This is the book I used when I took combinatorics in college!
Please help! As stated in another comment, this does not work for #36 (or at least I think it doesnt). The trick tells me that there should be 9 factors when I can find 10. Please explain if and where I am going wrong
Joycinha Borges The 9 factors of 36 are 1,2,3,4,6,9,12,18, 36 Six shows up twice put we only count it once when we list all the factors. Factors will repeat with all perfect square numbers.
Glad you think so! I remember learning this counting factors technique in a college math class and thinking this would be great to learn in the earlier grade levels.
LOVE this TRICK!!! I had some GREAT math teachers growing up but can we just clone you across the US bc these math teachers in the NW are dry and boring and they’re trying to reinvent the wheel! Stop making math so difficult for our kids! Just teach to reach! Thank you for this, I’ll pass it along to my boys.
Is there a name for the trick? Is there a sound mathematical explanation on why this works? I’m very interested. If someone could tell me that would be great 🙏
this probably won't get answered, but why 1? why add 1 to the exponent to come to the conclusion with how many factors total there are for the given number? I also got lost at the trick but I understood it through. just, why the 1??
This trick is an application of the multiplication principle from combinatorics. This is the same concept as "Vin has 4 different T-shirts and 3 different pairs of pants. How many outfits can he make? Ans: 4*3 = 12" For this number trick, let's use 12 as an example. 12 = 2^2 * 3^1, so there are (2+1)(1+1) = 6 factors of 12. Any factor is built up of a combination of a 2 or a 3 (similar to a t-shirt and pants from the last example). However, for one of the factors involving 2, we could use 2^0, 2^1 or 2^2; that exponent 0 case adds one extra potential choice for each factor, that is why we add 1. For 3, there is 3^0 or 3^1, so there are two choices for the factor involving 3. The first factor 1, should be thought of as 1 = 2^0 3^0, and 12 can thought of as 12 = 2^2 3^1. This is a tricky concept to type, but I hope this explanation helps!
Mr. D, when I tried this trick for factors of 100, my product of prime numbers was 2 squared and 5 squared and my exponents were 2x2x2x2=16. I thought I would be looking for 16 factors of 100. The factors I came up with were not 16 factors but 9 factors and they are listed as 1, 2, 4, 5, 10, 20, 25, 50, 100. Please tell me where I went wrong. Same thing with factoring for the number 27. My exponents indicate 8 factors for 27. What am I doing wrong? Please help.
Question here - consider '100' as the context number. The prime factorization of 100 is 2^2 x 5^2. Using your method when calculating the number of factors, (2 + 1) * (2 + 1) = 9. Obviously the number of factors must be even, so this can't be right. Do you add one ( 9 + 1 = 10) in these odd-number situations?
Perfect squares will have an odd number of factors because one pair of factors repeats. 100 = 10x10, but when you list all the factors of 100, you only list 10 once.
First year college student here, majoring in CS and I slacked off in math throughout school. This channel is helping me catch up so much its a lifesaver
I am happy to help! I am glad to hear that you are challenging yourself with a major like CS. The struggle is definitely worth it!
Same reason why I’m here rn 😂
I’ve never experienced an original plot in my life
8 years later...
STILL HELPING PEOPLE!
I am happy to help! I appreciate you supporting my old videos!
That's an amazing trick. I'm going to show this to my son, hopefully it will help him. Thanks for posting it.
Good luck to him and thanks for watching!
@@vinteachesmath oo
@@vinteachesmath good luck your channel will improve
After multiple professors and RUclips videos, this is the first technique and explanation I have finally been able to understand and use. Thaaaank you!
This is a game changer for my 10yr old :)
I am happy to help out the next generation. This trick was casually mentioned in one of my college classes and I remember thinking how I wish I had this back in the day.
I wish your the 10 year old the best with this new trick!
Thank you so much sir, this was immensely useful for me. Even as a 12th grader taking Maths, I never actually thought there was a way where you could find all factors of a number, but I had to learn. And no video on RUclips was as clear as yours, not only that, but also, very easy to do. Unlike these complex methods on RUclips I found that I honestly could not understand at all or memorize. But this, this is it! Thank you again. (This would've been so useful to learn when I was younger, too. But at least I know now. )
I am very happy to hear that this was so useful! I learned this trick in my last year of college and would have loved to learn this trick earlier!
Hello Vin,
Currently I'm studying in class 5th and tomorrow is my Maths test so I searched that easiest way to find all factors of a number and after watching this video, I think there will be no doubt in my mind that how to list factors of a number and learnt few new thing from here that exponents and all that thing. It was really helpful no mater this video is uploaded 7 years ago.
Hope the test went well! I appreciate the support, thank you for keeping my old videos relevant!
Best wishes with the rest of your school year.
Thank you so much dude because I take a long time listing factors and miss the sneaky ones so this video will benefit me a lot :)
Glad it helped!
Now this was one amazing and clear video that helped my problems. Thank you for your help.
Glad it helped! Best wishes with the rest of your school year!
This was a lifesaver. The first video I found that gives me a method that makes sense. Well explained. Thank you.
Glad it helped!
So interesting that math works like that sometimes. Best method I've found, thank you!
Oh my gosh this trick is a life saver. I can’t thank you enough
Happy to help! Hope your math class is going well.
This was really helpful. Appreciate your help.
@santassurprise7614 glad it was helpful! I hope your school year is going well
Thx for this trick it helped me win a math competition
Congratulations on winning the math competition! That is a huge accomplishment! I am glad this trick was useful.
I have always struggled with factoring numbers lol, but this trick has really helped!! Thank you so much :)
Happy to help! I love this trick! This should be shown in every elementary school! I wish I knew it sooner!
you used to be my old math teacher in CSI back in like 2018-2019 you were mad cool and would always tell us to watch the ads I don't think this video had nowhere near this many views at the time. good luck bro and keep it up
Great to hear from you and I appreciate your kind words, Anthony! I enjoyed that time at CSI, I can't believe how fast the years are going. What career path did you choose?!
@@vinteachesmath i graduated with a soc/ant degree but that never was my passion I'm a sales guy so I have always been interested in shoes and have been reselling full time and enjoy making connections with people
best math teach i ever seen bro.
I appreciate the vote of confidence!
Thank you so much! still helps till today!
Glad this video is still helpful! Looking back, this is one of my favorite videos.
BRILLIANT tutorial. Thanks so much!
I am happy to help! This is one of my favorite math tricks.
This guy needs an award!
Thanks for the vote of confidence!
@@vinteachesmath my pleasure!
That's fantastic! Thanks a lot for this video!
Glad you liked it!
That is exceptionally cool! I had no idea - thanks for sharing :)
😎
THIS IS SOO HELPFUL AND I HAVE A FINAL MATH EXAM COMING UP!!!
How did the final go?!
@@vinteachesmath GREAT!!
Love your accent! Are you from Jersey or PA? Great video it really helped me understand.
NY! I'm glad you liked this video. I wish I knew this technique back in 5th grade
Thanks for the clear lesson.
Glad it was helpful! Thank you for supporting one of my oldest videos. I hope your school year is going well.
Huge help... who would 'dislike' this!?!?!
Thanks for the upvote! There will always be haters out there...
Thanks mate! That was really helpful.
Glad it helped! I love this trick, it still blows my mind.
This might be late but THANK YOU you’ve made finding factors soooooo much easier
Well done ,keep going.Your video was very helpful .you are best
Glad it helped!
Amazing video, thanks a lot!
Glad you liked it! I am very happy that this video did well! I want everyone to know this awesome trick for counting factors.
Very cool... I did get lost on the explanation of “The Trick”
Love this!! But I tried this method with the number 48 and it said 18 but I only got 10 factors? I don't know if something wrong or I messed up..
48 = 2^4 * 3^1.. So do (4+1)*(1+1).. Then there are 10 factors of 48.
Wonderful explanation, sir. Is there a name for the 'trick' formula, so I can look up how it's derived?
I found this technique in a college combinatorics textbook. It was mentioned somewhat casually but I thought it was very interesting. I remember thinking that this topic should be taught in every elementary school!
Thank you after watching this video I clear prime factors concept.
Glad it helped and thanks for supporting one of my original videos!
The easiest way to _find_ the factors of a larger number, is to complete the prime factorization, then we can simply list all the unique prime factors as factors., then multiply every possible combination of that number. Note that where you have more than one of the same factor, for example, if you have 3 x 3 in your prime factorization, those should be multiplied together, however, if you have 2 x 3 x 3, you don't need to multiply 2 x 3, and the 2 x 3 again. Only unique combinations count.
For 90, as an example, the prime factorization as we know is 2 x 3 x 3 x 5
This therefore means that 2, 3 and 5 must be factors on their own. We can also list 1, obviously.
Now if we take every possible combination of numbers from there, and multiply them we get:
2 x 3 = 6
2 x 5 = 10
3 x 3 = 9
3 x 5 = 15
2 x 3 x 3 = 18
2 x 3 x 5 = 30
3 x 3 x 5 = 45
2 x 3 x 3 x 5 = 90
Then, if we simply sort these out in order, we get: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90
But what if you have a really large number? Well, it's simple. The less prime factors, the less multiplications you have to do, and vice versa
For example, take 30030.
I have chosen this number specifically, because its prime factorization is 2x3x5x7x11x13
Now, since there are no repeated factors, this makes it _even_ easier.
Now firstly, we can obviously list 1, 2, 3, 5, 7, 11, 13, and then 30030 as factors
Now let's multiply every combination:
2 x 3 = 6
2 x 5 = 10
2 x 7 = 14
2 x 11 = 22
2 x 13 = 26
3 x 5 = 15
3 x 7 = 21
3 x 11 = 33
3 x 13 = 39
5 x 7 = 35
5 x 11 = 55
5 x 13 = 65
7 x 11 = 77
7 x 13 = 91
11 x 13 = 143
2 x 3 x 5 = 30
2 x 3 x 7 = 42
2 x 3 x 11 = 66
2 x 3 x 13 = 78
2 x 5 x 7 = 70
2 x 5 x 11 = 110
2 x 5 x 13 = 130
2 x 7 x 11 = 154
2 x 7 x 13 = 182
2 x 11 x 13 = 286
3 x 5 x 7 = 105
3 x 5 x 11 = 165
3 x 5 x 13 = 195
3 x 7 x 11 = 231
3 x 7 x 13 = 273
3 x 11 x 13 = 429
5 x 7 x 11 = 385
5 x 7 x 13 = 455
5 x 11 x 13 = 715
7 x 11 x 13 = 1001
2 x 3 x 5 x 7 = 210
2 x 3 x 5 x 11 = 330
2 x 3 x 5 x 13 = 390
2 x 3 x 7 x 11 = 462
2 x 3 x 7 x 13 = 546
2 x 3 x 11 x 13 = 858
2 x 5 x 7 x 11 = 770
2 x 5 x 7 x 13 = 910
2 x 5 x 11 x 13 = 1430
2 x 7 x 11 x 13 = 2002
3 x 5 x 7 x 11 = 1155
3 x 5 x 7 x 13 = 1365
3 x 5 x 11 x 13 = 2145
3 x 7 x 11 x 13 = 3003
5 x 7 x 11 x 13 = 5005
2 x 3 x 5 x 7 x 11 = 2310
2 x 3 x 5 x 7 x 13 = 2730
2 x 3 x 5 x 11 x 13 = 4290
2 x 3 x 7 x 11 x 13 = 6006
2 x 5 x 7 x 11 x 13 = 10010
3 x 5 x 7 x 11 x 13 = 15015
2 x 3 x 5 x 7 x 11 x 13 = 30030
And so, if we list those in order, we get:
1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 21, 22, 26, 30, 33, 35, 39, 42, 55, 65, 66, 70, 77, 78, 91, 105, 110, 113, 143, 154, 165, 182, 195, 210, 231,273, 286, 330, 385, 390, 429, 455, 462, 546, 715, 770, 858, 910, 1001, 1155, 1365, 1430, 2002, 2145, 2310, 2720, 3003, 4290, 5005, 6006, 10010, 15015, 30030
Of course you can list them as pairs of factors:
1 | 30030
2 | 15015
3 | 10010
5 | 6006
6 | 5005
7 | 4290
10 | 3003
11 | 2720
13 | 2310
14 | 2145
15 | 2002
21 | 1430
22 | 1365
26 | 1155
30 | 1001
33 | 910
35 | 858
39 | 770
42 | 715
55 | 546
65 | 462
66 | 455
70 | 429
77 | 390
78 | 385
91 | 330
105 | 286
110 | 273
113 | 231
143 | 210
154 | 195
162 | 185
I don't know about you but honestly, I don't find it easy at all. Multiplying all possible combinations etc etc like how can I possibly do that, and sometimes there's even a certain order for it, it's just very complicated and it's easy to make mistakes doing it. I like the method shown in this video much more better. But you should do whatever you're comfortable with. What matters is that in the end the result is one, no matter how different a person's methodology might be.
@@_Dreamer_9306
Well, it kind of depends on how many unique prime factors there are.
For example, with a number like 50, you already know 1 and 50 are factors.
Then when you work out the prime factorization, you get 2 x 5 x 5.
So, there's two more factors, 2 and 5.
Now, when you multiply all unique combinations of these prime factors, you get 2 x 5 = 10, and 5 x 5 = 25.
Those are the only ones you need to do.
This gives you the factors of 50: 1, 2, 5, 10, 25 and 50.
However, if you have to do something like 510,510 then even though you'll know that 1 and 510,510 are factors, when you work out the prime factorization, you'll end up with 2 x 3 x 5 x 7 x 11 x 13 x 17.
While this does give you 7 more factors, i.e. 2, 3, 5, 7, 11, 13 and 17, you'll then need to multiply every unique combination of these 7 factors.
In total there are 21 pairs, 35 sets of 3, 35 sets of 4, 21 sets of 5 and 7 sets of 6.
This makes a total of 119 more multiplications. Quite a lot more than just 2.
VERY COOL TRICK SIR! THANKS FOR SHARING THE TRICK. GOD BLESSES YOU ALWAYS FOR HAVING A KIND-HELPFUL-MIND!
I am glad you appreciate the trick! This needs to be in every elementary and middle school!
OMG it actually works,my maths exam is tomorrow and i am always confused in finding the factors of 216,Oh My Gosh!!!!, Thank you so so so much 😊😃
Glad it helped! Thanks for supporting my first generation videos!
Thank you for this amazing technique. It's so enlightening. Can you do some videos on LCMs and GCFs, please.
Hi, I just wanted to know the proof for this trick..... Where can I find it?
Introductory Combinatorics Book by Richard A. Brualdi, 5th edition. I believe there is a pdf copy of the book somewhere online. This is the book I used when I took combinatorics in college!
@@vinteachesmath thank you so much
This Is sooooo Gonna help me in GRE preparation
Its the easiest trick of all, I will use this trick to calculate. THANK YOU VERY MUCH.
Happy to help! I love this trick and I wish I knew it back in grade school!
Nice and clear Good job! 👍
thanks man now im a pro 😎you earned a sub 😎😎😎😎😎😎😎😎
Thanks for the sub and the emojis!
Trick is very good. It gives different approach
I love this trick! Thanks for watching.
Nice and easy method. Thanks
Most welcome!
This trick is so wonderful
Thanks very much for sharing this trick
Glad you think so! I remember learning this in college and thinking it should be taught as early as possible!
Thanks that was so helpful👍
Glad to hear it! Best wishes with the rest of the school year.
Thank you so much for the trick
I'm happy to share this trick! Thanks for watching
wow i've trying to relate it for a long time but with ur help its done so thank u
Glad it helped!
@@vinteachesmath yeah thanks
You saved my day 🙂Thnq....
Happy to help!
Awesome! Helpful for gmat!
Glad it helped!
Thank You Very Much Sir for putting such informative video. May God Bless you so much happiness and knowledge!!!
Nice one i like it so much Thank You!!!!!!!!!
I'm glad you like it!
Please help! As stated in another comment, this does not work for #36 (or at least I think it doesnt). The trick tells me that there should be 9 factors when I can find 10. Please explain if and where I am going wrong
Joycinha Borges
The 9 factors of 36 are 1,2,3,4,6,9,12,18, 36
Six shows up twice put we only count it once when we list all the factors. Factors will repeat with all perfect square numbers.
vinteachesmath thank you. Yes I finally worked it out when I counted the facrors and when I counted 6 twice!
amazing educator much thanks
Glad you think so!
Sir salute you, I have my exams on 12th October, please wish me luck🤞🍀
Good luck! I hope your exams go well!
This is incredible
I love this trick! Thanks for supporting one of my older videos!
That is a VERY cool trick.
Glad you think so! I remember learning this counting factors technique in a college math class and thinking this would be great to learn in the earlier grade levels.
now I am going to do good on my test
THANKS!!!!!!!!!
I am glad you enjoyed the trick!
I don't really get the trick. How did you get 2 times 3 times 2 from those exponents?
He added 1 to each of them
LOVE this TRICK!!! I had some GREAT math teachers growing up but can we just clone you across the US bc these math teachers in the NW are dry and boring and they’re trying to reinvent the wheel! Stop making math so difficult for our kids! Just teach to reach! Thank you for this, I’ll pass it along to my boys.
I appreciate the vote of confidence! I hope the boys find it helpful!
Amazing trick 👍👌
I am glad you enjoyed it! Thanks for watching.
Thank you 😊 bro 👍 it did really help me
Welcome 😊
@@vinteachesmath will you please share us tricks on cube root and square root
Wow I tried it now and I got the answer easier than my brother did thanks
2:02 why should we add the 1, were does it comes from?
Is there a name for the trick? Is there a sound mathematical explanation on why this works? I’m very interested. If someone could tell me that would be great 🙏
It was very helpful
Glad it helped!
Amazing and realy helpfull so much !
Glad you think so!
Sir, this trick can be apply on all numbers ?😕💭💭
Any whole number greater than 1... Basically, any number that has prime factorization.
Extreamly.. Hyperbolically.. Thermodynamically.. Helpful.. 😶😋😁
I dont know english even then i understood trick because of your teaching so thanks
this probably won't get answered, but why 1? why add 1 to the exponent to come to the conclusion with how many factors total there are for the given number? I also got lost at the trick but I understood it through. just, why the 1??
This trick is an application of the multiplication principle from combinatorics. This is the same concept as "Vin has 4 different T-shirts and 3 different pairs of pants. How many outfits can he make? Ans: 4*3 = 12"
For this number trick, let's use 12 as an example. 12 = 2^2 * 3^1, so there are (2+1)(1+1) = 6 factors of 12. Any factor is built up of a combination of a 2 or a 3 (similar to a t-shirt and pants from the last example). However, for one of the factors involving 2, we could use 2^0, 2^1 or 2^2; that exponent 0 case adds one extra potential choice for each factor, that is why we add 1. For 3, there is 3^0 or 3^1, so there are two choices for the factor involving 3.
The first factor 1, should be thought of as 1 = 2^0 3^0, and 12 can thought of as 12 = 2^2 3^1.
This is a tricky concept to type, but I hope this explanation helps!
Wow!!! Awesome
Glad you like it!
Thats a key right there
cool math:)
Glad you think so! I love this trick!
super good method
Hello other struggling math students
The struggle is part of the game! That means you care.
@@vinteachesmath Thank you! This video helped out a lot as well
hello
Very helpful video!!! 😯
wOw dude you r soo talented!!! Super Helping stuff!
Happy to help! Thanks for watching.
So helpful! Thank you so much!!
I am glad the video helped! This trick should be taught in every school. It definitely makes it easier to verify that you have found all the factors.
Super Cool Trick
such type tricks are available in indian channals also available in english
hello sir pls when will be your next class
Thx so much it helped😊
Glad it helped! I love this trick.
Wow. Really helped me with my exam tomorrow 👌 very helpful, thanks!
why didnt you break 45 into 3x15 ? Im new to this after a 12 year break from Maths since class 10
90 = 2x9x5 = 2x3x3x5
90 = 2x3x15 = 2x3x3x5
Your method gets to the same result!
Wow nice idea
Glad you liked it!
It's really helped me very much 😊☺️😊
Now , my doubt is clear about finding factor of any number 🤗🤗
I am glad the video helped! Thanks for watching.
love the cool trick!!!
Mr. D, when I tried this trick for factors of 100, my product of prime numbers was 2 squared and 5 squared and my exponents were 2x2x2x2=16. I thought I would be looking for 16 factors of 100. The factors I came up with were not 16 factors but 9 factors and they are listed as 1, 2, 4, 5, 10, 20, 25, 50, 100. Please tell me where I went wrong. Same thing with factoring for the number 27. My exponents indicate 8 factors for 27. What am I doing wrong? Please help.
Question here - consider '100' as the context number. The prime factorization of 100 is 2^2 x 5^2. Using your method when calculating the number of factors, (2 + 1) * (2 + 1) = 9. Obviously the number of factors must be even, so this can't be right. Do you add one ( 9 + 1 = 10) in these odd-number situations?
Perfect squares will have an odd number of factors because one pair of factors repeats. 100 = 10x10, but when you list all the factors of 100, you only list 10 once.
You are very correct... the number of factors is 8.... so this trick doesn't work for all numbers
2:00 Can someone please explain why this trick works this way? Thanks.
This didn’t work for me with 3235 it says I should have 16 pairs and there’s only 4 pairs?
Great 👍
Glad you think so!
Good work kid
Nice one.
Thank you! Cheers!
Pure gold. Thank you for your work.
I am glad you enjoyed! Thanks for watching.
In my school they call it the factor tree method
thank you so much!
Pretty cool that was awsome
Thanks for the support!