Alternative rules for 4, 8 and 16: 4: If after division by 2, the number is still even, the number is divisible by 4. 8: If after division by 2 twice, the number is still even, the number is divisible by 8. 16: If after division by 2 thrice, the number is still even, the number is divisible by 16.
32: If after division by 2 four times, the number is still even, the number is divisible by 32. 64: If after division by 2 five times, the number is still even, the number is divisible by 64. 128: If after division by 2 six times, the number is still even, the number is divisible by 128. And it goes on forever.
If this shows an example, in those tables- I'll give you an example, the table of 8 right? Must skip numbers that are multiplied by odds in the table of 4
Here I have alternative rules for 4, 8 and 16: 4: The 10s digit is even and the last digit is 0, 4 or 8, or the 10s digit is even and the last digit is 2 or 6 8: The 100s digit is even and the last 2 digits are 00 or a multiple of 8, or the 100s digit is odd and the last 2 digits are the result of multiplying 4 by an odd number 16: The 1,000s digit is even and the last 3 digits are 000 or a multiple of 16, or the 1,000s digit is odd and the last 2 digits are the result of multiplying 8 by an odd number
256: If after division by 2 seven times, the number is still even, the number is divisible by 256 512: If after division by 2 eight times, the number is still even, the number is divisible by 512 1024: If after division by 2 nine times, the number is still even, the number is divisible by 1024 2048: If after division by 2 ten times, the number is still even, the number is divisible by 2048 4096: If after division by 2 eleven times, the number is still even, the number is divisible by 4096...
Divisiblity of 17- -multiply the last number with 5 -then subtract with remaining number -if the result is multiple of 17 and it may also negative or positive and 0 For ex- 204,4×5=20, 20-20=0.zero 187,7×5=35, 18-35=-17.negative 765,5×5=25, 76-25=51.positive
21: follow rules for 7 and 3 22: follow rules for 2 and 11 24: follow rules for 8 and 3 26: follow rules for 2 and 13 28: follow rules for 4 and 7 30: follow rules for 3 and 10 or 2 and 15 33: follow rules for 3 and 11
Ghxxuxykxykfsts you too baby and early on Friday night out the first time in a s and early on lutdulygdstsyfagksdyzydadysfsyfsoyrayayrayrararysiastwouet jlzfzyofsyftydyfsyxtcggi😍🤔😑😋😚😎😘😚😘😋😎😚😎😋😚😋😃😘
Rather than list out all the numbers one by one, here are generalised rules, the positive integers split into 4 categories: Powers of 2 (2^n): check last n digits Powers of 5 (5^n): same as above Relatively prime with 10 (p): multiply last digit by 10^[-1] mod p (modular multiplicative inverse) and add remaining truncated digits (repeat as necessary) Other composites: apply rules of greatest factors belonging in each of the above categories (if present)
@SHELDON PHAM 1. How to mind makes no sense. 2. If you think someone is salty, make it literal. (jk) 3. It's "Ctrl+C" and "Ctrl+V", not "C+Ctrl" and "V+Ctrl". 4. Eat- NO!
Hey Felicity, l was also about to write a comment on that but I have to agree( and I do ) that this vedio proved to be very very very very very very very helpful for me.
Trouble spelling the word "multiply" for segments covering 13, 17, and 19. And the "add" wording for the 17 segment should say, "subtract." Otherwise, a stunning video we love!
alternate test for 4: 1: remove all the digits except for the last two 2: divide the last digit by 2 3: add it with the first digit 4: if the sum is even, the whole number is divisible by 4
50: Last digits must be 50 or 00. 60: Apply rules for 3 and 20. 101: Multiply the last dight by 10 and subtract it from the remaining digits 103: Multiply the last digit by 31 and add it to the remaining digits. 104: Apply rules for 13 and 8. 105: Apply rules for 21 and 5 106: Apply rules for 53 and 2. 125: Last three digits must be 125, 250, 375, 500, 625, 750, 875 or 000. 151: Multiply the last digit by 15 and subtract it from the remaining digits.
Did this channel just blow up with that paper video out of nowhere? How does that happen? How does one video blow up suddenly after 3 months of being uploaded? (Not commenting on that video because my comment wouldn't be seen)
Easier way to determine whether a number is a multiple of 4. If the tens digit is even, it’s only a multiple of 4 if the ones digit is 0, 4, or 8. If the tens digit is odd, it’s only a multiple of 4 if the ones digit is 2 or 6.
I would adjust the rules for the powers of 2 (only 4, 8, and 16 were shown in this video) so they check that power's place digit for even/odd, then do the check for the next power of 2 down with the lower digits. If odd, add that lower power of 2 to the lower digits before checking. When we reach 4, even should have 0, 4, or 8 in the ones place and odd should have 2 or 6. This avoids any long division and is extendable to all powers of 2. Is 10,848 divisible by 16? 0 is even so, is 848 divisible by 8? 8 is even so, is 48 divisible by 4? 4 is even so, we're left with 8 and yes, 10,848 is divisible by 16. Is 10,928 divisible by 16? 0 is even so, is 928 divisible by 8? 9 is odd so, is 28 + 4 = 32 divisible by 4? 3 is odd so, we're left with 2 and yes, 10,928 is divisible by 16. Is 125,518,848 divisible by 512 (2^9)? 1 is odd so, is 25,518,848 + 256 = 25,519,104 divisible by 256? 2 is even so, is 5,519,104 divisible by 128? 5 is odd so, is 519,104 + 64 = 519,168 divisible by 64? 5 is odd so, is 19,168 + 32 = 19,200 divisible by 32? 1 is odd so, is 9,200 + 16 = 9,216 divisible by 16? 9 is odd so, is 216 + 8 = 232 divisible by 8? 2 is even so, is 32 divisible by 4? 3 is odd so, we're left with 2 and yes, 25,518,848 is divisible by 512!
Rules from 101-200 (not completed) 101: take the last digit, multiply it by 10 and subtract it from the remaining digits 102: apply rules of 51 and 2 103: take the last digit, multiply it by 31 and add it to the remaining digits 104: apply rules of 13 and 8 105: apply rules of 21 and 5 106: apply rules of 53 and 2 107: 108: 109: take the last digit, multiply it by 11 and add it to the remaining digits 110: apply rules of 10 and 11 111: take the last digit, multiply it by 11 and subtract it from the remaining digits 112: 113: 114: 115: 116: 117: 118: 119: 120: apply rules of 10 and 12 121: take the last digit, multiply it by 12 and subtract it from the remaining digits 122: take the last digit, multiply it by 6 and subtract it from the remaining digits 123: 124: 125: last three digits must be a multiple of 125 (125, 250, 375, 500, 625, 750, 875 and 000) 126: 127: 128: last seven digits must be a multiple of 128 129: 130: apply rules of 10 and 13 131: 132: apply rules of 66 and 2 133: 134: 135: 136: 137: 138: apply rules of 23 and 6 139: 140: apply rules of 10 and 14 141: 142: apply rules of 71 and 2 143: 144: apply rules of 48 and 3 145: 146: 147: 148: 149: 150: apply rules of 10 and 15 151: 152: 153: 154: apply rules of 77 and 2 155: 159: 160: apply rules of 10 and 16 161: 162: 163: 164: apply rules of 82 and 2 165: 166: 167: 168: 169: 170: apply rules of 10 and 17 171: 172: apply rules of 86 and 2 173: 174: 175: 176: 177: 178: 179: 180: apply rules of 10 and 18 181: take the last digit, multiply it by 18 then subtract it from the remaining digits 182: 183: 184: 185: apply rules of 37 and 5 186: 187: 188: 189: 190: apply rules of 10 and 19 191: take the last digit, multiply it by 19 and subtract it from the remaining digits 192: 193: 194: 195: 196: apply rules of 98 and 2 197: 198: apply rules of 33 and 6 199: 200: last three digits must be 200, 400, 600, 800 and 000 BONUS: 0.5: last decimal digit must be .0 or .5
104: Apply rules for 13 and 8. 105: Apply rules for 21 and 5 106: Apply rules for 53 and 2. 109: Multiply the last digit by 11 and add it to the remaining digits. 111: Multiply the last digit by 11 and subtract it from the remaining digits. 121: Multiply the last digit by 12 and subtract it from the remaining digits. 122: Multiply the last digit by 6 and subtract it from the remaining digits. 131: Multiply the last digit by 13 and subtract it from the remaining digits. 141: Multiply the last digit by 14 and subtract it from the remaining digits. 151: Multiply the last digit by 15 and subtract it from the remaining digits. 161: Multiply the last digit by 16 and subtract it from the remaining digits. 162: Multiply the last digit by 8 and subtract it from the remaining digits. 171: Multiply the last digit by 17 and subtract it from the remaining digits. 181: Multiply the last digit by 18 and subtract it from the remaining digits. 182: Multiply the last digit by 9 and subtract it from the remaining digits. 191: Multiply the last digit by 19 and subtract it from the remaining digits. By the way, change the rule for 125, it's too vague. Instead, use "Last three digits must be 125, 250, 375, 500, 625, 750, 875 or 000"
10 factorial 1 - 3628800 is a number, so it is divisible by 1 2 - 3628800 362880 - 0 -> 362880 -> 36288 - 0x2 -> 36288 - 0 -> 36288 -> 3628 - 8x2 -> 3628 - 16 -> 3612 -> 361 - 2x2 -> 361 - 4 -> 357 -> 35 - 7x2 -> 35 - 14 -> 21. 21 is divisible by 7 so 3628800 is divisible by 7 8 - 907200 / 2 = 453600, so 3628800 is divisible by 8 9 - going back to 3, shows that the final number is 9, so 3628800 is divisible by 9 10 - 3628800
Tip for divisibility of 11 only can be used in a 3 digit number The middle digit should be the SUM of first and last digit For example 253 2+3=5, 253 is divisibile by 11
21-40 21: If both 3 and 7’s rules apply 22: If both 2 and 11's rules apply 23: Take the last digit off and multiply it by 7, add it to the truncated number and the number if be divisible by 23 the original number will be divisible by 23 24: If both 3 and 8's rules apply 25: If the last 2 digits are ether 00, 25, 50 and 75 26: If both 2 and 13's rules apply 27: LONG DIVISION ALERT! If you can divide the number by 3 3 times (and not a decimal) 28: If both 4 and 7's rules apply 29: Take the last digit off and times it by 3 and add it to the truncated number and if the number is divisible by 29 the original number will be divisible by 29 30: If both 5 and 6's rules apply 31: Take the the last number and times it by 3 and add it on the truncated number and if the number is divisible by 31 the original number will be 31 32: If the last 5 digits are a multiple of 32 33: If both 3 and 11's rules apply 34: If both 2 and 17's rules apply 35: If both 5 and 7's rules apply 36: If both 3 and 12's rules apply 37: Take the first digit and add it to the digit 3 places in, if the number is divisible by 37 the original number will be 37 or if the number is 37 or 74 38: If both 2 and 19's rules apply 39: If both 3 and 13's rules apply 40: If both 5 and 8's rules apply
For dividing 3 digit numbers by 11, subtract the second and last digits together. 132; 3-2=1; two ones make eleven, so 132 is divisible by eleven The more you know
For 7, 11, 13 make uae of the fact that 7*11*13=1001 to never have to deal with numbers above 1000. Example: 2639. Divide into groups of 3 and do the alternating sum, just like for 11. In this case you have to test 2-639=-637. Now you can check for divisibility by 7, 11 and 13 of this number. It's obviously divisible by 7, as 637=91*7 and 91 is divisible by 13, thus your original number is divisible by 7 and 13.
Divisible by 21:Repeat unusual-7 and 3 Divisible by 22:Repeat unusual-11 and 2 Divisible by 23:Every number digits is: 2*(x*10)+3*x,x=random number 24-26 will make it next sunday
I have more: 27 - get last digit, multiply by 8, subtract the remaining, result must equal to a multiple of 27 28 - repeat unusual 4 and 7 30 - repeat unusual 3 and 10 or 2 and 15 33 - repeat unusual 3 and 11 34 - repeat unusual 2 and 17 36 - repeat unusual 4 and 9 37 - get last digit, multiply by 11, subtract the remaining, result must equal to a multiple of 37 38 - repeat unusual 2 and 19 40 - repeat unusual 4 and 10 or 5 and 8 42 - repeat unusual 2 and 21 44 - repeat unusual 4 and 11 45 - repeat unusual 5 and 9 46 - repeat unusual 2 and 23 47 - take last digit, multiply by 14, subtract the remaining, result must equal to a multiple of 47 50 - repeat unusual 2 and 25
In the divisibility rule of 17th, you have mentioned that "Multiply the last digit by 5, Add the remaining digits, the result must be divisible by 7, But in the example you subtracted. Why? 🤔
He doesn’t take the odd and even digits, but the odd and even NUMBERED digits. So, in 23657, 2, 6, and 7 are in the 1st, 3rd, and 5th places (odd) while 3 and 5 are in the 2nd and 4th places (even).
I found one for 4: - Look if the last number is even - If the last number is 4, 8 or 0 and the penultimate number is a even number the number is divisible by 4 - Instead If the last number is a 2 or a 6 and the penultimate number is odd the number is divisible by 4 Examples: 348 -> 34(8) even -> 3(4)8 even = 348 is divisible by 4 176 -> 17(6) even -> 1(7)6 odd = 176 is divisible by 4 414 -> 41(4) even -> 4(1)4 odd = 414 is NOT divisible by 4
1: it’s anything 25: rules with 5x5 30:Last digit is a 0 and the rest is divisible by 3 40:same explanation with 30 50:ends in 00 or 50 60:same thing with 30 and 40 75:use 25 and 3 to help,If both visible then it’s visible by it 100:Two zeros at the end 150:15x anything,but with a 0 200:same applies with 100,two zeros at end but with a twist.Third digit has to be even 300:30 and 10 apply 500:Ends in 500 or 000 1000:Three zeros 10000:Four zeros 100000:Five zeros Million:six zeros And so on…….
21: Rules with 3 and 7 22: Rules with 2 and 11 23: 💀 24: Rules with 2 and 12 25: Ends in 25,50,75, or 00 26: Rules with 2 and 13 27: Rules with 3 and 9 28: Rules with 2 and 14
70: Divisible by 7 and last digit is 0 80: Divisible by 8 and last digit is 0 90: Divisible by 9 and last digit is 0 Heres a bonus: 1250: Divisible by 2 and 625.
In divisibility rule of 17 u have wrote that after multiplying last digit by 5 we have to add the product to the remaining digits, but at the same time u have done subtraction instead of addition ( in example).............. Plzz make sure whether there will be subtraction or addition.....
Hey just check the 17th divisibility....there is a mistake where you said to add the product of last digit and 5 to the remaining digits in the theory....but...you subtracted it from the remaining digits....which one to follow????
another way of saying something is divisible by 8 So every 25 multiples is 200 If the first 3 digits are in one of the 25 multiples + 8x(any number that is dividable by 25) So 272 is 8x(25+9)
2,157,683 Isn’t Just Only Divisible By 11! (2 + 5 + 6 + 3) = 16 - (1 + 7 + 8) = 16 ______________________ 0 (11 Goes Into 2,157,683... 196,153 Times! And 196,153 Is Divisible By 53 Itself, By Going In 3,701 Times Those Are One Of The Bigger Examples Of Using 11’s Divisibility Rule...
Video was suggested by RaGHAFazKa MuHaMMaD
Yenji Jem This is great!
tysm for the vid
Props to that man. This video is great and short enough to understand everything.
RaGHAFazKaMuHaMMaD
Hey idot fool in my coming bad words
Much better than what my school taught.
IKR
Same here Jake😜
Same
Which is ur school
@@nishanthb.l7500 Next to... wait, it's private!
Alternative rules for 4, 8 and 16:
4: If after division by 2, the number is still even, the number is divisible by 4.
8: If after division by 2 twice, the number is still even, the number is divisible by 8.
16: If after division by 2 thrice, the number is still even, the number is divisible by 16.
32: If after division by 2 four times, the number is still even, the number is divisible by 32.
64: If after division by 2 five times, the number is still even, the number is divisible by 64.
128: If after division by 2 six times, the number is still even, the number is divisible by 128.
And it goes on forever.
If this shows an example, in those tables- I'll give you an example, the table of 8 right? Must skip numbers that are multiplied by odds in the table of 4
Here I have alternative rules for 4, 8 and 16:
4: The 10s digit is even and the last digit is 0, 4 or 8, or the 10s digit is even and the last digit is 2 or 6
8: The 100s digit is even and the last 2 digits are 00 or a multiple of 8, or the 100s digit is odd and the last 2 digits are the result of multiplying 4 by an odd number
16: The 1,000s digit is even and the last 3 digits are 000 or a multiple of 16, or the 1,000s digit is odd and the last 2 digits are the result of multiplying 8 by an odd number
256: If after division by 2 seven times, the number is still even, the number is divisible by 256
512: If after division by 2 eight times, the number is still even, the number is divisible by 512
1024: If after division by 2 nine times, the number is still even, the number is divisible by 1024
2048: If after division by 2 ten times, the number is still even, the number is divisible by 2048
4096: If after division by 2 eleven times, the number is still even, the number is divisible by 4096...
Divisiblity of 17-
-multiply the last number with 5
-then subtract with remaining number
-if the result is multiple of 17 and it may also negative or positive and 0
For ex- 204,4×5=20, 20-20=0.zero
187,7×5=35, 18-35=-17.negative
765,5×5=25, 76-25=51.positive
Thanks a lot.
Yes, it is given wrong in the video.
Plzzzzzz can you try it for 289 I cant get the answer
Yes this also makes my mind confused😢
@@bijunaac795 it's divisible by 17
21: follow rules for 7 and 3
22: follow rules for 2 and 11
24: follow rules for 8 and 3
26: follow rules for 2 and 13
28: follow rules for 4 and 7
30: follow rules for 3 and 10 or 2 and 15
33: follow rules for 3 and 11
23: Multiply last digit by 7 then add to remaining digits. Resulting number must be a multiple of 23.
25: Last 2 digits must end in 25, 50, 75, or 00
AAAAAAAAHAISNAJZWKJMKSKSM!SLMX DJDS NS
Ghxxuxykxykfsts you too baby and early on Friday night out the first time in a s and early on lutdulygdstsyfagksdyzydadysfsyfsoyrayayrayrararysiastwouet jlzfzyofsyftydyfsyxtcggi😍🤔😑😋😚😎😘😚😘😋😎😚😎😋😚😋😃😘
@Aiden Miller bro r u working on every primes divisibility rules
@Aiden Miller Could you give an example?
Rather than list out all the numbers one by one, here are generalised rules, the positive integers split into 4 categories:
Powers of 2 (2^n): check last n digits
Powers of 5 (5^n): same as above
Relatively prime with 10 (p): multiply last digit by 10^[-1] mod p (modular multiplicative inverse) and add remaining truncated digits
(repeat as necessary)
Other composites: apply rules of greatest factors belonging in each of the above categories (if present)
Ehh to hard
I didn't understood a thing from this comment i am already trying hard with maths and this bro just .....
this music makes me feel like a spy
Me also
🥸😎
You're spying on the structure of numbers 😂
@@peasant7214ye late
short and crisp. loved it and thank you for making such a wonderful video
.
Nice
Precise. Very helpful. Though there's a mistake in 17 divisibility rule, apart from that it's very time saving. Appreciate your hard work.
the music beat is so good
Why does the 17 divisibility rule stating that it should be added, but in the example it was subtracted? Which should be done? Add or subtract?
Sorry yeah, it was supposed to say subtract
@SHELDON PHAM Says the person who pressed Ctrl+C then Ctrl+V.
@SHELDON PHAM
1. How to mind makes no sense.
2. If you think someone is salty, make it literal. (jk)
3. It's "Ctrl+C" and "Ctrl+V", not "C+Ctrl" and "V+Ctrl".
4. Eat- NO!
Hey Felicity, l was also about to write a comment on that but I have to agree( and I do ) that this vedio proved to be very very very very very very very helpful for me.
@@sumanlata4955 oh, yah it was v helpful haha i was simply, genuinely confused because i wasn't familiar at all with rules above 10 xD
For divisibilty rule of 7, you can also take out the last digit, multiply the remaining by 3 and then add the digit you removed to the result.
LaGG335 SBD3 You mean multiply by five...
No, i dont mean that. It should be this gonna edit it for that:
315 ----> 31x3=93 ----> 93+5=98
98 ----> 9x3=27 ----> 27+8=35 which is divisible by 7
LaGG335 SBD3 9
OK, you meant multiply the first part by three, which would seem to be harder than multiplying the last digit by five.
Amazing video. How fast, effective, and cool edits to make us understand. You really deserve much more for this video. Thank you
Trouble spelling the word "multiply" for segments covering 13, 17, and 19. And the "add" wording for the 17 segment should say, "subtract." Otherwise, a stunning video we love!
For 4, 8 and 16, if ur lazy and don't wanna do long division, just see if can divide by 2 twice, thrice and four times respectively. Works.
0:22
11 is not divisible by 3.
3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36...
Geremy Gonzales the Rarest GoAnimator why did u like your own comment
IT SAY IT IS NOT@@geremygonzalessleepwalkerr2900ARE YOU DUMB
WARNING LONG DIVISION
The long division was found: Divisible by 8.
@@MarylandballProductionscry
my favorite math
@@MarylandballProductions4yo moment
and dor 16
2:22 if you apply that rule for 17, then the result would be 34, and if you apply that rule for 34, then the result would be 17
Thanks for your help wish you all good healthy
alternate test for 4:
1: remove all the digits except for the last two
2: divide the last digit by 2
3: add it with the first digit
4: if the sum is even, the whole number is divisible by 4
50: Last digits must be 50 or 00.
60: Apply rules for 3 and 20.
101: Multiply the last dight by 10 and subtract it from the remaining digits
103: Multiply the last digit by 31 and add it to the remaining digits.
104: Apply rules for 13 and 8.
105: Apply rules for 21 and 5
106: Apply rules for 53 and 2.
125: Last three digits must be 125, 250, 375, 500, 625, 750, 875 or 000.
151: Multiply the last digit by 15 and subtract it from the remaining digits.
105: I admit for 105 I usually combinate the rules of 3, 5 and 7 and I don't use. It's kinda hard not to use the rules of 3 and 7
Every divisibility rule is a combination of other divisibility rules unless you’re testing divisibility for a perfect power of a prime number.
I just wanted to know that in the divisibility test of 20 it must end with 20,40,60,80 and 00 or it will end with all multiples of 20?
Did this channel just blow up with that paper video out of nowhere? How does that happen? How does one video blow up suddenly after 3 months of being uploaded? (Not commenting on that video because my comment wouldn't be seen)
UA-cam recommendation algorythm is a hell of a strange place...
It would be easier to divide any number by 13 by following the mentioned steps until a three digit number or two digit number comes 🎉🎉❤
Easier way to determine whether a number is a multiple of 4. If the tens digit is even, it’s only a multiple of 4 if the ones digit is 0, 4, or 8. If the tens digit is odd, it’s only a multiple of 4 if the ones digit is 2 or 6.
Thanks bro , it helped me in my project!!
I must say I'm somehow impressed with 7's rules (including some other primes). Honestly.
perfect video..Supab explain🙏
1:08 But what if the number itself is only 3 digits?
divisible by 8 if the number is divisible by 8
I would adjust the rules for the powers of 2 (only 4, 8, and 16 were shown in this video) so they check that power's place digit for even/odd, then do the check for the next power of 2 down with the lower digits. If odd, add that lower power of 2 to the lower digits before checking. When we reach 4, even should have 0, 4, or 8 in the ones place and odd should have 2 or 6. This avoids any long division and is extendable to all powers of 2.
Is 10,848 divisible by 16? 0 is even so, is 848 divisible by 8? 8 is even so, is 48 divisible by 4? 4 is even so, we're left with 8 and yes, 10,848 is divisible by 16.
Is 10,928 divisible by 16? 0 is even so, is 928 divisible by 8? 9 is odd so, is 28 + 4 = 32 divisible by 4? 3 is odd so, we're left with 2 and yes, 10,928 is divisible by 16.
Is 125,518,848 divisible by 512 (2^9)? 1 is odd so, is 25,518,848 + 256 = 25,519,104 divisible by 256?
2 is even so, is 5,519,104 divisible by 128?
5 is odd so, is 519,104 + 64 = 519,168 divisible by 64?
5 is odd so, is 19,168 + 32 = 19,200 divisible by 32?
1 is odd so, is 9,200 + 16 = 9,216 divisible by 16?
9 is odd so, is 216 + 8 = 232 divisible by 8?
2 is even so, is 32 divisible by 4?
3 is odd so, we're left with 2 and yes, 25,518,848 is divisible by 512!
What is the font used?
Awesome video... Smart, clear & time saving... Except for the mistake of divisibility rule 17.. ( it must be subtracted)
Better way to divide by 8
Divide by 2
Again
Then again
If u can successfully do that without decimals, it’s divisible by 8
(no shit really)
Rules from 101-200 (not completed)
101: take the last digit, multiply it by 10 and subtract it from the remaining digits
102: apply rules of 51 and 2
103: take the last digit, multiply it by 31 and add it to the remaining digits
104: apply rules of 13 and 8
105: apply rules of 21 and 5
106: apply rules of 53 and 2
107:
108:
109: take the last digit, multiply it by 11 and add it to the remaining digits
110: apply rules of 10 and 11
111: take the last digit, multiply it by 11 and subtract it from the remaining digits
112:
113:
114:
115:
116:
117:
118:
119:
120: apply rules of 10 and 12
121: take the last digit, multiply it by 12 and subtract it from the remaining digits
122: take the last digit, multiply it by 6 and subtract it from the remaining digits
123:
124:
125: last three digits must be a multiple of 125 (125, 250, 375, 500, 625, 750, 875 and 000)
126:
127:
128: last seven digits must be a multiple of 128
129:
130: apply rules of 10 and 13
131:
132: apply rules of 66 and 2
133:
134:
135:
136:
137:
138: apply rules of 23 and 6
139:
140: apply rules of 10 and 14
141:
142: apply rules of 71 and 2
143:
144: apply rules of 48 and 3
145:
146:
147:
148:
149:
150: apply rules of 10 and 15
151:
152:
153:
154: apply rules of 77 and 2
155:
159:
160: apply rules of 10 and 16
161:
162:
163:
164: apply rules of 82 and 2
165:
166:
167:
168:
169:
170: apply rules of 10 and 17
171:
172: apply rules of 86 and 2
173:
174:
175:
176:
177:
178:
179:
180: apply rules of 10 and 18
181: take the last digit, multiply it by 18 then subtract it from the remaining digits
182:
183:
184:
185: apply rules of 37 and 5
186:
187:
188:
189:
190: apply rules of 10 and 19
191: take the last digit, multiply it by 19 and subtract it from the remaining digits
192:
193:
194:
195:
196: apply rules of 98 and 2
197:
198: apply rules of 33 and 6
199:
200: last three digits must be 200, 400, 600, 800 and 000
BONUS: 0.5: last decimal digit must be .0 or .5
Please help me find the other Divisibility Rules.
104: Apply rules for 13 and 8.
105: Apply rules for 21 and 5
106: Apply rules for 53 and 2.
109: Multiply the last digit by 11 and add it to the remaining digits.
111: Multiply the last digit by 11 and subtract it from the remaining digits.
121: Multiply the last digit by 12 and subtract it from the remaining digits.
122: Multiply the last digit by 6 and subtract it from the remaining digits.
131: Multiply the last digit by 13 and subtract it from the remaining digits.
141: Multiply the last digit by 14 and subtract it from the remaining digits.
151: Multiply the last digit by 15 and subtract it from the remaining digits.
161: Multiply the last digit by 16 and subtract it from the remaining digits.
162: Multiply the last digit by 8 and subtract it from the remaining digits.
171: Multiply the last digit by 17 and subtract it from the remaining digits.
181: Multiply the last digit by 18 and subtract it from the remaining digits.
182: Multiply the last digit by 9 and subtract it from the remaining digits.
191: Multiply the last digit by 19 and subtract it from the remaining digits.
By the way, change the rule for 125, it's too vague. Instead, use "Last three digits must be 125, 250, 375, 500, 625, 750, 875 or 000"
@@emmazepeda7383 Use those I said
@@Henrique.tRo_Real ok
108: Apply rules for 2 and 54
10 factorial
1 - 3628800 is a number, so it is divisible by 1
2 - 3628800 362880 - 0 -> 362880 -> 36288 - 0x2 -> 36288 - 0 -> 36288 -> 3628 - 8x2 -> 3628 - 16 -> 3612 -> 361 - 2x2 -> 361 - 4 -> 357 -> 35 - 7x2 -> 35 - 14 -> 21. 21 is divisible by 7 so 3628800 is divisible by 7
8 - 907200 / 2 = 453600, so 3628800 is divisible by 8
9 - going back to 3, shows that the final number is 9, so 3628800 is divisible by 9
10 - 3628800
Thanks it was really good 🙂
Sir/mam,In divisibility rule of 17,you said to add but in example you subtracted.
Subtracting is correct.
Thanks for this :)
this is my dad's acc btw hehe..
This really helped me :)
I'm on Grade 5 now
stay safe everyone😊
I am in Grade 5 too
Who else expected divisibility of 20 to be like " apply 4 &5 "
Well, that could be,
1240 yes 4/5/
1235 no 4X5/
1232 no 4/5X
0:46 or you can multiply 1's digit by 5 then add that to the remaining example: 123 is 12+(3x5)
or just do the same as 17
2:22 - You said "Add" but I think meant "Subtract".. 7x5=35 MINUS 1 = 34, / 17 = 2.
Yes
For 4, you can divide by 2 twice, 8 divide by 2 thrice, and 16 divide by 2 4 times
Tip for divisibility of 11 only can be used in a 3 digit number
The middle digit should be the SUM of first and last digit
For example 253 2+3=5, 253 is divisibile by 11
Thank you for consuming our Time:)
This makes it so much better I had no idea before... Thanks!!!!!😀
Nice video today I learnt all 1 to 10 divisibility rules
This video is very short...
But with required knowledge
You told divisibility of 17 second step add to remaining you subtract bro ??
Ha yah
The video is very fast i cant see properly
21-40
21: If both 3 and 7’s rules apply
22: If both 2 and 11's rules apply
23: Take the last digit off and multiply it by 7, add it to the truncated number and the number if be divisible by 23 the original number will be divisible by 23
24: If both 3 and 8's rules apply
25: If the last 2 digits are ether 00, 25, 50 and 75
26: If both 2 and 13's rules apply
27: LONG DIVISION ALERT! If you can divide the number by 3 3 times (and not a decimal)
28: If both 4 and 7's rules apply
29: Take the last digit off and times it by 3 and add it to the truncated number and if the number is divisible by 29 the original number will be divisible by 29
30: If both 5 and 6's rules apply
31: Take the the last number and times it by 3 and add it on the truncated number and if the number is divisible by 31 the original number will be 31
32: If the last 5 digits are a multiple of 32
33: If both 3 and 11's rules apply
34: If both 2 and 17's rules apply
35: If both 5 and 7's rules apply
36: If both 3 and 12's rules apply
37: Take the first digit and add it to the digit 3 places in, if the number is divisible by 37 the original number will be 37 or if the number is 37 or 74
38: If both 2 and 19's rules apply
39: If both 3 and 13's rules apply
40: If both 5 and 8's rules apply
😢💔
which ediiting platform do you use
Nice video, very tricky
Very nice video
But why u did not give examples for 14 15 16
It would be very helpful
Mst h yaar
Rule number 17 isn't apply for 51
1*5=5
5-5=0
51 is divisible by 17
1:12 Or,
If the last 2 digits are divisible by 8,
And the hundreds place is even.
(Because 100 is not, but 200 is, 300 is not, 400 is, etc.)
For dividing 3 digit numbers by 11, subtract the second and last digits together.
132; 3-2=1; two ones make eleven, so 132 is divisible by eleven
The more you know
Thanks man! it was helpful
and your video making skills are really appreciating
For 7, 11, 13 make uae of the fact that 7*11*13=1001 to never have to deal with numbers above 1000. Example: 2639. Divide into groups of 3 and do the alternating sum, just like for 11. In this case you have to test 2-639=-637. Now you can check for divisibility by 7, 11 and 13 of this number. It's obviously divisible by 7, as 637=91*7 and 91 is divisible by 13, thus your original number is divisible by 7 and 13.
Divisible by 21:Repeat unusual-7 and 3
Divisible by 22:Repeat unusual-11 and 2
Divisible by 23:Every number digits is:
2*(x*10)+3*x,x=random number
24-26 will make it next sunday
They never made it...
OK.
I can make 24 - 26 for you.
24 - repeat unusual 3 and 8
25 - last 2 numbers are divisible by 25
26 - repeat unusual 2 and 13
I have more:
27 - get last digit, multiply by 8, subtract the remaining, result must equal to a multiple of 27
28 - repeat unusual 4 and 7
30 - repeat unusual 3 and 10 or 2 and 15
33 - repeat unusual 3 and 11
34 - repeat unusual 2 and 17
36 - repeat unusual 4 and 9
37 - get last digit, multiply by 11, subtract the remaining, result must equal to a multiple of 37
38 - repeat unusual 2 and 19
40 - repeat unusual 4 and 10 or 5 and 8
42 - repeat unusual 2 and 21
44 - repeat unusual 4 and 11
45 - repeat unusual 5 and 9
46 - repeat unusual 2 and 23
47 - take last digit, multiply by 14, subtract the remaining, result must equal to a multiple of 47
50 - repeat unusual 2 and 25
@@ashtonplays2613 *u n u s u a l*
Nice tricks
2:24 you said you add
2:29 you subtracted
We have to subtract
No king (opposite of yes queen)
@@ROMAN.OTC-1 2:24 it says ADD. What does ADD spell?
@@ROMAN.OTC-1 I thought you were All In One. Why do you appear to be Roshan Empire?
U are better than my teacher
How do I use the rules for 49 being divisible by 7? 49-18 =31...that is not a multiple of 7...
9×2=18
4-18=-14 which is divisible by 7
You subtract from the remaining digits, not the whole number.
I hope you understand
It is amazing It help me a lot
I scored 70/70in my test
Thanks for making this video
But there is 17 divisibility is wrong
So please correct it
Congratulations! Great Effort bro!
Can you do Divisiblity Rules 21 to 40?
bro this video helped me this is better than my techer explaining i realize the divisibility rules are super easy thanks for your help! 😄
In the divisibility rule of 17th, you have mentioned that "Multiply the last digit by 5, Add the remaining digits, the result must be divisible by 7, But in the example you subtracted. Why? 🤔
51.
Must have been a typing error
Good
Brother in 11 divisibility rule how you take 8 as a odd number and 3 as a even number??
He doesn’t take the odd and even digits, but the odd and even NUMBERED digits. So, in 23657, 2, 6, and 7 are in the 1st, 3rd, and 5th places (odd) while 3 and 5 are in the 2nd and 4th places (even).
*digits in odd and even NUMBERED PLACES
I found one for 4:
- Look if the last number is even
- If the last number is 4, 8 or 0 and the penultimate number is a even number the number is divisible by 4
- Instead If the last number is a 2 or a 6 and the penultimate number is odd the number is divisible by 4
Examples:
348 -> 34(8) even -> 3(4)8 even = 348 is divisible by 4
176 -> 17(6) even -> 1(7)6 odd = 176 is divisible by 4
414 -> 41(4) even -> 4(1)4 odd = 414 is NOT divisible by 4
1: it’s anything
25: rules with 5x5
30:Last digit is a 0 and the rest is divisible by 3
40:same explanation with 30
50:ends in 00 or 50
60:same thing with 30 and 40
75:use 25 and 3 to help,If both visible then it’s visible by it
100:Two zeros at the end
150:15x anything,but with a 0
200:same applies with 100,two zeros at end but with a twist.Third digit has to be even
300:30 and 10 apply
500:Ends in 500 or 000
1000:Three zeros
10000:Four zeros
100000:Five zeros
Million:six zeros
And so on…….
21: Rules with 3 and 7
22: Rules with 2 and 11
23: 💀
24: Rules with 2 and 12
25: Ends in 25,50,75, or 00
26: Rules with 2 and 13
27: Rules with 3 and 9
28: Rules with 2 and 14
70: Divisible by 7 and last digit is 0
80: Divisible by 8 and last digit is 0
90: Divisible by 9 and last digit is 0
Heres a bonus:
1250: Divisible by 2 and 625.
🥰🥰🥰🥰😎😎fantastic, fabulous, amazing............👏👏👏👏
can you do 20 - 40
You don't need to do long division for rule 8. Just do 4.
@Kevin YU no
The guide in my classroom only went up to 10 and didn't show 7 but I bet no one asked for this comment
Nice teaching and nice examples
The video is nice but the divsidibilty of seven should be subtracted
If is odd a remainder of 1 and is a rational by 2
Kid's after learning decimals:- All of that was a lie!?
0:05 *Every whole number is divisible by 1.
Great video, very useful
The test of divisibility of 17 has been subtracted. It must be added according to test of divisibility. Please recover your mistake.
In divisibility rule of 17 u have wrote that after multiplying last digit by 5 we have to add the product to the remaining digits, but at the same time u have done subtraction instead of addition ( in example).............. Plzz make sure whether there will be subtraction or addition.....
An integer x is divisible by another integer y if x is divisible by the other factors of y.
12,580 is not divide by 10.
10, 20. Not number 12,580!
It's divide by 0,01.
(12 580 divide by 10)
(realizes the comma is a thousands mark) Still Divisible by 10.
Nice video mad knowledge too.💗💗💗💗💗
When your favorite UA-cam becomes your favorite teacher
Hey can you show divisibility of 14398 by 11
2:23
You mean subtract from remaining digits, right?
Ummm the divisibility rule of 17 on 2:31 is subtract but on the steps is to add to the remaining digits
Hey just check the 17th divisibility....there is a mistake where you said to add the product of last digit and 5 to the remaining digits in the theory....but...you subtracted it from the remaining digits....which one to follow????
At 2:22 you said 'add to the remaining digits' but then u subtracted
This is a perfect video, specially when you don't have much time to study it
This was really helpful thanq
another way of saying something is divisible by 8
So every 25 multiples is 200
If the first 3 digits are in one of the 25 multiples + 8x(any number that is dividable by 25)
So 272 is 8x(25+9)
Since 11,060.5 is a multiple of 22 but the closest INTEGER MULTIPLE are 11,044 and 11,066 and thus, 11,060.5 isn't a integer multiple of 22.
2,157,683 Isn’t Just Only Divisible By 11!
(2 + 5 + 6 + 3) = 16
- (1 + 7 + 8) = 16
______________________
0 (11 Goes Into 2,157,683... 196,153 Times! And 196,153 Is Divisible By 53 Itself, By Going In 3,701 Times
Those Are One Of The Bigger Examples Of Using 11’s Divisibility Rule...
Umm
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