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That particular rule works due to the fact that the number can be divided down into numbers which are divisible by 4.
Mathdoc's rule is that in a 5-digit number (abcde [which is actually 10000a+1000b+100c+10d+e] ),
1000a+100b+10c+d+4e=13V
(13V just represents a number which is divisible by 13.)
So if we have our number 10000a+1000b+100c+10d+e=13K which is supposedly divisible by 13, and the equation 1000a+100b+10c+d+4e=13V should be giving an answer which is divisble by 13 (according to Mathdoc's rules).
Therefore, we must try and make the equations equal the same.
So a start would be to multiply Mathdoc's equation by ten so it is closer to our number:
10000a+1000b+100c+10d+40e=13V
Now we have 40e instead of just e, so we take away 39e from both sides of the equation:
10000a+1000b+100c+10d+e=13V-39e
Guess what?
The left side of the equation is our number and the right side must be divisible by 13.
Since the left side of the equation equals the rightside, the Mathdoc formula works.
What else can I help you with?
Is 223 a prime number show your work?
the square root of 223 is less than 15, so we only need to test its divisibility by 2,3,5,7,11 and 13. If you try dividing by any of these you do not get a whole number, so 223 is a prime.
Divisibility rule for 13?
13 is a prime number - it can only be divided by 1 and itself (for an integer answer).
What is the divisibility rule for 13?
There is no divisibility rule for 13 because it is a prime number. If you are thinking: why is there a divsibility rule for 2 and 3 then. Well, i don't know so go look it up on google.
What is the divisibility rule for 143?
The number must be divisible by 13 and by 11.
What is the divisibility for 91?
All Factors of 91:1, 7, 13, 91
Does 39 divide the product of 3 X 5 X 7 X 11 and if not what is the smallest missing factor that would allow for divisibility of the product?
no it does not and the smallest missing factor that would allow for divisibility is 13
What are the divisibility rules to find at least four factors of the number 312?
A number is a multiple of 312 if it's a multiple of 3, 8 and 13 at the same time
What is 13156 divisble by?
To determine the divisibility of 13156, we can check for divisibility by common factors. It is divisible by 2 since it is an even number. Additionally, it can be divided by 4 (as the last two digits, 56, form a number divisible by 4) and by 13 (since 13156 ÷ 13 = 1012). Other factors can be found through further factorization.
What is the divisibility rule for thirteen?
Subtract nine times the last digit from the rest; repeat if necessary. The result must be divisible by 13.
What is the divisibility rule for 65?
For a number to be divisible by 65 it must be divisible by 5 and by 13.To be divisible by 5 it must end in 5 or 0.Checking for divisibility by 13 is a bit harder:split the number into two parts: X = the number formed by all but the last digit and U = the last digit.test X + 4U for divisibility by 13.if necessary split the answer in two again and test.if you don't know your 13 times table well then keep going until the answer is 13, 26 or 39 (once you get there then X + 4U will simply repeat).For example, to check 1767675 for divisibility by 65.The number ends in 5 so it is divisible by 5 and it remains to be seen if it is divisible by 13.X = 176767 and U = 5 so X + 4U = 176787: a number that is too big so have another go.X = 17678 and U = 7 so X + 4U = 17706: still too big, so another go.X = 1770 and U = 6 so X + 4U = 1794: still too big, so another go.X = 179 and U = 4 so X + 4U = 195: not really too big but you could have another goX = 19 and U = 5 so X + 4U = 39 Bingo!if you repeat,X = 3, U = 9 and X + 4U = 39.So the number is divisible by 13.Divisibility by 5 and 13 implies divisibility by 65.
Why does divisibility rule for 13 work?
Depends which rule you are talking about.You can make a divisibility rule by making a balanced formula or factorising.Say you wanted to make a divisibility test for 13 for a 3-digit number.Let's call that number abc.In actual fact, the number is100a+10b+c=13V(V just represents the idea that 100a+10b+c= a number which is divisble by 13).So if the right side of the equation (13V) is divisble by 13, so is the left side.First we will use factorising to make a rule. We can split the numbers on the left:100a+10b+c=13V will be the same as:91a+9a+13b-3b+c+c=13VWe know that 91a and 13b are divisible by 13, so we can take them out of the equation.9a-3b+c=13VSo to test a 3-digit number we can use this.Take the number 117We can put it through the equation and get:9-3+7=1313 is divisible by 13, so 117 is too.Bigger equations can be made for numbers with more digits.Another method is to create numbers of a/b/c which are divisible by 13 on the other side.A 3-digit number example is:100a+10b+c=130V-91cthis is equal to:100a+10b-90c=130VNow we can divide the whole thing by ten since it won't make a difference:10a+b-9c=13VAnd that's our formula.Let's take a different 3-digit number546.50+4-54=00 counts as being divisible by 13 so it works!
Which of the Divisibility Rules go on 741?
A number is a multiple of 741 if it's a multiple of 3, 13 and 19 at the same time 741 = 3 x 13 x 19
