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Chapter 1
when you are trying to simplify fractions, chances are you are having a hard time, you don't know what to factorize. Here are some of the techniques to know if a number is divisible by a simple factor, say from 2 to 9.
Note that we are going to deal only with base-10.
These tests can be divided into 3 types : Dividing Carries, Digit Sums & Complex Equations
Type 1: Dividing Carries
the main idea is that some factors can divide some power of ten perfectly, so we don't need to worry about how many that power of ten we have, instead, can we divide what is not that power of ten ?
these factor numbers are:
2,4,5 and 8
[dividing by 2]
since 10/2 = 5, we know that, no matter how many 10s we have, they always can be divided by 2, but it is unsure for the ones.
Divisible test:
if the ones can be divided by 2 (a.k.a. if the ones is 0,2,4,6 or 8),
then the original number can be divided by 2.
Example:
1356 --(look at ones)--> 6 (can be divided by 2) ----> divisible by 2
1357 --(look at ones)--> 7 (can't be divided by 2) ----> not divisible by 2
[dividing by 4]
since 100/4 = 25, we know that, no matter how many 100s we have, they always can be divided by 4, but it is unsure for the tens and ones
Divisible test:
if the tens and ones can be divided by 4,
then the original number can be divided by 4.
Example:
1316 --(look at tens and ones)--> 16 (can be divided by 4) ----> divisible by 4
1357 --(look at tens and ones)--> 57 (can't be divided by 4) ----> not divisible by 4
[dividing by 5] // the simplest factor
since 10/5 = 2, we know that, no matter how many 10s we have, they always can be divided by 5,
but it is unsure for the ones
Divisible test:
if the ones can be divided by 5 (a.k.a. if the ones is 0 or 5)
then the original number can be divided by 5.
Example:
1355 --(look at ones)--> 5 (can be divided by 5) ----> divisible by 5
1370 --(look at ones)--> 0 (can be divided by 5) ----> divisible by 5
2689 --(look at ones)--> 9 (can't be divided by 5) ----> not divisible by 5
[dividing by 8]
since 1000/8 = 125, we know that, no matter how many 1000s we have, they always can be divided by 8, but it is unsure for the hundreds, tens and ones
Divisible test:
if the hundred, tens and ones can be divided by 8,
then the original number can be divided by 8.
Example:
24000 --(look at hundred, tens and ones)--> 000 (can be divided by 8) ----> divisible by 8
24832 --(look at hundred, tens and ones)--> 832 (can be divided by 8) ----> divisible by 8
24257 --(look at hundred, tens and ones)--> 257 (can't be divided by 8) ----> not divisible by 8
Type 2: Digit sums
the main idea is assuming a certain criterion then we can work out something from that.
these factors numbers are : 3 and 9
[dividing by 3]
say we call each digit d0 (ones),d1 (tens),d2 (hundreds),d3 (thousands), ... dn
example : for the number 4869, d3=4, d2=8, d2=6, d0=9
let's set the criteria: sum of the digits is a multiple of 3 (divisible by 3)
then
formular1: d0 + d1 + d2 + d3 + d4 + ... = 3n
we also know that
9d1 + 99d2 + 999d3 + 9999d4 + ... is a multiple of 3, say 3m
then
formular2: 9d1 + 99d2 + 999d3 + 9999d4 + ... = 3m
when we add both formulas together, we would have
d0 + 10d1 + 100d2 + 1000d3 + 10000d4 + ... = 3n + 3m = 3 (n + m) (tada, a multiple of 3),
where d0 + 10d1 + 100d2 + 1000d3 + 10000d4 + ... is the original number we are testing
Divisible test:
if the sum of the digits is a multiple of 3, then the original number can be divided by3
Example:
1368 ----> 1+3+6+8 ----> 18 (divisible by 3) ----> divisible by 3
[dividing by 9]
the proof is similar, but instead of setting the sum of the digits is a multiple of 3, this time the sum should be a multiple of 9, i.e.
formular1: d0 + d1 + d2 + d3 + d4 + ... = 9n
and for sure we know that
formular2: 9d1 + 99d2 + 999d3 + 9999d4 + ... = 9m
add both equations together we have
d0 + 10d1 + 100d2 + 1000d3 + 10000d4 + ... = 9n + 9m = 9 (n + m) (tada, a multiple of 9),
where d0 + 10d1 + 100d2 + 1000d3 + 10000d4 + ... is the original number we are testing
Divisible test:
if the sum of the digits is a multiple of 9, then the original number can be divided by 9
Example:
1368 ----> 1+3+6+8 ----> 18 (divisible by 9) ----> divisible by 9
Type 3: Complex Equations
These factors are combinations of previous factors OR individual factors which is totally different from the others, these factors are 6 and 7
[dividing by 6]
since dividing by 6 can be treated as dividing by 2 then by 3
so it is actually the combination of 2 and 3
Divisible test:
if the ones can be divided by 2 (a.k.a. if the ones is 0,2,4,6 or 8) AND
if the sum of the digits is a multiple of 3,
then the original number can be divided by 6.
Example:
1368 --(look at ones)--> 8 (divisible by 2) ----> divisible by 2
1368 ----> 1+3+6+8 ----> 18 (divisible by 3) ----> divisible by 3
and finally, weirdest of all, the number 7
[dividing by 7]
first let's say the number of tens we have is d1, and the digit of ones is d0, and the original number is K.
example : if K = 8641969, then d0 = 9, and d1 = 864196
let's set a criterion : K is a multiple of 7
so we know that
K = 10d1 + d0 = 7n, where n is an interger
10d1 + d0 - 21d0 = 7n - 21d0
10d1 - 20d0 = 7n - 21d0
10(d1 - 2d0) = 7(n - 3d0)
since 10(d1 - 2d0) is a multiple of 7, then 10(d1 - 2d0) must somehow contain the factor 7, so d1-2d0 must contain the factor 7, i,e,
d1 - 2d0 = 7m
where m is an integer
Divisible test:
if the ((quantity of tens) minus the (double of ones)), is a multiple of 7, then the number K is divisible by 7, or
if d1 - 2d0 = 7n
the number is divisible by 7
Example:
8641969
-> 864196 - 18 = 864178
-> 86417 - 16 = 86401
-> 8640 - 2 = 8638
-> 863 - 16 = 847
-> 84 - 14 = 70
-> 7 - 0 = 7 (divisible by 7)
so to gather up what we have so far, here's a table:
[Added on 2021-10-22 and 2021-10-23]
Chapter 2
Extra Numbers I have tried 2 new numbers : 13, and 11, turns out, we can use almost the same method to find the determinating equation, so let's get started:
say the original number is K,
d1 is the number of tens we have and
d0 is the number of ones.
note that d1 ≥ 1
and d1 ∈ W, d0 ∈ W as well.
[dividing by 13]
then let's set the criterion as : when K is a multiple of 13, say 13 n
then we know that:
K = 10d1 + d0 = 13n
then we can also work out the following :
10d1 + d0 = 13n
10d1 + d0 - 91d0 = 13n - 91d0
10d1 - 90d0 = 13(n-7d0)
10(d1 - 9d0) = 13(n - 7d0)
since 10(d1-9d0) is a multiple of 13, then the term 10(d1-9d0) must contain the factor 13,
therefore:
d1 - 9do = 13m, m ∈ W
Divisible test:
if the ((quantity of tens) minus the (nine times of ones)), is a multiple of 13, then the number K is divisible by 13, or
if d1 - 9d0 = 13n
the number is divisible by 13
Example:
59538232
-> 5953823 - 2x9 = 5953805
-> 595380 - 5x9 = 595335
-> 59533 - 5x9 = 59488
-> 5948 - 8x9 = 5876
-> 587 - 6x9 = 533
-> 53 - 3x9 = 26 (divisible by 13)
[dividing by 11]
surprisingly, this is much simpler.
let's set the criterion as: when K is a multiple of 11, say 11 n
then we know that:
K = 10d1 + d0 = 11n
then we can also work out the following :
10d1 + d0 = 11n
10d1 + d0 - 11d0 = 11n - 11d0
10d1 - 10d0 = 11(n-d0)
10(d1 - d0) = 11(n-d0)
since 10(d1-d0) is a multiple of 11, then the term 10(d1-d0) must contain the factor 11,
therefore:
d1 - do = 11m, m ∈ W
Divisible test:
if the ((quantity of tens) minus the (quantity of ones)), is a multiple of 11, then the number K is divisible by 11, or
if d1 - d0 = 11n
the number is divisible by 11
Example:
54745394
-> 5474539 - 4 = 5474535
-> 547453 - 5 = 547448
-> 54744 - 8 = 54736
-> 5473 - 6 = 5467
-> 546 - 7 = 539
-> 53 - 9 = 44 (divisible by 11)
Chapter 3:
Similarly, we can use the same technique to come up with different algorithms to check if a number is divisible by something... therefore we an try to generalize them:
say the number we want to test is "K",
the number system base we are dealing with is "b",
the numbers of bases we have is d1,
and the numbers of ones we have is d0
and the number K will be divided by "p"
where K,b,d1,d0,p ∈ W
so we would have:
K = d1b + d0 = pq, where q ∈ W
by adding "- (βb + 1)d0" on both sides, where β ∈ W, we would have
d1b + d0 - (βb + 1)d0 = pq - (βb + 1)d0
d1b - βbd0 = pq - (βb + 1)d0
from the examples from the previous parts, we know that (βb + 1) must be a multiple of p,
so let
(βb + 1) = pr, where r ∈ W
β = (pr - 1)/b
so
d1b - βbd0 = pq - (βb + 1)d0
(d1 - βd0)b = pq - prd0
(d1 - (pr-1)/b)b = p(q-rd0)
welp, this is still a draft though working is still in progress... I will add more details and content when more progressions are made.
basically, we have the following formula... for now:
βb +1 = pr
I hope this helps you a lot, if you are a primary school maths teacher or a tuition teacher, why not share this with your students ? if you have other division tests, share it in the comment section below.
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