What my Elementary Math Teacher Didn't Teach Me: Divisibility by 7
THE SCENARIO...
When I was in my elementary days, I was stumbled upon by this lesson in mathematics called "divisibility rules". In that lesson, the reference book states the divisibility rules of 2,3,4,5,6,8,9,10, and even 11 and 12. I was wondering why there is no rules in case of seven.
After some time of thinking, I asked my math teacher about the matter and he told me that "Just follow what is written on the book. Is there a rule about divisibility rule of seven? No right? then there's no rule about it." This actually sparks my interest in mathematics as I grew older and always thinking about why there is no seven.
Years later when I turned seventh grade, I somehow remembered this divisibility question that I always wondered when I was a kid and did a little research about it. It turns out there exist a rule that a number can be divided by seven.
DIVISIBILITY RULES OF 7
1) Take a number, lets say 546.
2) Get the last digit, multiply it by 2, (6 * 2 = 12)
3) Subtract the answer to the remaining digit (54 - 12 = 42)
Check the resulting number. If it is divisible by 7, then the number is divisible by 7.
The steps are iterative, meaning you can repeat steps 2 and 3 until you can check if the resulting number is divisible by 7.
By iterating 2 and 3 from the result of 42:
4) 2 * 2 = 4
5) 4 - 4 = 0
Another indicator of being divisible by seven is when the number ended up to zero.
Therefore, 546 is divisible by 7.
Lets try another example...
1) Take a number, lets say 1234
2) 4 * 2 = 8
3) 123 - 8 = 115
Iterating steps 2 and 3 from the result 115:
4) 5 * 2 = 10
5) 11 - 10 = 1
Since 1 is not divisible by 7...
Therefore, 1234 is not divisible by 7.
Last example...
1) Take a number, lets say 34853
2) 3 * 2 = 6
3) 3485 - 6 = 3479
Iterating steps 2 and 3 from result of 3479:
4) 9 * 2 = 18
5) 347 - 18 = 329
Iterating steps 2 and 3 from result 329:
6) 9 * 2 = 18
7) 32 - 18 = 14
Since 14 is divisible by 7...
Therefore, 34853 is divisible by 7.
PROOF
For those who are interested, here is the proof that I got and simplified from mathandmultimedia.com
let N = number we need to check if it is divisible by 7,
let A = the rest of the digit that starts from the tenths place and:
let B = the digit at unit digit.
We know that N and 10A + B are the same number.
N = 10A + B
Now, to imply that 10A + B can be divided by 7, we let k = a natural number
10A + B = 7k
We will subtract 21B from both sides (this will make sense later.)
10A + B - 21B = 7k - 21B
10A - 20B = 7k - 21B
Now, factoring both sizes:
10 (A - 2B) = 7 (k - 3B)
Since these are equal, we know that:
A - 2B is Divisible by 7 and;
k - 3B is Divisible by 10
Now, by looking at A - 2B, it is the steps 2 (doubling the last digit B) and step 3 (Subtracting the result from the remaining digit A) thus, completing the proof of divisibility.
CONCLUSION
Divisibility rules of 7 is a thing. Do not just base your study on what school teach you or you will be limited on what school curriculum is to offer. Do your own research. Now that I am a college student, I can safely say that I can apply this piece of advice that really helped me on my college journey and in the future. Stay curious folks!
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