Magic of 3?

Thanks for "Please don't forget who invented the zero!"

As I know, zero was invented in India..... long long ago. I am proud of it.

Maybe it is applicable, if You take as true that ZERO is divisible by any number equally, regardless that result is 0/7 or again zero.

Can we modify the rule?

Test for divisibility by 7. Double the last digit and subtract it from the remaining leading truncated number. If the result is divisible by 7 or the result is 0, then the original number would be divisible by 7.

I am not sure, but we have to check this rule for many more numbers.

NO!! Because zero is already divisible by 7, as previously explained.

Furthermore, trying this out on a whole bunch more numbers is not a valid test. A full proof would recognize what I said before. By successively doubling the last two digits and truncating them, you are successively comparing the original number to 98, 9800, 980,000, etc., all of which are divisible by 7. That's why it always works.

GA. There seems to be a lot of my math classes that I missed during those naps. BTW, what grade were you at when you had the number theory class?

We learned about "casting out nines" and the easier of those divisibility tests in about the 4th or 5th grade. My number theory class was much later on, in college. Number theory has been jokingly called the queen of mathematics because it's all pure, but useless. That's not true, of course, but I don't know of too many real-world applications for, say, Fermat's Last Theorem, or Goldbach's Conjecture, or the Archimedes Cattle Problem. Then again, I may be way out of date on that....

Thanks. My wife was in the public school system in NYC, while I was in private schools in same. She remembered very little of her math, while I was able to help the daughters (and now grandsons) with math regularly. But she was the one that had the tricks with the 9s, not me.

It ia all there in the great wiki .

Yes, it is something that every third grader should know but doesn't. If the sum of the digits is divisible by 3, then the number is divisible by three, so it doesn't matter what order the digits are in. It also works for nine.

There is a test for divisibility by seven as well, which I didn't learn about until I was 56, so don't feel bad about not knowing the divisibility test for three.

It's probably not taught at any grade level because there are bigger more practical and applicable concepts which make demands of teacher's time. Other than whimsical curiosity when's the last time you used the "magic of 3" in a practical matter? Having said that, this is an enjoyable exploration.

I like to play "Slug bug" and "Bump A head" when I am on long trips.

Or occasionally "Asleep at the Wheel" but so few people pay attention to other drivers when they are driving now a days no one gets it.

divisibility rules of 3 Add up all the digits in the number. Find out what the sum is. If the sum is divisible by 3, so is the number For example: 12123 (1+2+1+2+3=9) 9 is divisible by 3, therefore 12123 is too!. similarly in u r case 8+5+1+1= 15 is div b 3

Numbers divisible by 3???y dis works is illustrated below.... Numerals whose sum of digits is divisible by 3 represent numbers divisible by 3: This one is different, because 3 does not divide any power of 10 evenly. That means we will have to consider the effect of all the digits. Here we use this fact: 10^k - 1 = (10 - 1)*(10^(k-1) + ... + 10^2 + 10 + 1) This is a fancy way of saying 9999...999 = 9*1111...111. So that our notation doesn't get way out of hand, let's make a definition: 10^k = 9*(10^(k-1) + ... + 10^2 + 10 + 1) + 1 = 9*a[k] + 1 We use this to rewrite our powers of 10 as 10^k = 9*a[k] + 1. Now if we let the digits of the number be d[0], d[1], d[2], and so on, the polynomial form of a numeral looks like this: d[k]*10^k + d[k-1]*10^(k-1) + ... + d[1]*10 + d[0] = d[k]*(9*a[k] + 1) + ... + d[1]*(9*a[1] + 1) + d[0] = 9*(d[k]*a[k] + ... + d[1]*a[1]) + d[k] + ... + d[1] + d[0] Now notice that 3 divides the first part, so the whole number is divisible by 3 if and only if the sum of the digits is. It is worth noting that once we have computed the sum of the digits to test for divisibility by three, we can apply the same test to this number to see if it is divisible by three! For example, 5697459746754985794849 is divisible by 3 if and only if 5+6+9+7+4+5+9+7+4+6+7+5+4+9+8+5+7+9+4+8+4+9 = 141 is, if and only if 1+4+1 = 6 is. Since 6 is a multiple of 3, so is 141, and so is 5697459746754985794849.