Why Do the Digits of a Number Divisible by 3 Sum Up to a Multiple of 3?
Have you ever wondered why the sum of the digits of a number divisible by 3 also turns out to be divisible by 3? Let’s break it down in a way that’s easy to understand!
#numbertheory #divisibilityrules #mathexplained
The Rule of Divisibility by 3
– When we add up the digits of a number, we are essentially looking at their individual values.
– A number divisible by 3 means that when divided by 3, the result is a whole number without any remainder.
– The sum of the digits of a number reflects the essence of the number itself.
– If the digits add up to a multiple of 3, it follows that the original number is also divisible by 3.
Explaining the Phenomenon
– Consider the number 123. The digits 1, 2, and 3 sum up to 6, which is a multiple of 3.
– Dividing 123 by 3 gives us 41, indicating that the number is indeed divisible by 3.
– This pattern holds true for all numbers that have digits summing up to a multiple of 3.
In Conclusion
Understanding why the sum of the digits of a number divisible by 3 also becomes divisible by 3 is like unlocking a hidden code in math! It’s a fascinating concept that showcases the beauty and intricacy of numbers.
Because 9 is divisible by 3.
If you have 10, you overshot 9 by 1. If you have 11-19, add this 1 to the second digit and you’ll see if the whole thing is divisible by 3.
If you have a multiple of 10, like 20, 30, 40… you overshot 9 2, 3, 4,… times. So add the number 2, 3, 4…. to the second digit.
It’s just a natural pattern of being the lowest non-1 odd value in a 3n+1 base system, which you can see counting upwards all multiples of 3.
* 03 – 06 – 09 … all divisible by 3
* 12 – 15 – 18 … notice that, you have +1 in the tens place, and -1 in the one’s place
* 21 – 24 – 27 … +2 in the tens plance, -2 in the one’s place
* 30 – 33 – 36 – 39 … and now the pattern repeats
This works in the base 10 system because it cycles cleanly over the ten’s place, always maintaining this relationship between the rightmost digits.
Now to further expand this, and you’ll note the hundreds and tens place follow the same (extended) pattern.
* 000 – 003 – 006 – 009
* …
* 030 – 033 – 036 – 039
* …
* 060 – 063 – 066 – 069
* …
* 090 – 093 – 096 – 099
* 102 – 105 – 108
* 111 – 114 – 117 … +1 hundred’s place, +1 ten’s place, -2 one’s place
* 120 – 123 – 126 – 129 … +1 hundred’s place, +2 ten’s place, -3 (or 0) one’s place
Thus, every increase in a value to the digit to the left, is accompanied by an decrease in the value of the one’s place. In this way, they balance out, so whatever’s left if you add them to the one’s place will *always* be a multiple of 3.
Here’s an algebraic explanation, to add to the others.
Let’s use T to represent the tens place of a two digit number, and O to represent the ones place.
So any two digit number can be written as:
10T + O
for example, 42 = 10 x 4 + 2
With me so far?
10 = 9 + 1 so we can write
10T + O = (9 + 1)T + O = 9T + T + O
Now we say this number is the result of multiplying 3 by another number, which we’ll call N
3N = 9T + T + O
We know 9T is divisible by 3 because 9 is divisible by 3. In order for the whole number to be divisible by 3, then T + O must also be divisible by 3. And T + O is just the sum of the digits!
This is for a two digit number, but you can extend it to any number of digits, because you can expand any number into powers of 10, which can then be expanded to (9 + 1)^n which expands to a bunch of numbers that are multiples of 9, plus a 1 at the end. Getting rid of the 9 terms, which we know are divisible by 3, it leaves us with a bunch of 1s which multiply the digits, resulting in the sum of the digits.
You want to check if you can take this number and make it into groups of three.
*If it’s a single-digit number (just units) it’s simple – check the total and see if it’s divisible by 3. If it’s 3, 6 or 9 it can be, otherwise no.
*If it’s a two-digit number (meaning it has some tens)…you can think of every ten as 9 (which takes care of itself; that’s divisible by three) plus one extra that adds to your pile of units.
*If it’s a three-digit number (so it has hundreds) every hundred consists of a 99 (which takes care of itself) plus one extra that adds to your units.
*Four digits / thousands? The 999 takes care of itself, then there’s one extra that adds to your units.
So you can see, tens, hundreds, thousands and so on up, are all “worth the same” when it comes to this. Each of them contains 9, 99, 999 or whatever which is automatically divisible, plus one extra unit. So the only real question is, does the total of all those units end up divisible by three, and the shortcut way to find that is to total all the digits, ignoring their place values.
Digits of a number are short hand for the notation. N0 * 1 + N1 * 10 + N2 * 100 … etc. The proof to your question will show that adding N0 to N(whatever) will sum to a number that has the same remainder as if you divided the original number by 3. It is called modulo arithmetic, and it is very useful for questions like the one you’re asking.
Essentially you just throw away anything other than the remainder if you divide by some number. In this case, 3. Have 5, divide by 3, remainder is two; therefore 5 mod 3 = 2. In short hand 5 % 3 = 2.
It’s pretty quick to see that 10 % 3 = 1. Then you can find that any power of 10 % 3 = 1. Using this, you can understand that if you add the individual digits of a number (in base 10), it is the same as taking the modulo of the entire number base 3.
That is to say 1 % 3 = 1, 2 % 3 = 2, 3 % 3 = 0, etc. and 10 % 3 = 1, 20 % 3 = 2, 30 % 3 = 0, etc. In the exact same way 100 % 3 = 1, 200 % 3 = 2, 300 % 3 = 0.
Since modulo arithmetic lets you sum in a modulus, then it’s a short jump to put the pieces together. The digits all behave the same, so adding the digits is the same as adding the modulo of the individual digits.
Adding to the other answers, we use base ten, so this property works for the number one less than ten, i.e., for 9. It also works for all factors of 9, i.e., 3 and 1, though 1 is not useful.
So, for the first few integer bases, it works with the following numbers greater than 1 (all values given in base ten):
* 2: nil;
* 3: 2;
* 4: 3;
* 5: 4, 2;
* 6: 5;
* 7: 6, 3, 2;
* 8: 7;
* 9: 8, 4, 2;
* 10: 9, 3;
* 11: 10, 5, 2;
* 12: 11;
* 13: 12, 6, 4, 3, 2;
* 14: 13;
* 15: 14, 7, 2;
* 16: 15, 5, 3.
Works for 3, but also works for all other factors of the base of the number – 1
We use base 10
10-1 = 9
If the digits add to 9 it is divisible by 9
Same for 3
In base 13 it would work for 6,3,2,4,12
are you asking why:
6 (divisible by 3) + 9 (divisible by 3) = 15 (also divisible by 3)
and/or similarly:
15 + 12 = 27 (all divisible by 3)
if I’m understanding your question, this is not isolated to the number 3.
I think I can explain it maybe clearly than others have if so, but not 100% sure I’m understanding what you’re asking based on the answers everyone else gave lol
edit: to further clarify, what you’re asking would also apply to numbers divisible by 4, 5, etc, and every other number?
Dividing the total number of bricks (1 + 2 + 6) by 3 is like separating them into equal groups of 3. You might have some leftover bricks depending on the total number. If there are no leftovers, the original number is divisible by 3.
Because using a base-10 number system.
ABCD = 1000xA + 100xB + 10xC + 1xD
you can rewrite as:
(999 + 1)A + (99 + 1)B + (9 + 1)C + D
(999A + 99B + 9C) + A + B + C + D
999A + 99B + 9C will always be divisible by three.
So if ABCD is divisible by 3, then A+B+C+D will be divisible by 3.
You’re just adding one group of threes to another group of threes. The sum is how many threes you have in total. 15 is a group of 5 threes, and 12 is 4 threes, so 15 added to 12 is 9 threes.
Because when you add 3 to a number, you have several cases:
(1) The ones place goes up by 3. So, if the original summation was divisible by 3, the new one will be also.
Example:Â 6 -> 9
(2) The tens place goes up by one and the ones place goes down by 7. So, you’ve added 1 and taken away 7. It’s as if you just took away 6, which is divisible by 3. So if the original summation was divisible by 3, the new one will be also.
Example: 18 -> 21
(3) The ones place goes down by 7, and there’s a 9 in one or more digits to the left. Those 9s all go to 0 (subtract 9, a multiple of 3) and you add 1 to the first place that doesn’t have a 9, just like in (2). Again, if the original summation was divisible by 3, so will the new one.
Example: 1998 -> 2001
Say, for example, you have a number with digits abc. This number is equal to the sum 100a+10b+c. If you subtract the sum a+b+c from this sum, then you get 99a+9b. 9 and 99 are both multiples of 3 (3 and 33 respectively), and a and b are both integers, so each term is divisible by 3. And because the sum of two multiples of a number is also a multiple of that same number, 99a+9b is divisible by 3.
Now, going back to our original number. Because subtracting a+b+c from abc gave us 99a+9b, we can say that abc=(99a+9b)+(a+b+c). And as we established before, if each term in a sum is divisible by some number, then the sum itself is also divisible by that number. In this case, we know that 99a+9b is divisible by 3. So if a+b+c is also divisible by 3, then the whole sum is divisible by 3. And since the original sum was abc, then that would make abc divisible by 3. And this logic can be applied to any integer, regardless of how many digits it has.
This logic only works in base 10, though. If we were working in base 12, then abc would be equal to 12^(2)a+12b+c, and subtracting a+b+c from that would give you 143a+11b. 143 and 11 are not divisible by 3, so the sum is not necessarily divisible by 3 either. But they are both divisible by 11, so this would still work as a divisibility check for 11. In fact, for any counting base n, this rule can be used as a divisibility check for any factor of n-1.
True ELI5? Because math.