A variation of divisibility rule of 3?

Note that $10 \equiv 1 \pmod{3}$ and consider

$N = 1\times 10^a+2\times10^b+3\times10^c+ \cdots + 9\times 10^i$ where $a,b,c,\cdots ,i$ are arbitrary positive integers and $N$ satisfy the requirement of the problem

$N \equiv 1\times1+ 2\times1 + 3\times1 + \cdots +9\times1 \equiv 0 \pmod{3}$

We proved $N$ is divisible by 3 and then the statement below is true

With any sum example:

    12345
+    6789
    19134

We can rewrite the sum as:

= 1*(10000) + 2*(1000) + 3*(100) + 4*(10) + 5*(1)
            + 6*(1000) + 7*(100) + 8*(10) + 9*(1)

= 1*(1+9999) + 2*(1+999) + 3*(1+99) + 4*(1+9) + 5
            + 6*(1+999) + 7*(1+99) + 8*(1+9) + 9

= 1 + 1*9999 + 2 + 2*999 + 3 + 3*99 + 4 + 4*9 + 5
            + 6 + 6*999 + 7 + 7*99 + 8 + 8*9 + 9

= 1*9999 + 2*999 + 6*999 + 3*99 + 7*99 + 4*9 + 8*9
            + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9

Since any number multiplied by *999~ is already a multiple of 3,

any combination will depend only on the sum of digits(1-9) = 45

45/3 = 15; 45%3 == 0

Therefore, it's always True

We just have to consider the following modular congruence for $m=3$ ,

$(a+b) \equiv (a \pmod m + b\pmod m) \pmod m$ ,

along with the following fact:

A number is divisible by $3$ if and only if the sum of its digits is divisible by $3$ .

Simple to overthink this question. Laws of Divisibility states that if the sum of digits of N is a multiple of 3, it is divisible by 3.

Sum of 1+2+3+.....+9 is 45, which is divisible by 3.

Thus, no matter how you add it, it is divisible by 3.

A number $N$ is divisible by 3, if and only if its digital sum is also divisible by 3. And the reverse is true. Since 123456789 has a digital sum of 45 which is divisible by 3, 123456789 is divisible by 3. How many and wherever " $+$ " signs are placed within 123456789 do not change the overall digital sum of 45, therefore all the "summed" numbers are divisible by 3.

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Proof that $N$ is divisible by 3 if and only if its digital sum is divisible by 3: Let $N=\displaystyle \sum_{k=0}^n 10^kd_k$ , where $d_k$ are the digits. Then we have $\displaystyle N \equiv \sum_{k=0}^n (9+1)^kd_k \equiv \sum_{k=0}^n d_k \text{ (mod 3)}$ . Therefore, $N$ is divisible by 3, iff its digital sum $\displaystyle \sum_{k=0}^n d_k$ is divisible by 3.

The sum of digits are same in any arrangement of 1,2,3,4,5,6,7,8,9. Is 45.

So they will be divisible in any arrangement

common sense way: if the sum of the digits of a real number equals three, then 3 is a factor of the number. add up all the digits (=45), then show that no possible placement of "+" can form a number, with the attribute that the sum of its digits is not a multiple of 3.

No matter where you put the '+' sign in between the digits, at last you will be adding up all the digits from 1 to 9 altogether. It will always work even if you jumble the digits. e.g.: The sum (51+24+93+67+8) is divisible by 9.

here, you see that, 1234568789/3 ; 123+456+789/3, cause all number last digit is 9, so all number is divisiable to 3. also we first work on division then work with sum, so on the sum section we only consider the last number, as 123+456+789/3=123+456+263; we only divide 789 by 3.